process flexibility design in unbalanced and...

Process Flexibility Design in Unbalanced and Asymmetric Networks

by

Tianhu Deng

A dissertation submitted in partial satisfaction of the

requirements for the degree of

Doctor of Philosophy

in

Engineering - Industrial Engineering & Operations Research

in the

Graduate Division

of the

University of California, Berkeley

Committee in charge:

Professor Zuo-Jun “Max” Shen , ChairProfessor Philip M. Kaminsky

Associate Professor Haiyan Huang

Spring 2013


Copyright 2013by

Tianhu Deng

1

Abstract


by

Tianhu Deng

Doctor of Philosophy in Engineering - Industrial Engineering & Operations Research

University of California, Berkeley

Professor Zuo-Jun “Max” Shen , Chair

In a multi-product multi-plant manufacturing system, process flexibility is both the abil-ity of each plant to produce a variety of products and the ability to produce products atmultiple plants. A balanced network has the same number of plants and products. In asymmetric network, all the plants have the same capacity and all the products have thesame demand. Several design guidelines and flexibility indices have been developed in theliterature to inform the design of flexible production networks. Yet little insights have beengained in unbalanced networks and asymmetric networks. In this thesis, we aim to providenew insights and results of unbalanced and asymmetric networks.

First, we propose additional flexibility design guidelines by refining the well-known Chain-ing Guidelines (Jordan and Graves, 1995). We study unbalanced and symmetric networkswhere each product is built at two plants. We also briefly discuss cases where (1) eachproduct is built at three plants and (2) some products are built at only one plant. An ex-tensive computational study suggests that our refinements work very well for finding flexibleconfigurations with minimum shortfall.

Then, instead of assuming a joint demand distribution for all products, we optimize theworst case over all the demand joint distributions that have the given marginal mean andvariance. We present two results. First, we show that most known results for the e↵ects ofplant capacity and product demand in symmetric networks can be extended to asymmetricnetworks. Second, the analysis sheds light on how to incorporate demand standard deviationinto the existing flexibility design guidelines and indices. We show that the proposed methodsignificantly enhances the performance of the Node Expansion Guideline.

i

To my wife, parents and grandparents

ii

Contents

Contents ii

List of Figures iii

List of Tables v

1 Introduction 1

2 Literature Review 92.1 Insights about flexibility configuration E . . . . . . . . . . . . . . . . . . . . 92.2 Marginal benefits of flexibility configuration . . . . . . . . . . . . . . . . . . 102.3 Marginal e↵ects of plant capacity, demand mean and standard deviation . . 102.4 Other settings where process flexibility design has been applied . . . . . . . . 11

3 Process Flexibility Design in Unbalanced Networks 123.1 Unbalanced Network Flexibility Design Guidelines . . . . . . . . . . . . . . . 133.2 Asymmetric Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Existing Design Guidelines and Indices . . . . . . . . . . . . . . . . . . . . . 253.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4 Robust Process Flexibility Design 334.1 Basic Model Setting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.2 Solution Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.4 A Marginal Distribution Model . . . . . . . . . . . . . . . . . . . . . . . . . 454.5 A Cross Moment Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

A Proofs 55

B Numerical examples for Figure 1.1 89

Bibliography 120

iii

List of Figures

1.1 Maximum Percentage Increase in Expected Shortfall over Total-Flexibility Con-figuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 An Illustration of the Unbalanced Network Flexibility Design Guidelines . . . . 31.3 The comparison of two flexibility configurations . . . . . . . . . . . . . . . . . . 5

2.1 Concepts of “distance” and “flow” in process flexibility design . . . . . . . . . . 10

3.1 Symmetric Configurations where k = m + 1 (“Configuration A > configurationB” means that configuration A is more flexible than configuration B.) . . . . . . 14

3.2 Expected Total Shortfalls of Configurations in Figure 3.1 (m = 10, k = 11,C

i

= 11, and E[Dj

] = 10. The horizontal line above the curve is the expectedtotal shortfall when product 11 is directly connected to only one plant. The hor-izontal line below the curve is the expected total shortfall of the total-flexibilityconfiguration.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.3 Unbalanced Network Flexibility Design Guideline (b) (“Configuration A > con-figuration B” means that configuration A is more flexible than configuration B.) 16

3.4 3-chain Configuration and Its Circular Representation . . . . . . . . . . . . . . . 183.5 A Configuration with 3k Edges . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.6 A Configuration where �(j) 6= 2 (“Configuration A > configuration B” means

that configuration A is more flexible than configuration B.) . . . . . . . . . . . . 193.7 Unbalanced Network Flexibility Design Guideline (b) (“Configuration A > con-

figuration B” means that configuration A is more flexible than configuration B.) 203.8 Inconstant k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.9 A Network in Jordan and Graves (1995) . . . . . . . . . . . . . . . . . . . . . . 243.10 Illustration of the Computation of the APL Metric . . . . . . . . . . . . . . . . 273.11 The Relationship Between Chaining Guideline and Flexibility Design Rules . . . 283.12 JG Index and Expansion Index of network (m = 10, k = 11) . . . . . . . . . . . 303.13 APL Metric Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.1 The shape of rj

(Dj

) when µj

= 100 . . . . . . . . . . . . . . . . . . . . . . . . . 404.2 Computation time as a function of network size . . . . . . . . . . . . . . . . . . 46

A.1 2-chain Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

iv

A.2 Case (i) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56A.3 Case (ii) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57A.4 Case (iii) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58A.5 Case (iii) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59A.6 Case (iii) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61A.7 Illustrations of the Structures under Di↵erent Signs of |�(⇤)|� |⇤| . . . . . . . . 62A.8 |�(⇤)| = |⇤| . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63A.9 |�(⇤)| < |⇤| . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64A.10 APL and Flexibility Design Rules . . . . . . . . . . . . . . . . . . . . . . . . . . 66A.11 An Equivalent Structure to Figure A.10 . . . . . . . . . . . . . . . . . . . . . . 66A.12 APL Metric Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67A.13 Illustrations of Lemmas 21 and 22 . . . . . . . . . . . . . . . . . . . . . . . . . . 68A.14 Illustrations of Lemma 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70A.15 p

1

> 1

2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71A.16 Three cases that cannot happen . . . . . . . . . . . . . . . . . . . . . . . . . . . 74A.17 Illustration of problems (A.26) and (A.27) . . . . . . . . . . . . . . . . . . . . . 83

v

List of Tables

1.1 Analytical sensitivity analysis in networks with more than two products . . . . . 51.2 Consistency between average and worst case performance . . . . . . . . . . . . . 71.3 Notation and symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3.1 Objective Functions of Flexibility Indices . . . . . . . . . . . . . . . . . . . . . . 29

4.1 Information on C, µ and � considered by di↵erent guidelines and indices . . . . 424.2 Comparison of (NE) and (NE- refinement) - fixed C and µ . . . . . . . . . . . . 444.3 Comparison of (NE) and (NE- refinement) - random C and µ . . . . . . . . . . 444.4 Computational aspects of the Total Flexibility Structure . . . . . . . . . . . . . 524.5 Computational aspects of the Chaining Structure . . . . . . . . . . . . . . . . . 534.6 Computational aspects of the Dedicated Structure . . . . . . . . . . . . . . . . . 53

B.1 k = 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89B.2 k = 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90B.3 k = 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92B.4 k = 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95B.5 k = 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98B.6 k = 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101B.7 19 k 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104B.8 k = 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104B.9 k = 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105B.10 k = 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107B.11 k = 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110B.12 k = 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113B.13 k = 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116B.14 Expected Total Shortfall of the Total-Flexibility Structure as a Function of k . . 118

vi

Acknowledgments

I would like to first thank my research advisor, Professor Zuo-jun Max Shen, for his con-stant encouraging and support. I thank him for making my Ph.D. full of fun and enjoyments.Working with him is a great pleasure. He has always been a source of power in my research.

I would also like to thank my other research committee members, Professor Philip M.Kaminsky, Professor Haiyan Huang, and former committee members (Professor J GeorgeShanthikumar and Professor Xuanming Su, who left Berkeley in the middle of my phd study),who have been very supportive in my research and who have been doing excellent researchin their fields. I am so blessed to have them in my research committee and be challengedby them. Part of my thesis was largely due to the help from Professor Chung Piaw Teoand Professor Stephen C. Graves. I am also challenged, generously helped, and inspired byProfessr Ying-ju Chen, Professor Dorit Hochbaum, and many other faculty members.

I treasure the time I spent with other IEOR graduate students. We all have had happyand di�cult times. We occasionally go hiking, play pocker and Xbox 360, and play sports.These times will be very memorable. I also thank my nice roommates who have beenvery generous and kind to me. When I was sick and depressed over di�culties in life, myroommates always cheer me up. I also want to thank all the good and meaning time I spentat International graduate student ministry.

I am very indebted to my parents and grandparents, who have sacrificed a lot for myspiritual and physical growth. I can always feel my parents’ love for me. I want to thankmy wife Yujia Chen for spending many di�cult times with me together. We had so manydi�cult times together and I am so glad that I will graduate soon.

There are many other people who help me in many ways which I cannot list here. Ithank all whom I have shared the past five years with. I would not be able to finish myPh.D without all of your helps.

1

Chapter 1

Introduction

2008 was a tough year for most automakers in the U.S. The slumping economy depressedoverall vehicle sales, and high gasoline prices spurred American consumers to buy small carsinstead of pickup trucks and SUVs. As a result, many Detroit automakers were forced toshut down their truck and SUV productions and cut employees. On the other hand, Hondaand Toyota were able to shu✏e production among di↵erent plants as well as make di↵erentvehicle models at one plant (e.g., LeBeau, 2008, Kent, 2009 and LeBeau, 2010). In thisregard, flexibility has widespread influence. For example, Ford announced (2010) that itsplant outside Detroit will be the first in the world to build standard internal combustion,hybrid, plug-in hybrid, and electric vehicles in one plant. It will be the first plant in theworld where all models with di↵erent power train systems are being built under the sameroof at the same time.

Many forms of flexibility have been studied in various contexts. We follow Jordan andGraves, 1995’s model setting, where “process flexibility” is both the ability of each plantto produce a variety of products and the ability to produce products at multiple plantsin a general multiproduct multiplant manufacturing system. Consider a bipartite graphrepresentation of process flexibility G = (V, V 0, E), where V is the set of plants (V ={1, 2, . . . ,m}), V 0 is the set of products (V 0 = {1, 2, . . . , k}), and every edge in E connects aplant node i 2 V to a product node j 2 V 0, indicating that plant i can produce product j.The capacity of plant i is C

i

and Dj

is the realized demand for product j. The total shortfallis the optimal value of the following problem:

MinX

j2V 0

sj

s.t.X

j:(i,j)2E

fi,j

Ci

, 8i 2 V,

X

i:(i,j)2E

fi,j

+ sj

= Dj

, 8j 2 V 0,

fi,j

� 0, 8(i, j) 2 E; sj

� 0, 8j 2 V 0,

CHAPTER 1. INTRODUCTION 2

where fij

, 8(i, j) 2 E are the amounts of product j produced by plant i and sj

is the shortfallfor product j. All of these variables are non-negative. Since demand is random, the expectedtotal shortfall is the average optimal value over all possible demand realizations D

j

, j 2 V 0.The expected total shortfall reflects the flexibility of a structure E.

The term “balanced networks” refers to networks where the numbers of plants and prod-ucts are equal. The term “symmetric networks”, by contrast, describes networks where allplants have the same capacity and di↵erent products’ demands are i.i.d. The central themeof this thesis is to extend existing work in the literature to unbalanced and asymmetricnetworks.

The most well known result in the literature is the Chaining Guideline (Jordan andGraves, 1995):

(a) Try to equalize the number of plants (measured in total units of capacity) to whicheach product is directly connected;

(b) Try to equalize the number of products (measured in total units of expected demand)to which each plant is directly connected; and

(c) Try to create a circuit(s) that encompasses as many plants and products as possible.

They show that the Chaining Guidelines can lead to flexible configurations that performalmost as good as the total-flexibility configuration where every plant is linked to all products.

Our analysis reveals that the Chaining Guidelines may not be su�cient in designingflexible configurations in unbalanced networks. Consider a symmetric network with m = 12plants. The number of products k varies from 13 to 24. We examine only configurationswhere each product is directly connected to two plants. Figure 1.1 shows that the ChainingGuidelines could lead to a configuration that has a considerable (e.g., 200%) increase fromthe expected shortfall of total-flexibility configuration. The simulation parameters are asfollows. There are m = 12 plants, each of which has capacity C

i

= 20, 81 i 12. Eachproduct has normally distributed demand with mean E[D

j

] = (C · m/1.1)/k and varianceV ar(D

j

) = (E[Dj

] · 0.4)2, 1 j k. Negative demand is replaced with 0 when generatingrandom demand. For each k, we randomly sampled more than 100 configurations. We employonly one set of parameters, so 200% may not be a typical percentage increase. Nonetheless,this example hints at the potential room for improvement for the Chaining Guidelines. Sincewe cannot sample all configurations that satisfy the Chaining Guidelines, the actual worst-case should be worse than the one obtained by simulation, and the numbers in Figure 1.1underestimate the actual percentage increase.

To gain intuition for the limitations of the Chaining Guidelines in unbalanced networks,consider a symmetric network with m = 12 plants and k = 15 products (see Configuration(i) in Figure 1.2). Each product is built at two plants. The Chaining Guideline (c) suggestsbuilding a circuit connecting all plants with the first m products (see Configuration (ii) inFigure 1.2). The Chaining Guidelines (a) and (b) suggest that 2 of the 12 plants shouldbe directly connected to product 13, that another 2 plants should be directly connected to


0.00%

50.00%

100.00%

150.00%

200.00%

250.00%

300.00%

350.00%

13 14 15 16 17 18 19 20 21 22 23 24

Chaining Guidelines

Chaining Guidelines and Unbalanced Network Flexibility Design Guidelines

Maximum % increase in expected shortfall over total-flexibility configuration

Number of Products (k)Number of Plants m=12

Figure 1.1: Maximum Percentage Increase in Expected Shortfall over Total-Flexibility Con-figuration

Plant Product

1

2

3

4

5

6

7

8

9

10

11

12

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

?

?

?

?

?

?

?

?

?

?

?

?

?

?

?

1

2

3

4

5

6

7

8

9

10

11

12

1

2

3

4

5

6

7

8

9

10

11

12

?

13

14

15

?

?

1

2

3

4

5

6

7

8

9

10

11

12 1

2

3

4

5

6 7

8

9

10

11

12

13

14 15

1 2

3

4

5

6

7

8

9

10

11

12 1

2

3

4

5

6 7

8

9

10

11

12

13

14 15

?

? ?

=

Unbalanced Network Flexibility Design Guidelines

Chaining

Guidelines

Configuration (i) Configuration (ii)

Configuration (iii) Configuration (iv)

Figure 1.2: An Illustration of the Unbalanced Network Flexibility Design Guidelines


product 14 and that another 2 plants should be directly connected to product 15. Becausethe 6 plants that are directly connected to products 13, 14 and 15 are connected to one morelink than the other plants, they are more heavily utilized. Therefore, we should not placethese more highly utilized plants around each other and should evenly mix the more utilizedproducts with the rest of the plants. The proposed Unbalanced Network Flexibility DesignGuidelines refine the Chaining Guidelines in this respect.

As our main contribution of Chapter 3, we lay out the Unbalanced Network FlexibilityDesign Guidelines. Given a symmetric network with m plants and k products, the proposedGuideline works as follows:(1) Build a chaining configuration between all the m plants and the first m products. Re-arrange the chaining configuration as a circle. As an illustration, see Figure 1.2 where thechain in Configuration (ii) is rearranged as a circle in Configuration (iii).(2) For products m+ 1 to k, the Unbalanced Network Flexibility Design Guidelines recom-mend:(a) connecting each of these products to plants on diametrically opposite ends of the circle;(b) connecting these products to plants that are evenly spaced on the circle, and(c) avoiding connecting the same pair of plants to more than one product. To illustrate, seeFigure 1.2.

In Chapter 4, we aim to provide a demand-distribution-free analysis of process flexibilitydesign. Instead of assuming a joint demand distribution for all products, we optimize theworst case over all the demand joint distributions that have the given marginal mean andvariance. There are four reasons that have led us to believe that demand-distribution-freeanalysis is particularly meaningful in the field of process flexibility design. First, little insighthas been gained over the role of demand distributions in process flexibility design. Oneimportant research question is that given two flexibility configurations, how can we easilytell which one is more flexible? Consider the two flexibility configurations in Figure 1.3.Configuration E

1

is more flexible than configuration E2

for one demand distribution. But aslight change in the demand distribution would reverse the preference. The example hints theimportance of specifying conditions on demand distributions when comparing two flexibilityconfigurations. In general, it is hard to identify a general set of demand distributions underwhich one can establish the preference relation between all configurations. A shortcut tothis challenge would be the demand-distribution-free analysis.

Second, it is prohibitively di�cult to obtain analytical results about e↵ects of plantcapacity and product demand. A network is symmetric if all plants are identical and allproducts have the same demands. As illustrated in Table 1.1, most previous analyticalresults are from symmetric networks. Compared with the flexibility setting in Bassamboo,Randhawa, and Mieghem, 2010, flexibility and capacity are two di↵erent decisions in Jordanand Graves, 1995’s setting of process flexibility. The decision of capacity investment cannotbe made without a clear understanding of the marginal e↵ects of plant capacity and flexibilityconfiguration. In particular, little is known about the causal e↵ects of demand average,demand standard deviation, and flexibility configuration on the marginal value of capacity.Demand-distribution-free analysis allows the sensitivity analysis for asymmetric networks.


Figure 1.3: The comparison of two flexibility configurations

Plant Product Product Plant

𝐸 𝐸

Note. Let C be each plant’s capacity. Configuration E1

is more flexible than configuration E2

when demand D equals (C, 0, C, 0) and (0, C, 0, C) with equal probability 0.5. The preference is

reversed when demand D equals (C,C, 0, 0) and (0, 0, C, C) with equal probability 0.5. See Jordan

and Graves, 1995, Chou et al., 2010 and Simchi-Levi and Wei, 2011 for detailed discussions.

Table 1.1: Analytical sensitivity analysis in networks with more than two products

Sensitivity analysis Symmetric networks Asymmetric networks

Flexibility Aksin and Karaesmen, 2007Simchi-Levi and Wei, 2011Bassamboo, Randhawa, andMieghem, 2010

this paper

Capacity (second or-der)

Aksin and Karaesmen, 2007Bassamboo, Randhawa, andMieghem, 2010

this paper

Demand mean & std this paper

Note. Please refer to Fine and Freund, 1990, van Mieghem, 1998 and Bish and Wang, 2004; Bish,

Muriel, and Biller, 2005 for analyses about process flexibility design in a two-product setting.


Third, the demand-distribution-free analysis provides decisionmakers with another crite-rion for flexibility design. Researchers have observed an important implication of flexibilitydesign - “the key point is that there is not one optimal flexibility plan; rather there aremany that are near-optimal (Jordan and Graves, 1995).” Deng and Shen, 2012 claim thatmany flexibility configurations are near-optimal when the number of products is more thanone and a half of the number of plants. These observations indicate that decisionmakershave freedom in choosing among near-optimal solutions. So it is meaningful to have moreselection criterion over flexibility design, especially in practical situations where only oneflexibility configuration can be implemented. The demand-distribution-free analysis maybeable to assist decisionmakers with a worst-case analysis of flexibility configurations that areclose in terms of expected total shortfall.

Fourth, flexibility design decisions are made in the planning stage of a production system,while ample and accurate demand date can only be obtained after the planning stage of theproduction system. A practical problem emerges as decisionmakers have to forecast demanddistributions in order to make flexibility design decisions. Yet information on demand distri-butions would be very hard to access before running the production system and selling theproducts. In this regard, a demand-distribution-free analysis is appealing as it relies only onthe mean and variance of demands. The demand-distribution-free analysis is advantageouswhen information on demand distribution is hard to obtain.

One major problem with worst case analysis is its potential inconsistency with thestochastic programming approach. Interestingly, we analytically justify that for the processflexibility design problem, the worst case analysis is consistent with the stochastic program-ming approach. As a numerical illustration, we ran a Monte Carlo simulation with 10,000replicated scenarios. Each scenario consists of three stages. In stage 1, plant capacity, de-mand mean and demand standard deviation were randomly selected from intervals [100,200],[90,180] and [20,50], respectively. In stage 2, two flexibility configurations were randomlygenerated. In stage 3, we compare the two flexibility configurations under both average andworst case performances. Table 1.2 suggests that the average performance and worst caseperformance are inconsistent for only 3.4% of the scenarios.

The contribution of chapter 4 is two-fold. First, the demand-distribution-free analysisdraws implications about the marginal value of plant capacity. We show that plant capacityhas diminishing marginal values, i.e., the optimal expected total shortfall is convex decreasingin each plant’s capacity. Furthermore, the analysis reveals that the marginal value of a plant’scapacity increases with the average demands that are directly connected to the plant. Theresult favors building capacity at plants that serve higher average demands than at plantsthat serve lower average demands. The e↵ect of demand standard deviation on the marginalvalue of plant capacity can go either direction, depending on whether or not the system isheavily utilized. At last, we provide an upper bound on the marginal value of a plant’scapacity.

Second, our analysis sheds light on the existing flexibility design guidelines and indices,most of which do not consider demand standard deviation. As indicated by our analysis,the e↵ects of demand standard deviation can go either direction. This adds substantial


Table 1.2: Consistency between average and worst case performance

Worst case (w)

Average (a) E1

>w

E2

E1

=w

E2

E1

<w

E2

TotalE

1

>a

E2

47.9% 0.0% 1.5% 49.4%E

1

=a

E2

0.0% 0.0% 0.0% 0.0%E

1

<a

E2

1.7% 0.0% 48.9% 50.6%Total 49.6% 0.0% 50.4% 100.0%

complexity to the task of incorporating demand standard deviation to existing flexibilitydesign guidelines and indices. Because the demand-distribution-free analysis has implicationsover the interaction between demand mean and standard deviation, it provides a naturalway of incorporating demand standard deviation into existing flexibility design guidelinesand indices. In particular, we demonstrate the potential benefits of incorporating demandstandard deviation into the Node Expansion Guideline (Chou et al., 2011).

We conclude this introduction with some notational conventions. Vectors will be inboldface. All vectors are column vectors by default, and primes denote transposes. Atilde (⇠) placed above a letter represents a random variable. E denotes the expectation.Throughout this paper, “increasing” and “decreasing” mean “non-decreasing” and “non-increasing”, respectively. Table 1.3 summarizes all the major notation and symbols.


Table 1.3: Notation and symbols

Exogenous parametersG = The graph representation.V = The set of plants.V 0 = The set of products.⇤ = A subset of products.�(⇤)= The subset of plants that is directly connected product set ⇤.m = The number of plants.k = The number of products.i = The index of a plant.j = The index of a product.E = A flexibility configuration.C

i

= Plant i’s capacity.D

j

= Product j’s demand.µj

= Product j’s average demand.�j

= Product j’s demand standard deviation.F = The demand joint distribution for all products.

Computed quantitiessj

= The shortfall of product j.fij

= Amount of product j produced by plant i.R = Total shortfall of a flexibility configuration.W = The worst-case expected total shortfall.D

j

= The amount of product j produced.rj

= Product j’s worst expected shortfall.Dual decision variables

✓ = The dual decision variable for the zeroth moment constraint.⇢ = The dual decision variable for the first moment constraint.� = The dual decision variable for the second moment constraint.x = The dual decision variable of plant capacity constraints.y = The dual decision variable of demand constrains.d = A demand realization.

9

Chapter 2

Literature Review

This chapter is divided into four parts. We start with reviewing the key insights aboutflexibility configuration E, followed by the analyses investigating the marginal benefits offlexibility configuration. Then, we review the literature mostly related to our research ques-tion - the e↵ect of plant capacity C, demand mean µ and standard deviation �. Finally, wereview the broad settings where process flexibility design has been applied.

2.1 Insights about flexibility configuration E

One important research question being studied in the flexibility design literature is that howcan one quickly tell a configuration’s flexibility? A number of researchers have identifiedproperties of E that govern flexibility under general plant capacity and product demand.The key insight is that more connected flexibility configurations are more flexible (Jordanand Graves, 1995). The idea is that two plants can share capacity if they are connectedin a configuration. So a flexibility configuration with higher connectivity can shift capacitybetween plants to satisfy more demands. The concept of “connectivity” is further extendedin at least two directions.

It is important not only to have two plants connected, but also to have them close toeach other. The idea of connectivity is further extended to ideas of “distance” and “flow”.In Figure 2.1, plants i

1

and i2

are connected in all three panels. To illustrate the concepts of“distance” and “flow”, loosely speaking, plants i

1

and i2

are closer to each other in panel (a)than they are in panel (b). The volume of flow between plants i

1

and i2

is larger in panel (c)than in panel (a). The work by Chou et al., 2011 and Deng and Shen, 2012 is along the lineof “distance”, while the work of Iravani, Oyen, and Sims, 2005; Iravani, Kolfal, and Oyen,2007 is related to “volume of flow”.

CHAPTER 2. LITERATURE REVIEW 10

Figure 2.1: Concepts of “distance” and “flow” in process flexibility design

plant product

𝑖

𝑖

𝑖

𝑖

𝑖

𝑖

plant product

(a) (b) (c)

plant product

2.2 Marginal benefits of flexibility configuration

Jordan and Graves, 1995 provided an interesting example of a configuration with n plantsand n products. They showed that the configuration’s flexibility increases notably when thenumber of links increases from n to 2n. Then, strikingly, they demonstrated that the config-uration with 2n links (the chaining configuration) is almost as flexible as the total-flexibilityconfiguration. Simchi-Levi and Wei, 2011 proved that the marginal benefit increases as thelong chain is constructed, and the largest benefit is always achieved when the chain is closed.Two properties of the marginal benefits of links have been found - supermodularity (Simchi-Levi and Wei, 2011) and concavity (Aksin and Karaesmen, 2007). For example, Aksin andKaraesmen, 2007 showed that increasing the total number of links from 2n to 3n has alarger impact on flexibility than increasing the total number of links from 4n to 5n. Notethat both results are based on the assumption that the network is symmetric, i.e., all plantshave the same capacity and all products have the same demand. In a another setting of pro-cess flexibility design, Bassamboo, Randhawa, and Mieghem, 2010; Bassamboo, Chu, andRandhawa, 2011 found similar results. Bassamboo, Randhawa, and Mieghem, 2010 provedthat flexibility exhibits decreasing returns. Bassamboo, Chu, and Randhawa, 2011 showedthat the correspondence that maps flexible resources to the set of demands that they canprocess is submodular.

2.3 Marginal e↵ects of plant capacity, demand mean

and standard deviation

Not many studies have aimed to study the e↵ects ofC, µ and � in Jordan and Graves, 1995’ssetting of process flexibility. Aksin and Karaesmen, 2007 analyzed how the marginal benefitof C changes with flexibility configuration E. They proved that when the total capacity islower than the total realized demand, the marginal benefit of C increases with flexibilityconfiguration E. Though the analysis of the e↵ects of µ and � is sparse in the literature,there are extensive discussions on the correlation between product demands. Jordan andGraves, 1995 show that it is only needed to have negatively correlated products in the same

CHAPTER 2. LITERATURE REVIEW 11

chain, not in the same plant. Fine and Freund, 1990, van Mieghem, 1998 and Bish andWang, 2004 discussed the e↵ect of demand correlation on flexibility. Note that capacity C isnot a decision variable in Jordan and Graves, 1995’s setting, whereas the process flexibilitysetting studied by Fine and Freund, 1990, van Mieghem, 1998 and many others explicitlyconsider the capacity investment decisions. So a sensitivity analysis of capacity C helpsconnect these two settings of flexibility design.

2.4 Other settings where process flexibility design has

been applied

Process flexibility design has been studied in many settings. For instance, Hopp, Tekin, andOyen, 2004 showed that the chaining strategy possesses strong capacity balancing and vari-ability bu↵ering properties in serial production systems with flexible workers. Iravani, Kolfal,and Oyen, 2007 studied flexibility in a call-center labor cross-training setting. Sheikhzadeh,Benjaafar, and Gupta, 1998 analyzed the chaining configuration for equipment flexibilityproblems. Lim et al., 2011 studied the chaining configuration with supply disruptions. Sev-eral di↵erent objectives have been studied in Jordan and Graves, 1995’s framework. Chouet al., 2011 and Chou, Chua, and Teo, 2010 analyzed the worst-case performance, range, andresponse of the flexibility system design. Chou et al., 2011’s worst-case analysis is di↵erentto ours in that they studied the worst-case performance over all possible demand realiza-tions given a particular demand distribution. Hopp, Iravani, and Xu, 2010 considered theplacement of vertical flexibility across multiple stages in a supply chain. Chou, Teo, andZheng, 2008 o↵ered a comprehensive review of several important issues in process flexibil-ity design. van Mieghem, 1998, Bish and Wang, 2004 and Bish, Muriel, and Biller, 2005provided outstanding analyses of flexibility in a two-product framework.

12

Chapter 3

Process Flexibility Design inUnbalanced Networks

In Chapter 3, we focus on minimizing expected total shortfall in unbalanced and symmetricnetworks that have 2k edges. Due to the complex combinatorial and stochastic character ofthe flexibility design problem, finding an optimal solution (i.e., which links to create) underan arbitrary resource constraint (i.e., the number of links |E|) is challenging. We insteadfocus on configurations with 2k edges. For configurations with fewer than 2k edges, at leastone product can only be produced by one plant, and the corresponding flexibility level canbe notably enhanced by connecting one more edge to that product. On the other hand,configurations with more than 2k edges are not systematically considered in this chapterfor two reasons. (i) It is accepted that chaining configuration (with 2k total edges) alreadyachieves most of the benefits of total-flexibility configuration. Some pertinent argumentsare available in Jordan and Graves, 1995, Aksin and Karaesmen, 2007 and Bassamboo,Randhawa, and Mieghem, 2010. (ii) Even in balanced and symmetric networks, the questionof the optimality of the l-chain configuration (l > 2) among all configurations with totalnumber of k · l edges is still open (e.g., Chou et al., 2011). An l-chain is an m by m bipartitegraph where each product j is linked to plants j, j+1, . . ., j+ l�1 (subtract m if exceedingm). The extension to configurations where |E| = 3k or |E| < 2k is briefly touched at theend of Chapter 3.

The rest of this chapter is organized as follows: Section 3.1 proposes a new flexibilitydesign guideline in symmetric networks. Section 3.2 discusses how the new design guidelinescan be implemented in asymmetric networks. In Section 3.3, we discusses the relationshipof the newly proposed design guideline and existing flexibility design guidelines and indices.Finally, Section 3.4 summarizes Chapter 3. Some of the results has been published in theManufacturing $ Serice Operations Management journal. The name of the article is processflexibility design for unbalanced network.

CHAPTER 3. PROCESS FLEXIBILITY DESIGN IN UNBALANCED NETWORKS 13

3.1 Unbalanced Network Flexibility Design

Guidelines

The ideal symmetric network is rarely encountered in practice. Yet, we find symmetricnetworks attractive for two reasons. First, the analysis for symmetric networks yields insightson flexibility for asymmetric networks. Second, as stressed by Iravani, Oyen, and Sims, 2005,accurate data and forecasts are usually unavailable for strategic and long-term decisions.

We use simulation to identify the most flexible configurations in this chapter. Throughoutthe chapter we use expected total shortfall as the sole gauge of a configuration’s flexibility,and low expected total shortfall is preferred. The expected shortfall is approximated bythe average shortfall of 10,000 randomly replicated demand scenarios. Negative demand isreplaced with 0 when generating random demand. In each scenario, we solve the capacityallocation problem using the randomly generated demand. We assume Normal demanddistributions and fix coe�cient of variation (hereafter cv) at 0.4. By definition of a symmetricnetwork all plants have the same capacity and all products face i.i.d. demand; therefore, wearbitrarily label the plants 1 to m and label the products 1 to k. Define �(⇤) = {i 2 V |9j 2⇤ s.t. (i, j) 2 E},⇤ ✓ V 0. �(⇤) is the set of plants that are directly connected to productset ⇤ ✓ V 0. There is a one-to-one correspondence between E and flexibility configurations.For instance, E = {(i, j)|i 2 V, j 2 V 0} denotes the total-flexibility configuration. There isalso a one-to-one correspondence between E and �(j), 8j 2 V 0.

Chaining Guideline (c) advocates building a chaining configuration between all plantsand m arbitrary products. By network symmetry, we assume without loss of generality thatthe chaining configuration is built out of the first m products. That is,

81 j < m,�(j) = {j, j + 1};�(m) = {1,m}.

Chaining Guideline (c) reduces the number of undetermined edges by 2m, and is the foun-dation of the Unbalance Network Flexibility Design Guidelines.

Unbalanced Network Flexibility Design Guidelines (a) and (b):m < k m+ bm2 cAccording to the Chaining Guidelines (a) and (b), m�mod(k,m) plants should all be directlyconnected to b k

m

c products, and mod(k,m) plants should all be directly connected to d k

m

eproducts. When k

m

is not an integer, some plants are directly connected to one more productthan other plants, showing that some plants are more utilized than others. Intuitively, plantsthat are more utilized should not be close to each other. In other words, we want to evenlymix the more utilized plants with the less utilized plants. How to define the distance betweenplants?

Recall that the chaining configuration can be rearranged as a circle. The circular repre-sentation of the chaining configuration provides a natural measure of the distance betweentwo plants. For example, in Figure 1.2, plants 1 and 2 are closer to each other than plants


Plant

1'

m'

m-1'

m-2'

3'

2'

m+1' 1

m

m-1

m-2

3

2

Product

»»º

««ª �21m

»»º

««ª �21m

»»º

««ª �21m

»»º

««ª �21m

Plant

1'

m'

m-1'

m-2'

3'

2'

m+1' 1

m

m-1

m-2

3

2

Product

»»º

««ª �21m

»»º

««ª �21m

»»º

««ª �21m

»»º

««ª �21m

Plant

1'

m'

m-1'

m-2'

3'

2'

m+1' 1

m

m-1

m-2

3

2

Product

»»º

««ª �21m

»»º

««ª �21m

»»º

««ª �21m

»»º

««ª �21m

Plant

1'

m'

m-1'

m-2'

3'

2'

m+1' 1

m

m-1

m-2

3

2

Product

»»º

««ª �21m

»»º

««ª �21m

»»º

««ª �21m

»»º

««ª �21m

Product Product Product Product

Figure 3.1: Symmetric Configurations where k = m + 1 (“Configuration A > configurationB” means that configuration A is more flexible than configuration B.)

1 and 3. We define the “distance” between two plants to be the minimum number of linksneeded to connect them in the chaining configuration.

For the simplest case where k = m + 1, Figure 3.1 lists all the distinct configurationsthat satisfy the Chaining Guidelines. Figure 3.2 reveals that the far left configuration inFigure 3.1 performs the best in terms of expected total shortfall. In addition, the network’sexpected total shortfall decreases from 7.4 to 6.14 when a link is properly added to product11 (m + 1). We suggest that all products should have some level of flexibility, which is oneof the many reasons why we compare configurations with 2k edges. Figure 3.2 indicatesthat the gap between the total-flexibility configuration and a configuration with 2k edges issignificant even when the 2k edges are properly added. Since the system capacity distributedover 10 plants is equal to the average total demand across 11 products, each plant has extracapacity that needs to be shared to satisfy demand of the last product, m+ 1.

Let i1j

and i2j

be the two plants that are directly connected to product j, 8m < j k.Unbalanced Network Flexibility Design Guideline (a):

8m < j k,�(j) = {i1j

, i2j

} s.t. bm+1

2

c � |i2j

� i1j

| � dm�1

2

e.

Unbalanced Network Flexibility Design Guideline (a) suggests connecting product j (m <j k) to a pair of plants that are far from each other. For example, in Configuration(iv) of Figure 1.2, product 13 is directly connected to plants 1 and 7. Unbalanced NetworkFlexibility Design Guideline (a) virtually now brings those segments of the circle that werefar away close together, which is beneficial not only for that additional product being linked,but also for the ones in the circle that can now easily reach capacity at the other end of thecircle.


6.96

6.606.32

6.17 6.14 6.176.32

6.60

6.96

7.40

6

6.5

7

7.5

8

pected�Total�Shortfall |ī(11)|=1|ī(11)|=1

5.32

5

5.5

2 3 4 5 6 7 8 9 10

Exp

The�second�plant�that�is�directly�connected�to�product�11

Product�11�is�first�directly�connected�to�plant�1

Figure 3.2: Expected Total Shortfalls of Configurations in Figure 3.1 (m = 10, k = 11,C

i

= 11, and E[Dj

] = 10. The horizontal line above the curve is the expected total shortfallwhen product 11 is directly connected to only one plant. The horizontal line below the curveis the expected total shortfall of the total-flexibility configuration.)

Unbalanced Network Flexibility Design Guideline (b) suggests connecting the last k�mproducts to plants that are evenly spaced on the circle denoting the chaining configuration.Unbalanced Network Flexibility Design Guideline (b) is comprised of two parts. We rearrangeplant indices i

1j

and i2j

, 8m < j k in ascending order, i(1)

i(2)

. . . i(2k�2m�1)

i(2k�2m)

. Then, the conditions i(l+1)

� i(l)

� b m

2(k�m)

c, 81 l < 2k � 2m and i(1)

+ m �i(2k�2m)

� b m

2(k�m)

c guarantee that plants i1j

and i2j

, 8m < j k are evenly spaced on thechain built by the Chaining Guidelines. The above idea is formally represented as the firstpart of Unbalanced Network Flexibility Design Guideline (b):

�(j) = {i1j

, i2j

}, 8m < j k, such that i(1)

i(2)

. . . i(2k�2m�1)

i(2k�2m)

-therearrangement of i

1j

and i2j

, 8m < j k in ascending order- satisfy i(l+1)

� i(l)

� b m

2(k�m)

c81 l < 2k � 2m and i

(1)

+m� i(2k�2m)

� b m

2(k�m)

c.

For example, in Configuration (iv) of Figure 1.2, we connect plants 1, 3, 5, 7, 9, and 11 toproducts 13, 14 and 15. Note that plants 1, 3, 5, 7, 9, and 11 are evenly spaced on thechaining configuration.

When k m+bm

2

c, the circle will be cut into 2(k�m) arcs after applying the UnbalancedNetwork Flexibility Design Guideline (b). When m

2(k�m)

is not an integer, mod(m, 2(k�m))

of these arcs each will contain d m

2(k�m)

e products, and other arcs each will contain b m

2(k�m)

cproducts. How to array these arcs in the circle? The second part of Unbalanced Net-work Flexibility Design Guideline (b) advises against placing the arcs that contain b m

2(k�m)

cproducts adjacent to each other.

To see why this arrangement makes sense, let us compare the two configurations inFigure 3.3. Both configurations satisfy Unbalanced Network Flexibility Design Guideline


>

Configuration (1) Configuration (2)

Figure 3.3: Unbalanced Network Flexibility Design Guideline (b) (“Configuration A > con-figuration B” means that configuration A is more flexible than configuration B.)

(a) and the first part of Guideline (b). Both configurations have m = 20 plants and k = 24products. All 20 plants and the first 20 products form a chaining configuration, whichtransforms to a circle in circular representation. The circle is cut into 8 circle arcs bythe 8 links connected to the last 4 products. Since 20/8 is not an integer, the first partof Unbalanced Network Flexibility Design Guideline (b) suggests 4 of these 8 circle arcseach contain 3 products, and other circle arcs each contain 2 products. The only di↵erencebetween configurations (1) and (2) in Figure 3.3 is that circle arcs containing 2 products arenot adjacent to each other in configuration (1). Simulation results indicate that configuration(1) performs notably better than configuration (2). When each plant has capacity C

i

= 20and each product has average demand E[D

j

] = (m · C/1.1)/k, the expected total shortfallsof configurations (1) and (2) are 2.7592 and 2.8085, respectively.

Conjecture 1. In a symmetric network with m plants, k products, m < k m+ bm

2

c, and|E| 2k, the most flexible configuration obeys the Chaining Guidelines and the UnbalancedNetwork Flexibility Design Guidelines (a) and (b).

Intuitively, the optimal configuration for symmetric networks should be symmetric. Per-fect symmetry is one of the many reasons why Unbalanced Network Flexibility Design Guide-lines (a) and (b) lead to flexible configurations. Figure 1.1 suggests that the combinationof the Chaining Guidelines and Unbalanced Network Flexibility Design Guidelines performsvery well when m < k m+ bm

2

c.Figure 1 displays a cyclical pattern in the Chaining Guidelines maximum percentage

increase in expected shortfall relative to the total-flexibility configuration. For example,the Chaining Guideline’s percentage increase is much lower for 18 products than for 15products. When k = 18, configurations that satisfy the Chaining Guidelines must also satisfy


Unbalanced Network Flexibility Design Guideline (b). However, when k=15, configurationsthat satisfy the Chaining Guidelines may violate Unbalanced Network Flexibility DesignGuideline (b). Hence, the Chaining Guidelines perform better for 18 products than for 15products.

Unbalanced Network Flexibility Design Guideline (c):m < k m(m� 1)/2

For k > m+bm

2

c, the diametrically-opposed, evenly spaced assignment of products to plantscannot be continued for the last k�(m+bm

2

c) products, since there is no pair of diametricallyopposed plants that do not have a third product assigned to each of them. Thus, we need anew guideline, (c), to guide the assignment of those last k � (m+ bm

2

c) products.Unbalanced Network Flexibility Design Guideline (c) advises against assigning two prod-

ucts to the same pair of plants. The key idea is to spread risk as much as possible. Assigningproducts to di↵erent pairs of plants provides a better hedge against the uncertain demand.Figure 1.1 indicates that when k � m + bm

2

c, the combination of Chaining Guidelines andUnbalanced Network Flexibility Design Guidelines guarantees that deviation from expectedshortfall of total-flexibility configuration is small. We hypothesize that Unbalanced NetworkFlexibility Design Guideline (c) and the Chaining Guideline adequately manage the lastk � (m+ bm

2

c) products.

Guidelines for Certain Configurations with |E| 6= 2k

The key to our analysis is the circular representation of flexibility configuration. By thiscircular representation, the distance between any two plants is well-defined. This circularrepresentation is meaningful as long as there exists a chaining configuration among all plants.It also has implications beyond networks where each product is directly connected to twoplants.

For symmetric and balanced networks where |E| = 3k, it is tempting to say that the3-chain configuration is optimal, e.g., see Theorem 1 in Iravani, Oyen, and Sims, 2005.Figure 3.4 plots the 3-chain configuration and its circular representation. Recall that theintuition behind Unbalanced Network Flexibility Design Guideline (a) is to connect diamet-rically opposite nodes. We follow this idea and create a configuration that has 3k edges inFigure 3.5. Numerical examples suggest that the configuration in Figure 3.5 consistentlyoutperforms the configuration in Figure 3.4. When each plant has capacity C

i

= 11 andeach product has average demand E[D

j

] = 10, the expected total shortfalls of configurationsin Figures 5 and 6 are 1.3934 and 1.3923, respectively. Although the di↵erence is practicallynegligible, it is statistically significantly di↵erent from 0 (p = 3.357e� 006, Wilcoxon).

The intuition behind the Unbalanced Network Flexibility Design Guidelines also appliesto networks where |E| = m+k. When only m+k links are available, we can build a chaining


Plant Product

circular representation

1

2

3

4

5

6

7

8

9

10

11

12

1

3-chain configuration

2

3

4

5

6

7

8

9

10

11

12

1

2

3

4

5

6

7

8

9

10

11

12

1

2

3

4

5

6 7

8

9

10

11

12

Figure 3.4: 3-chain Configuration and Its Circular Representation

�

1

2

3�

4�

5�

6

7

8�

9�

10�

11�

121

2

3�

4�

5

67

8�

9�

10�

11�

12

Figure 3.5: A Configuration with 3k Edges


1

2

3

4

5

6

7

8

9

10

11

12 1

2

3

4

5

6 7

8

9

10

11

12

13

14 <

Configuration (1) Configuration (2)

1

2

3

4

5

6

7

8

9

10

11

12 1

2

3

4

5

6 7

8

9

10

11

12

13 14

Figure 3.6: A Configuration where �(j) 6= 2 (“Configuration A > configuration B” meansthat configuration A is more flexible than configuration B.)

configuration1 connecting all m plants and m products, and have one link for each productm+1, m+2, . . ., k. Unbalanced Network Flexibility Design Guideline (b) suggests that linksfor products m + 1 to k should be evenly spread. Consider the two configurations in Fig-ure 3.6. Both configurations satisfy the Chaining Guidelines, but configuration (1) violatesthe intuition behind Unbalanced Network Flexibility Design Guideline (b) because the circleis cut into two arcs of di↵erent lengths (in the circular representation). Simulation confirmsthat configuration (2) performs better than configuration (1). When each plant has capacityC

i

= 20 and each product has average demand E[Dj

] = (20 · m/k)/1.1, configurations (1)and (2) have expected total shortfalls of 9.3996 and 6.5047, respectively.

Figure 3.7 plots another example in networks where |E| = m + k. Both configurationshave 12 plants and 20 products. All 12 plants and the first 12 products form a chainingconfiguration, which is a circle in the circular representation. The circle is cut into 8 circlearcs by the 8 links connected to the last 8 products. In both configurations, 4 of the 8circle arcs each contain 2 plants and the remaining 4 circle arcs each contain 1 plant. Thedi↵erence between these 2 configurations is the link for product 16. Configuration (2) isbetter because its links for products 13-20 are chosen to alternate the circle arcs containingtwo products with the groups containing one. Simulation indicates that configuration (2)performs notably better than configuration (1). When each plant has capacity C

i

= 20 andeach product has average demand E[D

j

] = (m · C/1.1)/k, the expected total shortfalls ofconfigurations (1) and (2) are 5.5529 and 5.1234, respectively.

1We would like to thank Prof. Stephen C. Graves for suggesting this configuration.


<

2

3

4

5

6

7

8

9

10

11

12 1

2

3

4

5

6 7

8

9

10

11

12

Configuration (2)

13 14

15

16 17

18

19

20

1

2

3

4

5

6

7

8

9

10

11

12 1

2

3

4

5

6 7

8

9

10

11

12

Configuration (1)

13 14

15

16 17

18

19

20

1

Figure 3.7: Unbalanced Network Flexibility Design Guideline (b) (“Configuration A > con-figuration B” means that configuration A is more flexible than configuration B.)

1

10

7

4

3

2

5

68

9

11

121

2

3

4

5

67

8

9

10

11

12

14

13

1

10

7

4

3

2

5

68

9

11

121

2

3

4

5

67

8

9

10

11

12

15

13

14

1

10

7

4

3

2

5

68

9

11

121

2

3

4

5

67

8

9

10

11

12

14

13

15

Figure 3.8: Inconstant k

Varying Number of Products

In a highly volatile market, the number of products k is likely to change over time, i.e.,k increases as new products are introduced to the market. The number of plants m mayalso change, i.e., m decreases as some plants are shut down. The analysis for uncertain kis analogous to that for uncertain m, so we focus on uncertain k. Consider a symmetricnetwork with 12 plants and 14 products. The most flexible structure for this network isshown in the left panel of Figure 3.8. Now imagine that a new product is introduced intothe market and that changing the existing edges is prohibitive. The best structure we canachieve without changing edges is shown in the middle panel of Figure 3.8. However, had theproduction manager known there would be 15 products in advance, he could have achieved


the more flexible structure shown in the right panel of Figure 3.82.On the other hand, some structures remain optimal over changes in network size. For

example, suppose that the original estimate of k is the same as m, k = m. In this case,the original optimal structure is a chaining structure. Even if the value of k increases tok0, there is always a chaining structure (Flexibility Design Rule (i)) which is a part of theflexible structure for unbalanced networks (k0 > m). So the most flexible structure can stillbe constructed without changing the existing edges. In general, we have the following result:

Theorem 2. For symmetric networks, if the number of products k (k < m+ dm

2

e) increasesto k0 (k0 m + dm

2

e), then the optimal (|E| = 2k) structure designed for k products can beused as part of the optimal (|E| = 2k0) structure for k0 products if (k0 �m)/(k �m) is aninteger.

3.2 Asymmetric Networks

The discussion for asymmetric networks serves two purposes. First, as symmetric networksyield insights that might not lend themselves to general conclusions for asymmetric networks,we present the evidence that Flexibility Design Rules drawn from symmetric networks arestill meaningful for asymmetric networks. Second, due to the complexity of asymmetricnetworks, there might exist systematic conditions under which the rules developed fromsymmetric networks do not work for asymmetric networks. We touch on some relevant ideasin the main text and explain them with more details in the Online Electronic Companion.

Symmetric Plants but Non-i.i.d. Demands

Non-i.i.d. demand poses serious technical challenges. We isolate the e↵ects of asymmetricdemand by assuming that all plants have identical capacity. To further simplify, we assumethat all products’ demands have identical coe�cient of variation. This assumption is notoverly restrictive - we will reiterate this point in greater details after the result is laid out.In addition, an approximation method for uniform coe�cient of variation is provided in theOnline Electronic Companion.

Jordan and Graves, 1995 studied product demand correlation and showed that it is notimportant to identify negatively correlated products and build them in the same plant. Itis only necessary to include negatively correlated products in the same chain. FlexibilityDesign Rule (i) guarantees that all products are encompassed in a chain. Thus, we do notneed to pay special attention to product demand correlation as long as Flexibility DesignRule (i) is obeyed. Subsequent analysis assumes independent demand.

Because product demand is not symmetric, di↵erent choices of the m products whenforming the chaining structure can result in di↵erent performance levels. Given uniform

2By the simulations using data C = 15, µ=12 and cv = 0.4, the expected total shortfalls of the middleand right structures in Figure 3.8 are 8.5869 and 8.1038, respectively.


coe�cient of variation, we suggest choosing the products with top m largest demands whenbuilding the chaining structure (Flexibility Design Rule (i)). This argument is bolstered bynumerical experimentation reported in the Online Electronic Companion and the followingtwo observations.

First, the Chaining Guideline advises practitioners to equalize the number of products(measured in terms of total expected demand) to which each plant is directly connected. TheChaining Guideline is more likely to be satisfied if the chaining structure is built out of theproducts with top m largest demands. For example, suppose that there are 5 products withaverage demands 1, 2, 3, 4 and 5 respectively. These 5 products will be built at 4 plants. Ifwe build the chaining structure out of demands 1, 2, 3 and 4, the last demand 5 will causehigh variations in the demands to which each plant is directly connected. Apparently, suchhigh variations could be avoided had we built the chaining structure out of demands 2, 3, 4and 5 in the first place.

Second, for a product with extremely large average demand, we may need to assign threeor more plants for production. For products with negligible demand, a single plant is enough.This argument is in line with the Node Expansion guideline (Chou07), but contradicts theChaining Guideline of equalizing the total capacity directly connected to each product. Ourprevious discussions assume that product demand is symmetric, so connecting each productto two edges is reasonable. Now with asymmetric product demand, it becomes reasonableto allocate more resources to products with larger average demand. We want to build thechaining structure out of the products with top m largest demands. By doing so, productswith large average demand will be linked to at least two edges.

Suggestions for non-i.i.d. demands. When implementing Flexibility Design Rule (i),build the chaining structure out of the products with top m largest demands.

Asymmetric Plants, i.i.d. Demands

Now assume that each product faces independently and identically distributed random de-mand, but plant capacity varies.

In general, a product should be directly connected to a plant with large excess capacity.Large excess capacity lowers the possibility of stockouts (Graves and Tomlin, 2003). Thisargument is well grounded numerical experiments that are reported in the Online ElectronicCompanion. Moreover, we contribute by highlighting the additional benefits of directlyconnecting products to plants that are close to plants with large excess capacity. Thus, incase not all products are able to connect to plants with large excess capacity, it is still bettero↵ connecting products to plants that are close to plants with large excess capacity.

Suggestions for asymmetric plants. Arrange all plants in such a way that products m+1to k are able to be assigned to plants that have large excess capacity or that are close toplants with large excess capacity.


Case Study

The Flexibility Design Rules are illustrated via the subsequent fictional case study. Thiscase study is based on a real set of vehicle and assembly plants in Jordan and Graves, 1995.Structure (1) in Figure 3.9 shows the plants and products in this example. The numbersbeside the plants and products are their capacities and expected demands3. They showthat by properly adding six more links to the original structure, the resulting structure’sperformance is significantly enhanced. Regardless, our main purpose with this example is toillustrate our guidelines rather than to compare our method with the Chaining Guideline4.We start with no initial edges and demonstrate the implementation of Flexibility DesignRules.

First, Flexibility Design Rule (i) and the suggestions for asymmetric demands lead us tostructure (2) in Figure 3.9, where a chaining structure is formed with all plants and eightproducts with larger demand. In this step we deliberately connect plants with larger capacityto products with higher expected demand, such as product A and plant 1. Next, we assignthe remaining eight products. The expected demand for these products is fairly small, sowe connect only one link to each of them5. Flexibility Design Rule (ii) does not apply here.Finally, according to Flexibility Design Rule (iii), each plant should only build exactly one ofthe remaining eight products. In this way, the circle that represents the chaining structureis cut evenly by the links connected to the remaining products. To follow the suggestionsfor asymmetric plants, each plant’s excess capacity is computed. Intuitively, subtract theaverage of the total expected demand served by each plant from its capacity. For example,plant 1’s excess capacity is 380�(320+150)/2 = 95. By the suggestion for asymmetric plants,we build products 9 to 16 at plants in descending order of excess capacity, see structure (3)in Figure 3.9.

In the preceding example, Flexibility Design Rules and the guidelines for asymmetricnetworks are easily implementable 6. Due to the inherent freedom of the suggestions forasymmetric network, the above procedure could lead to many structures. Most of thesestructures are near-optimal7. For example, the expected total shortfall for structure (3) inFigure 3.9 is 166, only slightly larger than the expected total shortfall of the total-flexibility

3We follow their assumption that product demands are truncated (±2�) normally distributed randomvariables with standard deviation � equal to 40% of expected demand. The sixteen products (vehicles) fallinto three groups: products A to F (compact cars), products G to M (full-sized cars), and products N to P(luxury cars). Demand for pairs of products in the same group is positively correlated with coe�cient 0.3.Demand for products in di↵erent groups is independent. In their example, there were 18 links before theproject.

4Chou07 made the comparison for this example and showed that the Node Expansion guideline is

slightly better.5There are only 24 links in the suggested structure from Jordan and Graves, 1995. We do not want to

exceed 24 links, and we already use 16 edges to build the chaining structure, so there are at most 8 morelinks to be added. We connect each remaining product to only one plant.

6It took the authors 5 minutes to produce this structure without the aid of a computer or calculator.7We strongly agree with Jordan and Graves, 1995 that “there is no one optimal flexibility plan; rather

there are many that are near-optimal.”


B

C

K

A

J

I

H

G

F

E

D

P

O

N

M

L

Product Plant

1

8

7

6

5

4

3

2

B

C

K

A

J

I

H

G

F

E

D

P

O

N

M

L

Product Plant

1

8

7

6

5

4

3

2

320

150

270

110

220

110

120

80

140

160

60

35

40

35

30

180

320

150

270

110

220

110

120

80

140

160

60

35

40

35

30

180

380

230

250

230

240

230

230

240

380

230

250

230

240

230

230

240

B

C

K

A

J

I

H

G

F

E

D

P

O

N

M

L

Product Plant

1

8

7

6

5

4

3

2

320

150

270

110

220

110

120

80

140

160

60

35

40

35

30

180

380

230

250

230

240

230

230

240

(1) (2) (3)

Figure 3.9: A Network in Jordan and Graves (1995)


structure (155). The expected total shortfall for structure (3) is lower than the expectedtotal shortfall of the final structure suggested by Jordan and Graves, 1995 (171).

3.3 Existing Design Guidelines and Indices

Besides the Chaining Guideline, many other guidelines/indices can deal with unbalancednetworks. It is necessary to address their relationships with the Flexibility Design Rules.The guidelines and indices are explained briefly in Sections 6.1 and 6.2, respectively. Weintroduce the necessary technical details and omit managerial implications. Then, Section6.3 presents the connections between di↵erent guidelines and indices and the FlexibilityDesign Rules.

Design Guidelines

1. Chaining GuidelineThe Chaining Guideline (Jordan and Graves, 1995) has received much attention in the

literature. They outlined a number of objectives to help design a flexible system:

(a). try to equalize the number of plants (measured in total units of capacity) to whicheach product is directly connected;

(b). try to equalize the number of products (measured in total units of expected demand)to which each plant is directly connected; and

(c). try to create a circuit(s) that encompasses as many plants and products as possible.

The Chaining Guideline leads to the chaining structure in a balanced symmetric network.2. Node Expansion GuidelineThe notion that a better connected structure is more flexible has led to the Node Ex-

pansion guideline (Chou07). The structure augments itself by adding links iteratively toimprove the Node Expansion ratio. The Node Expansion ratio of product node j 2 V 0 is

�j

,P

i2V :(i,j)2E Ci

E[Dj

], j 2 V 0.

Similarly, the Node Expansion ratio of plant node i 2 V is

�i

,P

j2V 0:(i,j)2E E[D

j

]

Ci

, i 2 V.

The Node Expansion guideline is to add an edge that is not in E yet to increase

� = min{mini2V

�i

,minj2V 0

�j

}

as much as possible.


Flexibility Indices

There are currently five major flexibility indices: (1) JG-Index (Jordan and Graves, 1995); (2)Structural Flexibility indices (Iravani, Oyen, and Sims, 2005); (3) WS-APL method (Iravani,Kolfal, and Oyen, 2007); (4) g-Measure (Graves and Tomlin, 2003) and (5) Expansion Index(Chou, Teo, and Zheng, 2008).

1. JG-IndexThe JG-Index, proposed by Jordan and Graves, 1995, is as follows. Choose any subset of

product nodes ⇤ ✓ V 0. The probability that unsatisfied demand for product set ⇤ exceedsunsatisfied demand for the total-flexibility structure is

Pr

{X

j2⇤

Dj

�X

i2�(⇤)

Ci

} � max{0,X

j2V 0

Dj

�X

i2V

Ci

}!

.

The JG-Index is the maximum of these probabilities over all subsets ⇤ ✓ V 0. A smallerJG-Index corresponds to a higher level of process flexibility.

2. Structural Flexibility IndicesIravani, Oyen, and Sims, 2005 defined the Structural Flexibility matrix M. The entry

(i, j) of M is the maximum number of non-overlapping paths from product node i to productnode j, i, j 2 V 0. M(i, i) is the degree of product node i. The Structural Flexibility indexis either the largest eigenvalue of M or the mean of all entries of M. Larger StructuralFlexibility indices correspond to greater process flexibility.

3. WS-APL MethodIravani, Kolfal, and Oyen, 2007 created a work sharing (WS) network model for which

its average shortest path length (APL) metric can predict flexibility. WS-APL stands forwork sharing- average shortest path length. The key is to convert the flexibility structure toa work sharing model for which the APL metric will be a useful indicator of flexibility. Theconversion consists of three steps.Step 1. Every plant i 2 V is represented by node i 2 V .Step 2. An undirected arc is placed between nodes i and j if there exists at least one productthat can be produced at both plants i and j.Step 3. The length of an arc connecting two nodes i and j is the reciprocal of the numberof products that can be produced at both plants i and j.The conversion is illustrated using a simple flexibility structure in Figure 3.10. Denote theminimum distance between nodes i and j in the WS network as Lmin

i,j

. The APL metric isthe average length of all paths,

APL =2

m(m� 1)

m�1

X

i=1

m

X

j=i+1

Lmin

i,j

.

Low APL values are preferred. The APL of the structure in Figure 3.10 is (1+ 1+2+0.5+1.5 + 1)/6 = 7/6.


�

�

flexibility�structure� Step�2Step�1�

Product�Plant�

1�

2�

3�

4�

1

2 3

4

1�

2� 3

4�

1� 1

1

0.5�

1�

2� 3

4�

Step�3�WS�network�

1�

2�

3�

4�

Figure 3.10: Illustration of the Computation of the APL Metric

4. g-Measureg-Measure (Graves and Tomlin, 2003) is based on the excess capacity available to any

subset of products ⇤ ✓ V 0, relative to an equal allocation of total capacity. Equal allocation

of total capacity provides each product with capacity C =P

i2V Ci

|V 0| . For any subset of products⇤ ✓ V 0, the relative excess capacity is

g(⇤) =

P

i2�(⇤)Ci

� |⇤|CC

.

The g-Measure is the minimum of g(⇤) over all ⇤ that do not have access to the totalproduction capacity,

g = min⇤✓V

0,|�(⇤)|<|V |

{g(⇤)}.

Larger g-Measures correspond to more flexible structures.5. Expansion IndexThe Expansion Index was proposed by Chou, Teo, and Zheng, 2008. It is defined as the

second smallest eigenvalue of the Laplacian matrix

L = T T T ,

where T is an (m + k) ⇥ |E| matrix. Consider a link l 2 E connecting node i 2 V (withcapacity C

i

) to j 2 V 0 (with mean demand µj

= E[Dj

]). The corresponding entry in columnl 2 E of T is

Til

= �p

Ci

µj

, Tjl

=p

Ci

µj

, and Tpl

= 0, 8p 6= i, j.

More flexible structures have larger Expansion Indices.


�

Chaining Guideline (c) (Chaining) Chaining Guideline (a) (b)

Flexibility Design Rules

All Possible Structures

Figure 3.11: The Relationship Between Chaining Guideline and Flexibility Design Rules

Flexibility Design Rules’ Relationship with ExistingGuidelines/Indices

The relationship between Chaining Guideline and our proposed Flexibility Design Rules isdepicted in Figure 3.11. The essence of the Chaining Guideline is the objective (c) - the chain-ing structure. The chaining structure is demonstrated to perform fairly well for symmetricnetworks. Chaining objectives (a) and (b) can be viewed as refinements of objective (c) forapplications to asymmetric networks. We suggest two new rules to refine chaining objective(c) when it is applied to unbalanced and symmetric network. The shaded area representsall the structures that satisfy Flexibility Design Rules and the suggestions for asymmetricnetworks. Our suggestions for asymmetric networks are probably the best we can achieve interms of robustness and conciseness- we basically tried all the possible interactions betweenFlexibility Design Rules and Chaining objectives (a) & (b).

The Node Expansion guideline can lead to a structure that satisfies Flexibility DesignRule (i) in unbalanced (m < k) symmetric networks where |E| = 2k. The advantage of theNode Expansion Guideline over the Chaining guideline is its capacity to take account explic-itly the demand and capacity information. We expect that the Node Expansion Guidelinewill play an important role in asymmetric networks.

For all the other possible combination of flexibility index and network, we are not ableto provide analytical results and instead rely on numerical studies. In general, all indicesreach consensus on networks where k = m + 1. But disagreements exist in networks wherek = m + 2 and asymmetric networks. Our opinion is that all indices yield insights aboutthe flexibility design problem, and no one can be replaced by the others. In the planningstage, the decision maker can first apply all the guidelines to narrow all possible structuresdown to fewer than 100 candidate structures. Then all the indices and simulations can be


Table 3.1: Objective Functions of Flexibility Indices

Index Research Objective FunctionJG-Index Jordan and Graves (1995) Expected Total Shortfallg-Measure Graves and Tomlin (2003) Expected Total ShortfallStructural FlexibilityIndices

Iravani et al. (2005) Throughput rate

WS-APL Iravani et al. (2007) Customer waiting timeExpansion Index Chou et al. (2008) Average and worst case per-

formance

applied further to pin down the final 5 to 10 candidate structures. Such scheme is feasible-according to our numerical experiences the whole process can be finished within an hourif all the related programs are coded beforehand. As in actual applications there might bemany unmodeled constraints, the ultimate decision may be based on those constraints andother factors. It should also be noted that existing indices were not developed under uniformobjective. Table 3.1 summarizes the di↵erent objective functions.

Generally, optimization of flexibility indices is prohibitively di�cult for various reasons.Some flexibility indices consider plant capacity and/or demand information, which addssignificant challenge to the corresponding optimization problem. As a result, we are onlyable to derive a limited number of analytical results. In summary, we show that (a) FlexibilityDesign Rule (i) is supported by Structural Flexibility indices, (b) Flexibility Design Rule(ii) leads to the optimal g-Measure and WS-APL metric for symmetric networks wherek = m + 1, and (c) Flexibility Design Rule (iii) is justified by g-Measure, and partiallyjustified by WS-APL metric in symmetric networks where Flexibility Design Rules (i) and(ii) are satisfied.

Next we report the details of each index’s relationship with Flexibility Design Rules. Wefocus only on relationships that are either analytically proven or supported by numericalexamples, e.g, the relationship between JG-Index and Flexibility Design Rule (ii). Otherresults, like the relationship between JG-Index and Flexibility Design Rule (i), are not men-tioned for the reason that we are not able to prove or numerically verify them.

Flexibility Design Rule (i)Flexibility Design Rule (i) is supported by the Structural Flexibility indices.

Theorem 3. Among all the unbalanced networks (m < k) with no more than 2k edges,the structures satisfying Flexibility Design Rule (i) and |�(j)| = 2, 8m < j k achieve themaximal attainable Structural Flexibility eigenvalue index of 2k and a mean index of 2.

The above theorem suggests that we can achieve the maximal attainable Structural Flex-ibility indices by applying Flexibility Design Rule (i).


0.157 0.157

0.188

0.16

0.18

0.2

10.65910.769

10.65910.6

10.8

11

x

|ī(11)|=10.157

0.1315

0.1088

0.0878

0 0679

0.0878

0.1088

0.1315

0.157

0.188

0.08

0.1

0.12

0.14

0.16

0.18

0.2

JG�Index

9.856

10.098

10.395

10.65910.769

10.659

10.395

10.098

9.8569 691

9.8

10

10.2

10.4

10.6

10.8

11

pansion�Index

|ī(11)|=10.157

0.1315

0.1088

0.0878

0.0679

0.0878

0.1088

0.1315

0.157

0.188

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

JG�Index

9.856

10.098

10.395

10.65910.769

10.659

10.395

10.098

9.8569.691

9.2

9.4

9.6

9.8

10

10.2

10.4

10.6

10.8

11

Expansion�Index

|ī(11)|=1

|ī(11)|=10.157

0.1315

0.1088

0.0878

0.0679

0.0878

0.1088

0.1315

0.157

0.188

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

2 3 4 5 6 7 8 9 10

JG�Index

The second plant that is directly connected to product 11

9.856

10.098

10.395

10.65910.769

10.659

10.395

10.098

9.8569.691

9

9.2

9.4

9.6

9.8

10

10.2

10.4

10.6

10.8

11

2 3 4 5 6 7 8 9 10

Expansion�Index

The second plant that is directly connected to product 11

|ī(11)|=1

|ī(11)|=10.157

0.1315

0.1088

0.0878

0.0679

0.0878

0.1088

0.1315

0.157

0.188

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

2 3 4 5 6 7 8 9 10

JG�Index


9.856

10.098

10.395

10.65910.769

10.659

10.395

10.098

9.8569.691

9

9.2

9.4

9.6

9.8

10

10.2

10.4

10.6

10.8

11

2 3 4 5 6 7 8 9 10

Expansion�Index


|ī(11)|=1

|ī(11)|=1

Figure 3.12: JG Index and Expansion Index of network (m = 10, k = 11)

Flexibility Design Rule (ii)Flexibility Design Rule (ii) is numerically grounded in the JG-Index and Expansion

Index in networks where k = m+ 1. The JG-Index and the Expansion Index are plotted inFigure 3.12 for the structures in Figure 3.1 when m = 10 (the plant and product parametersare the same as those used in the simulation). The x-axis in Figure 3.12 identifies thesecond plant that is directly connected to product 11 (m + 1). Lower values of the JG-Index and larger values of the Expansion Index indicate higher flexibility. The horizontallines in each figure are the index values when product 11 is directly connected to onlyone plant. Comparing these two figures with Figure 3.2, we conclude that both the JG-Index and Expansion Index accurately assess the flexibility of the structures in Figure 3.1.The JG-Index for the total-flexibility structure is always 0. The Expansion Index for thetotal-flexibility structure is 1100, which is much larger than the Expansion Indices for thestructures in Figure 3.1.

Larger g-Measure indicates greater flexibility. The g-Measure accurately compares therelative flexibility levels of the di↵erent structures in Figure 3.1.

Theorem 4. For symmetric networks satisfying k = m+ 1, Flexibility Design Rule (i) anddenote �(k) = {k, k}, the g-Measure increases with min{|k � k|,m� |k � k|}.

Indeed, Theorem 4 suggests that even when demands are not i.i.d., Flexibility DesignRule (ii) is still valid since g-Measure does not consider demand information.

The WS-APL method supports Flexibility Design Rule (ii) when m is large.

Theorem 5. For any network satisfying k = m+ 1, Flexibility Design Rule (i), and denote�(k) = [k, k], the APL value decreases with min{|k�k|,m�|k�k|} when m is asymptoticallylarge.

Smaller APL values mean greater flexibility. The APL metric of a WS network is essen-tially the expected minimum distance between two uniformly distributed points in the WSnetwork. For any networks satisfying k = m + 1 and Flexibility Design Rule (i), the APL


1

1

1

1

111

1

1

1

1

1

1 1

1

11

0

p

1-p

fom

Figure 3.13: APL Metric Calculation

value is the expected minimum distance between two randomly chosen points from the leftpanel of Figure 3.13. When m is infinite, it is equivalent to assume that all points are uni-formly distributed on the circumference of the circle and that the weight/length of the line is0 (see the right panel of Figure 3.13). The perimeter of the circle is 1, and that the line cutsthe circle into two segments of lengths p and 1� p. The APL value is commensurate to theexpected minimum distance between two uniformly distributed points on the circumferenceof the circle. Following routine computations8 we find that the expected minimum distanceis p3/4+(1�p)3/4+p(1�p)/2. This expression is minimized at p = 1/2, and it is decreasingon p 2 (0, 1/2]. Theorem 5 is valid when m is infinitely large. Note that when m is small,the WS-APL metric may not be so accurate. For example, when m = 6, the most flexiblestructure in the far left of Figure 3.1 does not have the smallest APL value.

Flexibility Design Rule (iii)Given that Flexibility Design Rules (i) and (ii) are satisfied, Flexibility Design Rule (iii)

is supported by g-Measure.

Theorem 6. Among all symmetric networks satisfying Flexibility Design Rules (i) and (ii),those with maximal g-Measure satisfy Flexibility Design Rule (iii).

For any asymptotically large networks satisfying k � m+ 2, Flexibility Design Rules (i)and (ii), APL metric accords with Flexibility Design Rule (iii). The supporting evidence arein the Online Technical Supplement.

3.4 Summary

This chapter proposes several new flexibility design guidelines for unbalanced network. Whenthe number of products is larger than the number of plants and each product is directlyconnected to two plants, we suggest the following.Build a chaining configuration between all plants and the same number of products. Viewthis chaining configuration as a circle. Then,

8The details are in the Online Electronic Companion.


(a) try to connect each remaining product to diametrically opposite plants on the circle;

(b) try to connect remaining products to plants that are evenly spaced on the circle;

(c) try to avoid connecting the same pair of plants to more than one product.

The optimal configuration of a symmetric network should exhibit strong symmetry. Asthe proposed Unbalanced Network Flexibility Design Guidelines are perfectly symmetric,we conjecture that the Unbalanced Network Flexibility Design Guidelines capture criticalelements of the flexibility design problem in symmetric networks. Even though we pri-marily focus on configurations where each product is directly connected to two plants, theproposed guidelines have implications in more general settings. Further, the UnbalancedNetwork Flexibility Design Guidelines serve as a test for future flexibility indices. Finally,our guidelines are easy to implement and can significantly reduce the number of candidateconfigurations in the planning stage of flexibility design.

This chapter builds on the essential intuition from Jordan and Graves, 1995’s ChainingGuidelines and subsequent flexibility indices, but fleshes out and extends these intuitionsto develop Guidelines for unbalanced networks. Since we focus only on Jordan and Graves,1995’s formulation, whether our results could be extended in recognizable form to moregeneral situations is a potential future research question.

33

Chapter 4

Robust Process Flexibility Design

In this chapter, we adopt a worst case analysis. The main reason is the that expected totalshortfall yields little analytical insights in asymmetric networks. We consider the worstexpected total shortfall given all demand joint distributions that have the marginal meanand variance. Section 4.1 covers the basic model. Section 4.2 explains the main solutionapproach. Section 4.3 presents the results. For the sake of completeness, Sections 4.4 and 4.5provide the marginal distribution model and the cross moment model, respectively. Section4.6 concludes this chapter.

4.1 Basic Model Setting

The total shortfall R(D) is random when demand D is random. Following Jordan andGraves, 1995, we use the expected total shortfall, E[R(D)], as the sole gauge of flexibility. LetF be the joint distribution function of demand D. E[R(D)] is a function of F . Informationon F is plausibly hard to obtain. The novel feature of the proposed model is that no specificdemand joint distribution is assumed. Instead, we assume only the marginal moments ofdemand D, i.e., E[D

j

] = µj

and E[D2

j

] = (µj

)2 + (�j

)2. Assume µ and � can be estimatedfrom historical demand data. LetW (C,µ,�) be the maximal expected total shortfall amongall demand joint distributions that have marginal mean µ and standard deviation �,

W (C,µ,�) = supF

ED[R(D)]

s.t.

Z

D2R|V 0|+

dF (D) = 1,

Z

D2R|V 0|+

Dj

dF (D) = µj

, 8j 2 V 0,

Z

D2R|V 0|+

(Dj

)2dF (D) = (µj

)2 + (�j

)2, 8j 2 V 0.

(4.1)

CHAPTER 4. ROBUST PROCESS FLEXIBILITY DESIGN 34

To avoid trivial cases, we have two assumptions on flexibility configuration E and demandstandard deviation �.

Assumption 7. Each product is directly connected to at least one plant, i.e., for all j 2 V 0,|{i : (i, j) 2 E}| � 1. Each plant is directly connected to at least one product, i.e., for alli 2 V , |{j : (i, j) 2 E}| � 1.

Assumption 8. All demands have a positive standard deviation, i.e., �j

> 0 for all j 2 V 0.

4.2 Solution Approach

There are four major steps to solve (4.1). We focus on the intuition and insights and relegatethe complete solution approach to the Appendices.

Step 1 - Simplified Dual Form of W (C,µ,�). The first step is to take dual of (4.1) andsimplify the dual formulation. Proposition 9 provides the simplified dual formulation.

Proposition 9. W (C,µ,�) equals the optimal value of the following optimization problem.

W (C,µ,�) = min✓,⇢,�

"

✓ +X

j2V 0

µj

· ⇢j

+X

j2V 0

✓

(µj

)2 + (�j

)2◆

�j

#

s.t. ✓ � max(x,y)2⇤

(

X

j2V 0

maxdj�0

(yj

� ⇢j

) · dj

� �j

· (dj

)2�

�X

i2V

Ci

· xi

)

,

(4.2)

where

⇤ =

⇢

(x,y) | yj

xi

, 8(i, j) 2 E; xi

� 0, 8i 2 V ; 0 yj

1, 8j 2 V 0�

.

The formal proof is provided in the Appendix. In Proposition 9, ✓, ⇢ and � are the dualdecision variables of the three constraints of (4.1), respectively. The constraint (x, y) 2 ⇤comes from the dual of flexibility design problem,

R(d) = max(x,y)2⇤

✓

X

j2V 0

dj

· yj

�X

i2V

Ci

· xi

◆

. (4.3)

Even though (4.2) is still complex, it tells the first order e↵ects of flexibility configurationE, plant capacity C

i

, and demand standard deviation �j

. W (C,µ,�) is decreasing in E asadding links to E results in a smaller feasible set of x and y. W (C,µ,�) is decreasing in C

i

because an increase in Ci

relaxes the constraint. W (C,µ,�) is increasing in �j

. The reasonis that the optimal �

j

must be non-negative. Suppose that the optimal �j

is negative, thenthe optimal d

j

would be positive infinity and (4.2) would be infeasible. (4.2) cannot tell thesecond order e↵ects.


Step 2 - Characterizations of the Optimal Solutions. The major tradeo↵s in (4.2) aremore evident if we view problem (4.2) as a Stackelberg Game. Imagine that there are threeplayers. Player 1 moves first and decides ✓,⇢,�. Player 2 moves next and decides (x,y) 2 ⇤.Player 3 moves last and selects d � 0. Substituting ✓ in the objective function of (4.2) yields

min⇢,�

"

max(x,y)2⇤

(

X

j2V 0

maxdj�0

(yj

� ⇢j

) · dj

� �j

· (dj

)2�

| {z }

Player 3’s optimization

�X

i2V

Ci

· xi

)

| {z }

Player 2’s optimization

+X

j2V 0

µj

· ⇢j

+X

j2V 0

✓

(µj

)2 + (�j

)2◆

�j

#

.

Now we focus on the other five decision variables, ⇢, �, x, y and d.Player 3 solves

X

j2V 0

maxdj�0

(yj

� ⇢j

) · dj

� �j

· (dj

)2�

(P3)

as a function of players 1 and 2’s decisions ⇢, � and y. Let player 3’s best response functionbe d⇤

j

(⇢j

,�j

, yj

). Player 2 solves

max(x,y)2⇤

(

X

j2V 0

(yj

� ⇢j

) · d⇤j

(⇢j

,�j

, yj

)� �j

· (d⇤j

(⇢j

,�j

, yj

))2�

�X

i2V

Ci

· xi

)

(P2)

as a function of player 1’s decisions ⇢ and �. Backward induction is assumed in (P2) asplayer 2 foresees player 3 selecting d⇤

j

(⇢j

,�j

, yj

). Let player 2’s best response function bex⇤i

(⇢,�) and y⇤j

(⇢,�). Finally, player 1 solves

min⇢,�

(

X

j2V 0

(y⇤j

(⇢,�)� ⇢j

) · d⇤j

(⇢j

,�j

, yj

)� �j

· (d⇤j

(⇢j

,�j

, yj

))2�

�X

i2V

Ci

· x⇤i

(⇢,�) (P1)

+X

j2V 0

µj

· ⇢j

+X

j2V 0

✓

(µj

)2 + (�j

)2◆

�j

)

as he foresees players 2 and 3 selecting (x⇤i

(⇢,�), y⇤j

(⇢,�)) and d⇤j

(⇢,�,y), respectively. Let⇢

⇤,�⇤ be player 1’s optimal decision. For notational simplicity, denote x

⇤ = x

⇤(⇢⇤,�⇤),y

⇤ = y

⇤(⇢⇤,�⇤) and d

⇤ = d

⇤(⇢,�,y). Therefore, (⇢⇤, �⇤, x⇤, y⇤, d⇤) represents an optimalsolution and an equilibrium point of the game. In case where any of (P1), (P2) and (P3)has more than one optimal solution, (⇢⇤, �⇤, x⇤, y⇤, d⇤) denotes any one of the optimalsolutions.

The three players’ incentives are not aligned. Player 1 would like to have (P3) as smallas possible, while player 3 would like to have (P3) as large as possible. For example, player


1 would like to take �j

as small as possible so that (P1) is minimized. However, once �j

is negative, player 3 would choose D⇤j

= 1 and (P3) goes to positive infinity. So player 1cannot have �

j

as small as possible. Player 1 has to take account of players 2 and 3’s optimalresponses, and player 2 has to take account of player 3’s optimal responses. As players 2and 3’s incentives are well aligned, one can combine players 2 and 3 as one player. For theease of exposition, we consider players 2 and 3 as two players. The following propositioncharacterizes the optimal solution ⇢

⇤, �⇤, x⇤, y⇤, and d

⇤.

Proposition 10. Several important characterizations of the optimal solution ⇢

⇤, �⇤, x⇤, y⇤,and d

⇤ are listed below. Under Assumptions 7 and 8,

(i) ⇢⇤j

< 1 and �⇤j

> 0 for all j 2 V 0,

(ii) d⇤j

= max{y⇤j

� ⇢⇤j

, 0}/(2�⇤j

) for all j 2 V 0,

(iii) x⇤i

= maxj:(i,j)2E

y⇤j

for all i 2 V .

Proposition 10 is useful as it reduces the freedom of the optimal solution. Part (ii) ofProposition 10 expresses player 3’s optimal decision as a function of players 1 and 2’s deci-sions. So we can ignore player 3 in the subsequent analysis. Part (iii) builds the connectionbetween x⇤

i

and y⇤j

and enables us to focus on y⇤j

. The major tradeo↵ of (P1) reduces to thetradeo↵ between y⇤

j

and (⇢,�). We examine player 2’s optimal behavior in the next step.Step 3 - The Optimal x⇤ and y

⇤. The following proposition presents a key characterizationof player 2’s optimal solution y

⇤.

Proposition 11. Under Assumptions 1 and 2, for any j 2 V 0, there must exist two distinctoptimal solutions (⇢⇤,�⇤,x⇤,y⇤,d⇤) and (⇢0⇤,�0⇤,x0⇤,y0⇤,d0⇤) where y⇤

j

= 1 and y0⇤j

= 0.

As Proposition 11 holds for any product j 2 V 0, it draws very strong implications abouty

⇤. Conventional wisdom from game theory helps explain Proposition 11. In principal-agentmodels, the optimal policy for the principal is often to o↵er the agent a contract so that theagent is indi↵erent between accepting and rejecting the contract. That is, both acceptingand rejecting the optimal contract is optimal to the agent. In our model, the optimal solution(⇢⇤,�⇤) for player 1 is such that both y⇤

j

= 1 and y⇤j

= 0 can be optimal for player 2.There are many possible cases hinted by Proposition 11. For example, player 2 may have

two optimal solutions

(y⇤1

, y⇤2

, . . . , y⇤|V 0|) = (0, 0, . . . , 0) or (1, 1, . . . , 1). (4.4)

It may also be that player 2 has |V 0| optimal solutions

(y⇤1

, y⇤2

, . . . , y⇤|V 0|) = (1, 0, . . . , 0) or (0, 1, . . . , 0), . . . , or (0, 0, . . . , 1). (4.5)

Proposition 11 does not distinguish between (4.4) and (4.5). The next proposition discussesthe interactions between di↵erent y⇤

j

’s.


Proposition 12. Under Assumptions 1 and 2, the optimal solution (⇢⇤,�⇤) must be suchthat the optimal solution (x⇤,y⇤) can be both (0,0) and (1,1).

Proposition 12 distinguishes between (4.4) and (4.5). To get the idea, consider a configu-ration with two plants and two products where all plants can produce all products. In sucha configuration, the objective value from (y

1

, y2

) = (0, 1) is

(max{�⇢⇤1

, 0})2

4�⇤1

+(1� ⇢⇤

2

)2

4�⇤2

� C1

� C2

,

which is dominated by the objective value from (y1

, y2

) = (1, 1),

(1� ⇢⇤1

)2

4�⇤1

+(1� ⇢⇤

2

)2

4�⇤2

� C1

� C2

.

So (y1

, y2

) = (0, 1) cannot be optimal. For the same reason, (y1

, y2

) = (1, 0) cannot beoptimal. The optimal solutions must be (y⇤

1

, y⇤2

) = (0, 0) and (y⇤1

, y⇤2

) = (1, 1) in this config-uration. The basic idea is that many solutions are inherently dominated by other solutions,and thus cannot be the optimal solution.

Proposition 12 has strong implications over the relationship between (⇢⇤,�⇤) and C. Inthe above example of two plants and two products, Proposition 12 implies that (x,y) = (0,0)and (x,y) = (1,1) have the same objective value,

X

j=1,2

(1� ⇢⇤j

)2

4�⇤j

�(max{�⇢⇤

j

, 0})2

4�⇤j

�

= C1

+ C2

.

IfX

j=1,2

(1� ⇢⇤j

)2

4�⇤j

�(max{�⇢⇤

j

, 0})2

4�⇤j

�

> C1

+ C2

.

(x,y) = (0,0) cannot be optimal. Similarly, if

X

j=1,2

(1� ⇢⇤j

)2

4�⇤j

�(max{�⇢⇤

j

, 0})2

4�⇤j

�

< C1

+ C2

.

(x,y) = (1,1) cannot be optimal. In the Appendix, we show that Proposition 12 leads tothe following result.

Corollary 13. The optimal ⇢⇤ and �

⇤ must satisfy the following conditions.

9fi,j

� 0, 8(i, j) 2 E such thatX

j:(i,j)2E

fi,j

= Ci

, 8i 2 V,

X

i:(i,j)2E

fi,j

=(1� ⇢⇤

j

)2

4�⇤j

�(max{�⇢⇤

j

, 0})2

4�⇤j

, 8j 2 V 0.


Step 4 - The Optimal ⇢⇤ and �

⇤. The last step is to solve player 1’s optimization problem(P1). Because of Proposition 12, we can get rid of (x,y) by replacing it with (0,0). Largelydue to Corollary 13, player 1’s objective function (P1) is equivalent to

W (C,µ,�) = minfij ,⇢j ,�j

X

j2V 0

⇢

µj

· ⇢j

+ ((µj

)2 + (�j

)2) · �j

+(max{�⇢

j

, 0})2

4�j

�

s.t.X

j:(i,j)2E

fi,j

= Ci

, 8i 2 V,

X

i:(i,j)2E

fi,j

=(1� ⇢

j

)2

4�j

� (max{�⇢j

, 0})2

4�j

, 8j 2 V 0.

fi,j

� 0, 8(i, j) 2 E.

⇢j

< 1,�j

> 0, 8j 2 V 0.

The final result is provided below.

Theorem 14. For any j 2 V 0, define

rj

(Dj

) =

(

µj

� Dj

(µj)2

(µj)2+(�j)

2 , when Dj

(µj)2+(�j)

2

2µj,

�1

2

(Dj

� µj

) + 1

2

p

(Dj

� µj

)2 + (�j

)2, when Dj

>(µj)

2+(�j)

2

2µj.

Under Assumptions 1 and 2,

W (C,µ,�) = minfij ,

¯

Dj

X

j2V 0

rj

(Dj

)

s.t.X

j:(i,j)2E

fi,j

= Ci

, 8i 2 V,

X

i:(i,j)2E

fi,j

= Dj

, 8j 2 V 0,

fi,j

� 0, 8(i, j) 2 E.

(4.6)

One important observation is that (4.6) and flexibility design problem are closely related.To see this, the flexibility design problem is equivalent to

R(D) = minfij ,

¯

Dj

X

j2V 0

max{Dj

� Dj

, 0}

s.t.X

j:(i,j)2E

fi,j

= Ci

, 8i 2 V,

X

i:(i,j)2E

fi,j

= Dj

, 8j 2 V 0,

fi,j

� 0, 8(i, j) 2 E.


We observe that (4.6) is equivalent to the flexibility design problem if rj

(Dj

) = max{Dj

�D

j

, 0}. The close resemblance between (4.6) and the flexibility design problem explains whythe average performance and the worst case performance are consistent for process flexibilitydesign problem, e.g., the results in Table 1.2. r

j

(Dj

) carries important information abouthow demand mean and variance a↵ect the worst-case performance. r

j

(Dj

) can be regardedas product j’s “worst case expected shortfall” when D

j

amount of product j is produced.(4.6) shows that the worst case performanceW (C,µ,�) equals a capacity allocation problemwhere the goal is to minimize the total “worst-case expected shortfall”.

4.3 Main Results

In this section, we provide results concerning (i) the marginal value of plant capacity C; (ii)the marginal value of flexibility configuration E; (iii) the implication on flexibility designguidelines and indices; and (iv) the computational aspects.

Result 1. First and second order e↵ects of C, µ and �.

(a) W (C,µ,�) is increasing in �C, µ and �.

(b) W (C,µ,�) is convex in Ci

for each i 2 V .

(c) For each i 2 V , �dW (C,µ,�)/dCi

is increasing in µj

for all j that satisfies (i, j) 2 E.

(d) |dW (C,µ,�)/dCi

| is bounded above by maxj:(i,j)2E µ2

j

/(µ2

j

+ �2

j

) for each i 2 V .

Part (a) of result 1 tells the first order e↵ects of plant capacity C, demand mean µ andstandard deviation �. As r

j

(Dj

) represents product j’s worst-case expected shortfall, lowervalues of r

j

(Dj

) are preferred. We infer that product j’s shortfall increases with the meanand standard deviation of its demand, but decreases with its total allocated capacity D

j

.This leads to the result that W (C,µ,�) decreases with plant capacity, but increases withdemand’s mean and standard deviation. The first order e↵ect of C is intuitive, as morecapacity reduces the likelihood of shortfall. The e↵ects of demand mean µ and standarddeviation � deserve more discussion. Though an increase in µ increases the amount ofdemands that are unsatisfied, it also raises the amount of demands that are satisfied. So thee↵ect of µ on total profit depends on the price and penalty cost di↵erentials. In contrastwith demand mean µ, an increase in demand standard deviation � raises the amount ofdemands that are unsatisfied and reduces the amount of demands that are satisfied. So asmaller demand standard deviation � is always preferred as long as the decisionmaker isrisk-neutral.

Part (b) suggests that plant capacity has diminishing returns. Figure 4.1 plots rj

(Dj

)

and �drj(¯

Dj)

d

¯

Djas functions of D

j

. Note that �drj(¯

Dj)

d

¯

Djmeasures the additional demand satisfied

by producing one more unit of product j. For example, producing one more unit of productj can satisfy 0.5 to 0.9 more unit of demand when D

j

= 60, while the number comes downto only 0.1 to 0.3 when D

j

= 150. Figure 4.1 confirms parts (a) and (b) of Result 1.


Figure 4.1: The shape of rj

(Dj

) when µj

= 100

(a) Product j’s expected shortfall

0 50 100 150 200 250 3000

10

20

30

40

50

60

70

80

90

100

allocated capacity Dj

expectedshortfallr j(D

j)

σ=40

σ=60

σ=80

σ=100

(b) Marginal demand satisfaction

0 50 100 150 200 250 3000

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

allocated capacity Djmarginalbenefit�

drj(D

j)

dD

j

σ=40

σ=60

σ=80

σ=100

Furthermore, we show that the expected total shortfall W (C,µ,�) is decreasing convex ineach C

i

. However, we are unable to show that W (C,µ,�) is convex in C. The analysisbecomes complex when at least two C

i

’s change at a time.Part (c) shows that the marginal value of plant capacity depends on the total average

demand to which the plant is directly connected. This result favors building capacity atplants that are directly connected to products with higher average demand. The e↵ect ofdemand mean on marginal value of plant capacity is intuitive. Plants that are connectedto more demands are heavily utilized. It brings more benefits to increase capacity at moreheavily utilized plants. The interesting aspect of part (c) is that it does not hold for demandstandard deviation. It should be cautious when using demand standard deviation as anindication of the marginal value of capacity.

Finally, part (d) provides an upper bound on the marginal value of capacity. The boundhas a simple form. As indicated in Figure 4.1, the bound works well when the system isheavily utilized. The bound provides the maximal marginal value of plant capacity. Theupper bound is easy to compute, and it is closely related to the coe�cient of variation µ

j

/�j

.In sum, result 1 is intuitive. It contributes to the literature because it provides analyticaljustification in asymmetric networks.

Result 2. First order e↵ect of flexibility configuration E. Conceptually, there is nofurther benefit of having one more link in E when current capacity is already fully utilized.Mathematically speaking, adding one more link in E would not reduce W (C,µ,�) if there


exist fij

� 0 for all (i, j) 2 E such that

X

j:(i,j)2E

fi,j

= Ci

, 8i 2 V,

X

i:(i,j)2E

fi,j

= µj

+ (X

i2V

Ci

�X

j2V 0

µj

) · �j

P

j2V 0 �j

, 8j 2 V 0.(4.7)

Flexibility has zero marginal value when (4.7) is satisfied. A central result from the processflexibility design literature is that the marginal value of flexibility diminishes quickly beyonda threshold flexibility value. Result 2 suggests that the worst case analysis captures thiscentral idea. One contribution of Result 3 is that it extends the idea to asymmetric networks.

Further, Result 2 sheds light on the role of a flexibility configuration. The role of aflexibility configuration is to allocate capacity to di↵erent products. The benefit of a flexi-bility configuration depends on how easily (4.7) can be satisfied. In (4.7), µ

j

+ (P

i2V Ci

�P

j2V 0 µj

)�j

/(P

j2V 0 �j

) can be viewed as “desired demand” when configuration E is total-flexibility configuration. So (4.7) requires a match between supply C

i

and the “desireddemand” µ

j

+ (P

i2V Ci

�P

j2V 0 µj

)�j

/(P

j2V 0 �j

). The match can be made as the totalsupply equals the total desired demand, i.e.,

X

i2V

Ci

=X

j2V 0

(

µj

+ (X

i2V

Ci

�X

j2V 0

µj

) · �j

P

j2V 0 �j

)

.

The match is di�cult if (i) some plants have larger capacity than other plants; (ii) someproducts have larger demand mean or variance than other products; and (iii) the numberof plants and products are not equal. Therefore, the need for flexibility can be measuredby network asymmetry. Result 2 delivers the idea that flexibility cannot totally substitutecapacity in hedging against demand risk. In a system where capacity is low and demand ishigh, building capacity is more appealing than building flexibility.

Result 3. Implication on Existing Flexibility Design Guidelines and Indices. To simplifythe process of process flexibility design, several easy-to-implement design guidelines andindices have been developed. Table 4.1 provides a summary based on the e↵ects of plantcapacity, demand mean and demand standard deviation. For example, the Node Expansionguideline considers plant capacity and product average demand, but not demand standarddeviation. So the Node Expansion guideline cannot capture the e↵ects of a change in demandstandard deviation. Intuitively, the ideal guideline/index should consider demand standarddeviation.

An interesting and unsolved question is that how does the e↵ect of demand standarddeviation interact with the e↵ects of plant capacity and demand mean? The question ishard for the following reason. Part (c) of Result 1 hints that the e↵ect of demand mean iscertain, but the e↵ect of demand standard deviation is uncertain - an increase in demandaverage raises the marginal value of plant capacity but an increase in demand standarddeviation may increase or decrease the marginal value of plant capacity. So demand mean


Table 4.1: Information on C, µ and � considered by di↵erent guidelines and indices

Design Guidelines & Indices Sources C? µ? �?

GuidelinesChaining Jordan and Graves, 1995 X XNode Expansion Chou et al., 2011 X X

Indices

J-G Jordan and Graves, 1995 X X Xg-Measure Graves and Tomlin, 2003 XStructural Flexibility Iravani, Oyen, and Sims, 2005WS-APL Iravani, Kolfal, and Oyen, 2007Expansion Chou, Teo, and Zheng, 2008 X X

Note. “X” means that the guideline/index is a↵ected by the information about plant capacity Ci

,

demand mean µj

or standard deviation �j

and standard deviation may have di↵erent e↵ects. This adds complexity as it is unclear howto combine µ and � in a flexibility design guideline/index.

We show that the idea of “desired demand” can help incorporate demand standard devia-tion information into flexibility design guidelines/indices. To illustrate the idea, we take theNode Expansion Guideline (Chou et al., 2011) as an example. Node Expansion Guideline isto add an edge in flexibility configuration E to increase

min

(

mini2V

P

j:(i,j)2E µj

Ci

,minj2V

P

i:(i,j)2E Ci

µj

)

(NE)

as much as possible. Demand standard deviation is not explicitly considered in the NodeExpansion Guideline. The intuition behind Node Expansion Guideline is to match plantcapacity C

i

with average demand µj

. Building on this intuition and the idea of “desireddemand”, we propose a refinement of the Node Expansion Guideline.

Definition 15. NE-refinement is a refinement of the Node Expansion guideline where prod-uct j’s average demand µ

j

is replaced with a combination of demand mean and standarddeviation

µj

+ (X

i2V

Ci

�X

j2V 0

µj

) · �j

P

j2V 0 �j

for all j 2 V 0.

We numerically compare (NE) and (NE-refinement) guidelines. Each comparison is basedon a Monte Carlo simulation of 1000 random scenarios. Each scenario consists of threestages. In stage 1, we determine parameters C, µ, � and the number of links |E|. In stage2, we follow (NE) and (NE-refinement) to get two flexibility configurations, respectively. In


stage 3, we compare the two flexibility configurations obtained in stage 2. We consider twoobjectives - average performance and worst case performance. The average performance isbased on Monte Carlo simulation with 2000 replications. The worst case performance isobtained from Theorem 14.

We consider a network with six plants and six products. We vary the number of links from|E| = 6 to |E| = 12. This allows us to compare (NE) and (NE-refinement) in both inflexibleand flexible systems. Table 4.2 provides the results when both C and µ are the same forall simulation scenarios. Only the demand standard deviation is randomly selected in eachscenario. Because both C and µ are fixed, (NE) suggests the same flexibility configurationfor all scenarios. In contrast, (NE-refinement) internalizes the e↵ect of demand standarddeviation and suggests di↵erent configurations for di↵erent scenarios.

Table 4.2 hints the potential improvement of (NE-refinement) over (NE). In Table 4.2,the second and sixth columns provide the percentage of scenarios where (NE-refinement) per-forms weakly better than (NE) under the average and worst case performances, respectively.The third and seventh columns provide the percentages of scenarios where (NE-refinement)performs strictly better than (NE). The fourth and eighth columns present the average rel-ative performance increment (reduction in expected total shortfall). The fifth and ninthcolumns indicate the maximum relative performance increment. We observe that for mostof the scenarios, (NE-refinement) performs better than (NE) and the relative performanceincrement is significant.

Compared with Table 4.2, Table 4.3 reports the results when C and µ are randomlygenerated. We observe that (NE-refinement) still performs significantly better than (NE).In summary, we argue and provide numerical evidence that properly incorporating demandstandard deviation can be beneficial. The performance increment is significant for a widerange of plant capacity C, demand mean µ and standard deviation �. The result is alsorobust to the number of links. Our analysis invites further systematic treatments of analyzingand improving existing flexibility design guidelines and indices.


Table 4.2: Comparison of (NE) and (NE- refinement) - fixed C and µ

|E| average performance worst case performance

percentim-proved

percentstrictlyim-proved

averagerelativeimprove

maxrelativeimprove

percentim-proved



maxrelativeimprove

12 84.3% 79.4% 29.2% 100.0% 100.0% 95.1% 2.2% 12.0%11 95.2% 94.3% 50.0% 100.0% 992.0% 98.3% 2.7% 12.0%10 99.4% 99.4% 62.2% 100.0% 100.0% 100.0% 3.3% 10.7%9 99.5% 99.5% 63.7% 99.1% 100.0% 100.0% 3.9% 12.6%8 99.7% 99.7% 58.0% 97.3% 100.0% 100.0% 4.3% 12.1%7 98.3% 94.8% 38.7% 92.4% 100.0% 96.5% 3.4% 8.8%6 100.0% 0.0% 0.0% 0.0% 100.0% 0.0% 0.0% 0.0%

Note. Ci

is 100 for all i 2 V . µj

is 60 for all j 2 V 0. �j

is randomly selected in between 12 and 37

for all j 2 V 0.

Table 4.3: Comparison of (NE) and (NE- refinement) - random C and µ

|E| average performance worst case performance

percentim-proved



maxrelativeimprove

percentim-proved



maxrelativeimprove

12 87.0% 72.7% 17.7% 71.2% 98.1% 83.8% 1.9% 9.8%11 90.6% 78.4% 24.5% 85.0% 97.0% 84.8% 1.9% 9.8%10 88.9% 81.3% 58.7% 100.0% 95.7% 88.1% 2.9% 17.1%9 90.2% 77.2% 36.3% 100.0% 96.6% 83.6% 3.2% 15.1%8 90.1% 72.8% 17.6% 75.6% 97.4% 84.1% 3.9% 14.3%7 91.9% 70.2% 25.9% 78.8% 98.7% 77.0% 6.7% 22.1%6 100.0% 0.0% 0.0% 0.0% 100.0% 0.0% 0.0% 0.0%

Note. Ci

is uniformly randomly selected in between 80 and 120 for all i 2 V . µj

is uniformly

randomly selected in between 50 and 70 for all j 2 V 0. �j

is randomly selected in between 12 and

37 for all j 2 V 0.


Result 4. The worst case performance W (C,µ,�) can be numerically solved e�ciently.W (C,µ,�) can be obtained by solving the following second order cone programming (SOCP).

W (C,µ,�)

= minfij ,

¯

D

�j ,

¯

D

+j

X

j2V 0

8

>

>

>

>

<

>

>

>

>

:

µj

�

=0 when

¯

D

�j = 0

z }| {

D�j

·µ2

j

µ2

j

+ �2

j

�µ2

j

+ �2

j

4µj

�D+

j

2+

r

(µ

2j+�

2j

2µj+ D+

j

� µj

)2 + �2

j

2| {z }

=0 when

¯

D

+j = 0

9

>

>

>

>

=

>

>

>

>

;

s.t.X

j:(i,j)2E

fij

= Ci

, 8i 2 V,

X

i:(i,j)2E

fij

= D�j

+ D+

j

, 8i 2 V,

0 D�j

µ2

j

+ �2

j

2µj

, 8j 2 V 0,

D+

j

� 0, 8j 2 V 0; fij

� 0, 8(i, j) 2 E.

Compared with the formulation in Theorem 14, in the above formulation, we replace Dj

with D�j

+ D+

j

, where D�j

is the amount of Dj

that is no greater than (µ2

j

+ �2

j

)/(2µj

), andD+

j

is the amount of Dj

that exceeds (µ2

j

+ �2

j

)/(2µj

). We show that such a transformationdoes not result in loss of optimality. The convexity of r

j

(Dj

) guarantees that D+

j

is zero aslong as D�

j

is less than (µ2

j

+ �2

j

)/(2µj

). The third constraint guarantees that D�j

cannotbe greater than (µ2

j

+ �2

j

)/(2µj

). It is easy to verify that rj

(Dj

) and the above objectivefunction are equivalent when (D�

j

(µ2

j

+ �2

j

)/(2µj

), D+

j

= 0) or (D�j

= (µ2

j

+ �2

j

)/(2µj

),D+

j

> 0). The only non-linear term of the above formulation is the last term of the objectivefunction. As the only non-linear term is a l

2

norm, the above formulation is a SOCP.The above SOCP formulation of W (C,µ,�) can be solved very fast by commercial

software. All the numerical examples of this chapter were solved using CVX, a packagefor specifying and solving convex programs (Grant and Boyd, 2008; CVX Research, 2012).Figure 4.2 illustrates the striking di↵erence in terms of computation time. Compared withthe demand-distribution-nonfree analysis where random sampling of demand is necessary,demand-distribution-free analysis delivers a more reliable result at a faster time.

4.4 A Marginal Distribution Model

In this section, we consider a marginal distribution model. Instead of assuming a demandjoint distribution, we optimize the worst case over all the demand distributions that have thegiven marginal distributions for all products. This section covers the following two topics- (1) the formulation of the marginal distribution model and (2) how the problem can bereformulated as a linear programming problem when F

j

’s are discrete.


Figure 4.2: Computation time as a function of network size

0

20

40

60

80

100

120

140

160

180

25 35 45 55 65 75 85 95

Computation time (sec)

Network size |V|=|V'|

distribution-nonfree distribution-free

Notes. The distribution non-free analysis is based on Jordan and Graves, 1995’s simulation method.

We run 10,000 demand scenarios to estimate the average performance. All plants have capacity

Ci

= 100. µj

and �j

are randomly generated from U [80, 120] and U [40, 60], respectively. The

configuration E being tested is the chaining configuration.

Let �(F1

, F2

, . . . , F|V 0|) be the set of demand joint distribution functions that have marginaldistributions F

j

. The worst expected total shortfall W 0(F1

, F2

, . . . , F|V 0|) is determined by

W 0(F1

, F2

, . . . , F|V 0|) = supF2�(F1,F2,...,F|V 0|)

ED

[R(D)].

Theorem 16. (i) W 0(F1

, F2

, . . . , F|V 0|) = mind2R|V 0|

+

n

R(d) +P

j2V 0 EDj [Dj

� dj

]+o

.

(ii) W 0(F1

, F2

, . . . , F|V 0|) is larger than W 0(F 01

, F 02

, . . . , F 0|V 0|) if Fj

is (stochastically) convexlylarger than F 0

j

for all j 2 V 0.

(iii)

W 0(F1

, F2

, . . . , F|V 0|) = minfij ,

¯

Dj

X

j2V 0

EDj [Dj

� Dj

]+

s.t.X

j:(i,j)2E

fi,j

= Ci

, 8i 2 V,

X

i:(i,j)2E

fi,j

= Dj

, 8j 2 V 0,

fi,j

� 0, 8(i, j) 2 E.


The definition of W 0(F1

, F2

, . . . , F|V 0|) involves optimizing over all the possible demandjoint distributions, whereas Theorem 16(i) suggests that we can instead optimize over a non-negative vector d. Theorem 16 is the key step towards linearizing the marginal distributionmodel. The idea behind Theorem 16(i) follows from the work by Meilijson and Nadas, 1979,Bertsimas, Natarajan, and Teo, 2004 and Natarajan, Song, and Teo, 2009; Natarajan, Shi,and Toh, 2012. Theorem 16(ii) tells the e↵ects of marginal distribution F

j

. The worst-caseshortfall is stochastically convex in the marginal distributions. Theorem 16(iii) provides afurther simplified formulation.

In fact, the marginal distribution model is closely related to the marginal moment modelwe studied earlier. The close resemblance between Theorem 14 and Theorem 16(i) hintsthat the marginal moment and marginal distribution models can be analyzed in a unifiedframework. In this paper, we focus on the marginal moment model, and some of the analysisapplies to the marginal distribution model as well. For example, Step 3 in Section 4.2 impliesthat the optimal d in Theorem 16(i) must satisfy R(d) = 0, which leads to Theorem 16(ii).Theorem 16(iii) provides a linear programming formulation when F

j

’s are discrete.

Proof. Proof of Theorem 16. For all D 2 R|V 0|+

and d 2 R|V 0|+

, we have

R(D) = max(x,y)2⇤

(

X

j2V 0

Dj

· yj

�X

i2V

Ci

· xi

)

= max(x,y)2⇤

(

X

j2V 0

(dj

+Dj

� dj

) · yj

�X

i2V

Ci

· xi

)

max(x,y)2⇤

(

X

j2V 0

dj

· yj

�X

i2V

Ci

· xi

)

+ max(x,y)2⇤

X

j2V 0

(Dj

� dj

) · yj

R(d) +X

j2V 0

[Dj

� dj

]+. (due to 0 yj

1)

Taking expectations over distribution F yields

EF

[R(D)] R(d) +X

j2V 0

EFj [Dj

� dj

]+,

where Fj

is the marginal distribution of Dj

. Because the above inequality holds for all

F 2 F(F1

, F2

, . . . , F|V 0|) and d 2 R|V 0|+

,

supF2F(F1,F2,...,F|V 0|)

EF

[R(D)] mind2R|V 0|

+

(

R(d) +X

j2V 0

EFj [Dj

� dj

]+)

.

The above inequality holds at equality (i.e., Theorem 16 holds) if we can provide a jointdistribution F ⇤ 2 F(F

1

, F2

, . . . , F|V 0|) such that

EF

⇤ [R(D)] = mind2R|V 0|

+

(

R(d) +X

j2V 0

EF

⇤j[D

j

� dj

]+)

, (4.8)


where F ⇤j

is the marginal distribution of F ⇤. The optimization on the right hand side of(4.8) is equivalent to

mint,d2R|V 0|

+

t+X

j2V 0

EFj [Dj

� dj

]+

t �X

j2V 0

Dj

· yj

�X

i2V

Ci

· xi

, 8(x,y) 2 ⇤.

Let �(x,y) be the dual variable associated with the constraint t �P

j2V 0 Dj

· yj

�P

i2V Ci

·xi

. Let t⇤,d⇤ and �⇤ be the primal and dual optimal solutions. Because the precedingoptimization is a convex optimization, the following KKT conditions apply.

StationarityX

(x,y)2⇤

�⇤(x,y) = 1; (K1)

X

(x,y)2⇤:yj=1

�⇤(x,y) = Pr(Dj

� d⇤j

), 8j 2 V 0; (K2)

Primal feasibilityX

j2V 0

d⇤j

· yj

�X

i2V

Ci

· xi

t⇤, 8(x,y) 2 ⇤; (K3)

Dual feasibility

�⇤(x,y) � 0, 8(x,y) 2 ⇤; (K4)

Complementary slackness

�⇤(x,y) ·

X

j2V 0

d⇤j

· yj

�X

i2V

Ci

· xi

� t⇤

!

= 0; 8(x,y) 2 ⇤. (K5)

(K1) and (K4) hint that �⇤(x,y) can be viewed as probabilities. Let (x, y) be a randomvector that takes value (x,y) with probability �⇤(x,y). Let f ⇤

j

and fj

be the marginal densityfunctions of F ⇤ and F , respectively. Construct f ⇤

j

as follows. Let I1

= {j 2 V 0 : d⇤j

> 0} andI2

= {j 2 V 0 : d⇤j

= 0}. For j 2 I1

, define

f ⇤j

(D|(x, y) = (x,y)) =

8

>

<

>

:

fj(D)·1{D�d⇤j }

1�Fj(d⇤j )

if yj

= 1,fj(D)·1{D<d⇤j }

Fj(d⇤j )

if yj

= 0.


For j 2 I2

, define f ⇤j

(D|(x, y) = fj

(D). Then for j 2 I1

,

f ⇤j

(D) =X

(x,y)2⇤

f ⇤j

(D|(x, y) = (x,y)) · Pr((x, y) = (x,y))

=X

(x,y)2⇤:yj=1

�⇤(x,y) ·fj

(D) · 1{D�d

⇤j}

1� Fj

(d⇤j

)+

X

(x,y)2⇤:yj=0

�⇤(x,y) ·fj

(D) · 1{D<d

⇤j}

Fj

(d⇤j

)

= fj

(D) · 1{D�d

⇤j} + f

j

(D) · 1{D<d

⇤j} = f

j

(D).

So we have f ⇤j

= fj

for all j 2 V 0. Finally, we verify that Theorem 16 holds at equality atF ⇤.

EF

⇤ [R(D)]

=X

(x,y)2⇤

�⇤(x,y) · EF

⇤ [R(D)|(x, y) = (x,y)]

=X

(x,y)2⇤

�⇤(x,y) ·(

EF

⇤ [X

j2V 0

Dj

· yj

|(x, y) = (x,y)]�X

i2V

Ci

· xi

)

=X

(x,y)2⇤

�⇤(x,y) ·(

EF

⇤ [X

j2V 0

(Dj

� d⇤j

) · yj

|(x, y) = (x,y)] +X

j2V 0

d⇤j

· yj

�X

i2V

Ci

· xi

)

=X

(x,y)2⇤

�⇤(x,y) ·(

EF

⇤ [X

j2V 0

(Dj

� d⇤j

) · yj

|(x, y) = (x,y)]

)

+X

(x,y)2⇤

�⇤(x,y) ·(

X

j2V 0

d⇤j

· yj

�X

i2V

Ci

· xi

� t⇤ + t⇤

)

=X

j2V 0

EF

⇤j[(D

j

� d⇤j

) · yj

]

+X

(x,y)2⇤

�⇤(x,y) ·(

X

j2V 0

d⇤j

· yj

�X

i2V

Ci

· xi

� t⇤

)

| {z }

=1 due to (K5)

+X

(x,y)2⇤

�⇤(x,y)

| {z }

=1 due to (K1)

·t⇤

=X

j2V 0

EF

⇤j[D

j

� d⇤j

]+ + t⇤

=X

j2V 0

EF

⇤j[D

j

� d⇤j

]+ +R(d⇤) due to (K3).

By definition, for all d 2 Rn, EF

0j[D

j

� Dj

]+ is larger than EFj [Dj

� Dj

]+ if F 0j

is stochas-tically convexly larger than F

j

. Part (ii) of Theorem 16 follows immediately. The proof ofpart (iii) follows Step 3 in Section 4.2. ⇤


4.5 A Cross Moment Model

In this paper, we aim to provide a demand-distribution-free analysis of process flexibilitydesign. The analysis assumes only the first and second moments of demand, but no specificdistributions of demand. We investigate the worst case over all the possible demand distribu-tions that have the given first and second moments. Suppose D is nonnegative with knownfirst moment vector µ = E[D] and second-moment matrix ⌃ = E[DDT ]. The problem is

(C) zC

= supD⇠(µ,⌃)

+

E[R(D)],

where D ⇠ (µ,⌃)+ denotes the set of feasible multivariate distributions supported on Rn

+

with first moment vector µ = E[D] and second-moment matrix ⌃ = E[DD

T ]. Matrix ⌃contains information about di↵erent product demands’ correlations. Such a model is calledcross moment model in the literature.

Preliminaries on Complete Positive Program (CPP)

Let Sn

denote the cone of n⇥n symmetric matrices. Let S+

n

denote the cone of n⇥n positivesemidefinite matrices. The cone of n⇥ n completely positive matrices is defined as

CPn

= {A 2 Sn

|9V 2 Rn⇥k

+

, s.t., A = V V T},

or equivalently,

CPn

= {A 2 Sn

|9v1

, v2

, . . . , vk

2 Rn

+

, s.t., A =k

X

i=1

vi

vTi

}.

The cone of n⇥ n copositive matrices is defined as

COn

= {A 2 Sn

|9v 2 Rn

+

, s.t., vTAv � 0}.

A ⌫cp

(⌫co

)0 indicates that the matrix A is completely positive (copositive). A copositiveprogram (COP) is a linear program over the convex cone of the copositive matrices. EachCOP is associated with a dual problem over the convex cone of completely positive matrices.Such a program is called a completely positive program (CPP).

The CPP Formulation

We provide a CPP formulation of problem (C). R(D) is expressed as a minimization problem.By taking the dual, R(D) becomes the optimal value of a maximization problem,

R(D) = maxx,y

D

T

y � c

T

x

s.t. xi

� yj

, 8(i, j) 2 E,

0 yj

1, 8j 2 V 0,

xi

� 0, 8i 2 V ; yj

� 0, 8j 2 V 0.


The above linear programming is transformed to the following standard form by introducingnew decision variables e

ij

and sj

.

R(D) = maxx,y,e,s

D

T

y � c

T

x

s.t. xi

� yj

� eij

= 0, 8(i, j) 2 E,

yj

+ sj

= 1, 8j 2 V 0,

x,y, e, s are all nonnegative.

For notational simplicity, let X = [x;y; e; s] be the new decision variable. X is of dimension1 + |V | + 3|V 0| + |E|. Let the row vector a

T

i,j

be such that constraint xi

� yj

� eij

= 0 is

equivalent to a

T

i,j

X = 0. Similarly, let the row vector bTj

be such that constraint bTj

X = 1 isthe same as y

j

+ sj

= 1. The above linear programming problem can be rewritten as

maxX

= [�c

T ,DT ,01⇥|E|,01⇥|V 0|]X

s.t. aT

i,j

X = 0, 8(i, j) 2 E;

b

T

j

X = 1, 8j 2 V 0.

X � 0.

Consider the following CPP,

(P) zP

= maxPX ,PDX ,PXX

[0|V 0|⇥|V |, I|V 0|,0|V 0|⇥(|E|+|V 0|)] • P T

DX

� [cT ,01⇥(2|V 0|+|E|)]P x

s.t. aT

i,j

P

X

= 0, 8(i, j) 2 E,

b

T

j

P

XX

b

j

= 0, 8j 2 V 0,

(PXX

)jj

= (PX

)j

, 8j = 1, 2, . . . , |V |+ 2|V 0|+ |E|,2

4

1 µ

T

P

T

X

µ ⌃ P

T

DX

P

X

P

DX

P

XX

3

5 ⌫cp

0.

A • B means the trace of ATB. The three decision variables are P

X

2 R(|V |+2|V 0|+|E|)⇥1

+

,

P

DX

2 R(|V |+2|V 0|+|E|)⇥|V 0|+

, and P

XX

2 R(|V |+2|V 0|+|E|)⇥(|V |+2|V 0|+|E|)+

.

Proposition 17. zC

= zP

.

Proposition 17 suggests that the worst case performance is the optimal solution to acomplete positive program. One di�culty of the complete positive program is that it is hardto check if a matrix A is complete positive, i.e., A ⌫

cp

0. Following the work of Natarajan,Teo, and Zheng, 2011 and Kong et al., 2012, we approximate the constrain A ⌫

cp

0 by twoconstraints A � 0 and A ⌫ 0. A � 0 means that all the elements of A are nonnegative.A ⌫ 0 means that matrix A is positive semidefinite.


Table 4.4: Computational aspects of the Total Flexibility Structure

|V 0| Primal Obj Dual Obj CPU Time/s No. of Variables No. of Constraints

3 22.7456943 22.7643289 3.46 506 3054 24.6979571 24.7969552 11.62 1122 6445 26.204827 26.4265383 68.29 2162 12026 27.3623946 27.7248698 247.82 3782 20577 653.728896 -183481333 877.81 6162 3299

Note. “Primal Obj” means primal objective value, and “Dual Obj” means the dual objective value.

Numerical results

All the numerical examples were solved using CVX, a package for specifying and solving convexprograms (Grant and Boyd, 2008; CVX Research, 2012). We are particularly interested inthe following three questions.

• What is the largest network size that can be numerically solved?

• How is the CPU computation time increases with the network size?

• How does the final problem size (i.e., numbers of variables and constraints) increasewith the network size?

We focus on networks where the number of plants equals the number of products. We testthree structures - the Total-flexibility Structure, the Chaining structure, and the Dedicatedstructure. In the Total-flexibility structure, all the plants are able to produce all the products,i.e., E = {(i, j)|i 2 V, j 2 V 0}. In the chaining structure,

E = {(1, 1), (2, 2), . . . , (|V |, |V |), (1, 2), (2, 3), . . . , (|V |� 1, |V |), (|V |, 1)}.In the Dedicated structure, each plant can only produce one product,

E = {(1, 1), (2, 2), . . . , (|V |, |V |)}.Tables 4.4, 4.5 and 4.6 provide the simulation results for the Total-flexibility structure, theChaining structure, and the Dedicated structure, respectively.

In Table 4.4, the algorithm fails to converge when |V 0| = 7 - the primal and dual objectivesare far from each other. The software returns either “steps too short consecutively” or “stepsin predictor too short”. We infer that the cross moment model can only solve networkswith very small sizes. For each table, we observe that even though the network is small,3 |V | 7, the numbers of decision variables and constraints are huge. This maybeone reason why only very small networks can be solved. A comparison between the CPUcomputation times reveals that the computation takes less time when the configuration issparser.


Table 4.5: Computational aspects of the Chaining Structure


3 27.060518 27.0717415 2.23 380 2334 32.8075555 32.8791677 3.42 650 3845 38.8118039 38.9283217 7.81 992 5726 45.1371842 45.202128 17.67 1406 7977 51.6979081 51.7891746 52.17 1892 1059


Table 4.6: Computational aspects of the Dedicated Structure


3 46.8465595 46.8465911 1.32 272 1704 62.4615132 62.4621899 2.11 462 2785 78.0692382 78.0841162 3.22 702 4126 93.6777211 93.6988086 5.37 992 5727 109.200758 109.332308 10.31 1332 758


4.6 Summary

In this chapter, we provide a demand-distribution-free analysis of process flexibility design.The analysis has a number of advantages and disadvantages. We take a first step towardsunderstanding the merits of demand-distribution-free analysis in process flexibility design.Compared with demand-distribution-nonfree analysis, the main advantage of the demand-distribution-free analysis is its tractability. Flexibility design problem is by nature a two-stage stochastic programming problem. The demand-distribution-nonfree analysis may haveto estimate 1016 di↵erent demand realizations in a network with sixteen products and eachdemand takes ten possible values. It would be di�cult to generalize a finding from these manycases. In contrast, the demand-distribution-free analysis is subject to no such limitationswhen o↵ering insights on the marginal value of plant capacity.

Our current work invites future research on the application of demand-distribution-freeanalysis of process flexibility design. This chapter can be extended in the following directions.First, it is also meaningful to investigate demand correlations between di↵erent products.Kong et al., 2012 study the appointment scheduling problem and investigate worst case


performance among all service time distributions that have the given mean and covariancematrix. Second, this chapter assumes all products have the same price and penalty cost.van Mieghem, 1998 argues in a two-product setting that price and cost di↵erentials playan important role. Important insights may be drawn from the price and cost perspective.Third, we study flexibility design in its simplest form. It would be interesting to applydistribution-free analysis to multi-period and multi-stage settings.

55

Appendix A

Proofs

Proof of Theorem 2. By the network symmetry assumption, we assume without loss ofgenerality that the optimal structure for k products takes the following form. �(i) = {i, i+1},81 i < m; �(m) = {1,m}; �(i) = {(i � m) · ✏, (i � m) · ✏ + bm/2c}, 8m < i k where✏ = b m

2(k�m)

c.The number of products k increases to k0. When (k0 � m)/(k � m) is an integer, the

optimal structure for k0 products can be achieved by adding the subsequent links. DenoteK = (k0�m)/(k�m). �(k+1) = {✏/K, ✏/K+bm/2c}, �(k+2) = {2 ·✏/K, 2 ·✏/K+bm/2c},. . ., �(k+K) = {(K+1) · ✏/K, (K+1) · ✏/K+ bm/2c}, �(k+K+1) = {(K+2) · ✏/K, (K+2) · ✏/K + bm/2c}, . . ., �(k0) = {[(k �m) ·K � 1] · ✏/K, [(k �m) ·K � 1] · ✏/K + bm/2c}.

Proof of Theorem 3. The proof consists of two steps. Step 1 shows that structuresthat satisfy Flexibility Design Rule (i) and |�(j)| = 2, 8m < j k achieve an eigenvalueindex of 2k and a mean index of 2. Step 2 proves that 2k and 2 are the maximal attainableeigenvalue and mean indices.

Step 1. Each product is attached to 2 links, soM(i, i) = 2, 8i 2 V . There cannot be morethan two non-overlapping paths connecting any two products, M(i, j) 2, 81 i 6= j k.We prove M(i, j) = 2, 81 i 6= j k by construction-identifying two non-overlapping pathsfor any pair of products 1 i, j k.

Without loss of generality, we assume the 2-chain structure (Flexibility Design Rule (i))is formed by the first m products (see the left panel in Figure A.1). For a better illustration,we adopt the circular representation in the right panel of Figure A.1. These two structuresare equivalent. For any products i and j (1 i 6= j k), we identify two non-overlappingpaths connecting them.

There are three cases: (i) 1 i 6= j m; (ii) 1 i m,m < j k or 1 j m,m <i k and (iii) m < i 6= j k.

Case (i) 1 i 6= j m:When both products are within the first m products, it is trivial to find two non-

overlapping paths from Figure A.2.Case (ii) 1 i m,m < j k or 1 j m,m < i k:

APPENDIX A. PROOFS 56

2

m

1

7

6

5

4

3

1

m

7

6

5

4

3

2

2

m

1

76

5

4

31

m

7

6

5

4

3

2

Plants Products

Figure A.1: 2-chain Structure

2

m

1

j-1i

5

4

31

m

j

j-1

i

4

3

2

route 1

route 2

Figure A.2: Case (i)


2

j2

1

j1i

5

4

31

m

j1i

4

3

2

route 1

route 2

j

Figure A.3: Case (ii)

Without loss of generality, we study the case 1 i m,m < j k. For convenience,denote the two plants directly connected to product j as j

1

and j2

. We can still find twonon-overlapping paths connecting them (see Figure A.3).

Case (iii) m < i 6= j k:When neither product is in the 2-chain structure, we can still find two non-overlapping

paths between them. See the two possible situations in Figures A.4 and A.5, where �(i) ={i

1

, i2

} and �(j) = {j1

, j2

}.In summary, M(i, j) = 2, 81 i, j k. The associated eigenvalue and mean indices are

2k and 2, respectively.Step 2. Next, we show that no structure with |E| 2k has an eigenvalue index or

mean index larger than 2k or 2. By the definition of structural flexibility matrix, M satisfiesM

ij

min{Mii

,Mjj

}, 81 i, j N .

Proposition 18. For any non-negative k by k matrix M with trace 2k and elements Mij

min{M

ii

,Mjj

}, 81 i, j k, the largest eigenvalue of M is no larger than 2k.

Proof. Proof. We prove by contradiction. Suppose there is an eigenvalue � of M satisfying� > 2k. By definition, there exists an eigenvector x = (x

1

, x2

, . . . , xk

)T ,x 6= 0, satisfies

Mx = �x.


2

j2

1

j1i2

i1

4

31

m

j1i2

i1

3

2

route 1

route 2

j

i

Figure A.4: Case (iii)

The above equation takes the form of

xi

=X

1jk,j 6=i

Mij

��Mii

xj

, 1 i k. (A.1)

Since x 6= 0, at least one xi

is either strictly negative or positive. Without loss of generality,assume at least one x

i

is strictly negative. We can always rearrange the xi

so that

x1

x2

. . . xk1 < 0 x

k1+1

. . . xn

, 1 k1

k.

Then, we have

x1

�X

2jk1

M1j

��M11

xj

�X

2jk1

M11

��M11

xj

=M

11

��M11

(X

1jk1

xj

� x1

).

The first inequality is due to (A.1) and xj

� 0 when j > k1

. The second inequality resultsfrom M

1j

M11

, xj

< 0 and � �M11

> 0 when j k1

. Adding M11��M11

x1

on both sides ofthe above inequality yields

x1

(1 +M

11

��M11

) � M11

��M11

X

1jk1

xj

,


2

j2

1

i2j1

i1

4

31

m

j1i2

i1

3

2

route 1

route 2

ji


which is equivalent to

x1

� M11

�

X

1jk1

xj

.

82 i k1

,

xi

� x1

� Mii

�

X

1jk1

xj

.

Summing up the above inequalities for 1 i k1

,

X

1jk1

xj

�P

1ik1M

ii

�

X

1jk1

xj

� 2k

�

X

1jk1

xj

. (A.2)

The last inequality holds becauseP

1ik1M

ii

P

1ik

Mii

= 2k andP

1jk1xj

is nega-tive. Finally,

P

1jk1xj

< 0, (A.2) reduces to

1 2k

�,

which contradicts to our assumption � > 2k. ⇤


The largest eigenvalue of M is no more than 2k. By checking the above inequalities atequality, the su�cient and necessary conditions for the largest eigenvalue of M to be 2k is,

91 k1

k, such that Mij

=2k

k1

81 i, j k1

and Mij

= 0 otherwise. (A.3)

(A.3) identifies all the possible matrices M with largest eigenvalue 2k. The elements of astructural flexibility matrix must be integers, so k

1

should be chosen so that 2k

k1is an integer.

By definition of the structural flexibility matrix, Mij

= 2k

k1 k, so k

1

� 2. In summary, theinteger k

1

must satisfy that (i) (2k)/k1

is an integer and (ii) 2 k1

k. After determiningthe structural flexibility matrix, we solve an integer programming problem to recover theprocess structure.

Although k1

could be less than k, in practice choosing k1

< k means we give up theproduction of certain products. Further, it obviously incurs a huge shortfall. Choosingk1

< k also violates the flexibility design guideline suggested by Jordan and Graves, 1995.Hence, k

1

= k is the only reasonable choice.The proof for the structural flexibility mean index is exactly the same as the proof for

Theorem 1 in Iravani, Oyen, and Sims, 2005. We refer the reader to their Online TechnicalSupplement for details. ⇤

Proof of Theorem 4. The theorem follows directly from the result that when the lastproduct m + 1 is directly connected to two plants that are at distance �, the g-measure ofthis unbalanced network is �+2

2m

.For any subset of products ⇤ ✓ V 0, we carry out separate analysis for the three cases (i)

|�(⇤)| > |⇤|; (ii) |�(⇤)| = |⇤| and (iii) |�(⇤)| < |⇤|.For case (i) |�(⇤)| = |⇤|, we have

g(⇤) =X

i2�(⇤)

Ci

C� |⇤| = m+ 1

m|�(⇤)|� |⇤| = m+ 1

m|�(⇤)|� |�(⇤)| = 1

m|�(⇤)| 1. (A.4)

For case (ii) |�(⇤)| > |⇤|, we have

g(⇤) =X

i2�(⇤)

Ci

C� |⇤| = m+ 1

m|�(⇤)|� |⇤| � m+ 1

m(|⇤|+ 1)� |⇤| > 1. (A.5)

Because the g-Measure is defined as g = min⇤✓V

0,|�(⇤)|<|V |

{g(⇤)}, case (ii) does not determine

g-Measure and therefore should be discarded. As we will show later that case (iii) shouldalso be discarded, the g-Measure in computed solely from case (i) as follows.

Suppose the last product m + 1 is directly connected to two plants i1

and i2

. The twonewly added edges, {m+ 1, i

1

} and {m+ 1, i2

}, will turn the chain that connects all plantsinto two chains that contain �

2

+1 and m� �

2

+1 plants respectively, where � is the distancebetween plants i

1

and i2

. By the definition of plant distance, � 2m � �. Thus, thesmallest set �(⇤) that satisfies |�(⇤)| = |⇤| contain �

2

+1 plants. The g-Measure is therefore(�2

+ 1)/m = �+2

2m

.


)(/* /

)(/*�V /�'V

V 'VPlants Products

(1)

(4)(3)

(2)


Finally, the reason why case (iii) should be discarded is that |�(⇤)| < |⇤| will necessarilylead to ⇤ = V 0, which will in turn result in |�(⇤) = V |- a case that is not considered in thecomputation of g-Measure.

We prove by contradiction. Suppose |⇤| < |V 0|, we use V 0�⇤ (set subtraction) to denotethe set of products which are not included in ⇤. Similarly, we use V � �(⇤) to denote theset of plants which are not directly connected to the product set ⇤. See Figure A.6 forillustration. For ease of exposition, let deg(i) denote the number of links attached to nodei, in terms of graph theory. Following the definition of �(⇤), there are no links directlyconnecting the product set ⇤ (region (1) in Figure A.6) to the set of plants V ��(⇤) (region(3) in Figure A.6). Hence, all the links directly connecting the V � �(⇤) (region (3) inFigure A.6) have an end-point in the set V 0 � ⇤ (region (4) in Figure A.6). Equivalently,the sum of the degrees of the nodes in V � �(⇤) (region (3) in Figure A.6) is no larger thanthat of V 0 � ⇤, (region (4) in Figure A.6)

X

i2(V��(⇤))

deg(i) X

j2(V 0�⇤)

deg(j). (A.6)

On the other hand, since |V | = |V 0|� 1, if |�(⇤)| < |⇤| then

|V � �(⇤)| � |V 0 � ⇤|.

Each product is attached to two links (deg(j) = 2, 8j 2 V 0) and each plant is attached to atleast two links (deg(i) � 2, 8i 2 V ). There exists at least one 2-chain structure, so

X

i2(V��(⇤))

deg(i) � 2|V � �(⇤)| � 2|V 0 � ⇤| =X

j2(V 0�⇤)

deg(j). (A.7)


�

�

Plant� Product�

|ȿ|<|ȳ(ȿ)|�|ȿ|>|ȳ(ȿ)||ȿ|=|ȳ(ȿ)|�

Figure A.7: Illustrations of the Structures under Di↵erent Signs of |�(⇤)|� |⇤|

Combining the results of (A.6) and (A.7),

X

i2(V��(⇤))

deg(i) =X

j2(V 0�⇤)

deg(j),

All the links directly connect to V 0 � ⇤ (region (4) in Figure A.6) have an end-point inV ��(⇤) (region (3) in Figure A.6). We conclude that ⇤ and V 0�⇤ are disconnected, whichcontradicts the assumption that there exists a 2-chain structure connecting all plants. So,case (iii) can be discarded. ⇤

Proof of Theorem 6. In the following, we first assume that m is an even number.Since the network is symmetric, all plants have the same capacity C. An equal allocation ofcapacity would provide each product with capacity C = m/k · C. The excess capacity forany product set ⇤ ✓ V 0 is

g(⇤) = |�(⇤)| km

� |⇤|.

Recall that computing g-Measure requires enumerating all the product sets that are notdirectly connected to all plants, i.e., |�(⇤)| < |V |. We proceed our analysis according to thesign of |�(⇤)|� |⇤|. |�(⇤)|� |⇤| = 0 implies that the product set ⇤ and plant set �(⇤) forma chain, or several disconnected chains. See the left structure of Figure A.7. |�(⇤)|� |⇤| < 0implies that there exists at least two chains that are connected. See the central structure ofFigure A.7. The right structure shows the case of |�(⇤)|� |⇤| > 0.

Case (i). When |�(⇤)| = |⇤|, g(⇤) = (k �m)/m|⇤|. Thus finding the minimum g(⇤)necessitates finding the minimum |⇤| that satisfies |�(⇤)| = |⇤|. Because flexibility designrule (ii) is satisfied, all the chains (i.e., |�(⇤)| = |⇤|) must be in the form depicted inFigure A.8.

In Figure A.8, plants i2j1�1

and i2j1 are the two plants that are directly connected to

product j1

, and plants i2j2�1

and i2j2 are the two plants that are directly connected to product

j2

. The chain is highlighted with bolded links and arcs. In this chain, |�(⇤)| = |⇤| = 2·(i2j2�


�

�

i2j1

i2j2-1

i2j2

1

m

i2j1-1

Figure A.8: |�(⇤)| = |⇤|

i2j1) + 2. When the positions of plants 1 and m are not known, we have |�(⇤)| = |⇤| = 2 ·mod (i

2j2 � i2j1 ,m) + 2. As m < j

1

, j2

k, 2m < 2j1

, 2j1

� 1, 2j2

� 1, 2j2

2k, we have

min⇤✓V

0,|�(⇤)|<|V |

|⇤| = min2m<j1 6=j22k

2 · mod (ij1 � i

j2 ,m) + 2.

Hence, finding the minimum |⇤| is the same as finding the two closest plants among plantset �({m+ 1,m+ 2, , k}). Since our goal is to maximize g-Measure, we want all plants thatare within plant set �({m + 1,m + 2, , k}) to be evenly spread out. Such idea is coined asFlexibility Design Rule (iii). The mathematical derivation of above idea is shown below-

maxi2m+1,i2m+2,...,i2k

{g} = maxi2m+1,i2m+2,...,i2k

{ min2m<j1 6=j22k

2 · mod (ij1 � i

j2 ,m) + 2}k �m

m.

The optimal ij

, 2m < j 2k should satisfy Flexibility Design Rule (iii).Case (ii). when |�(⇤)| > |⇤|,

g(⇤) = |�(⇤)| km

� |⇤| � (|⇤|+ 1)k

m� |⇤|

= |⇤|k �m

m+

k

m� k �m

m+

k

m

=2k � 2m

m+ 1 � 2k � 2m

m+ b m

k �mck �m

m.

It concludes that the case |�(⇤)| > |⇤| always has larger g(⇤) than the case |�(⇤)| = |⇤|.Thus, we can discard case (ii) |�(⇤)| > |⇤|.

Case (iii). |�(⇤)| < |⇤|. Figure A.9 shows two adjacent chains: i2j1�1

�! i2j2�1

�!


12 1�ji

12 2 �ji

12 ji

22 ji 12 3�j

i

32 ji

Figure A.9: |�(⇤)| < |⇤|

i2j2 �! i

2j1 �! i2j1�1

(chain 1) and i2j2�1

�! i2j3�1

�! i2j3 �! i

2j2 �! i2j2�1

(chain 2).Both of these chains satisfy |�(⇤)| = |⇤|. Because they overlap with each other (they sharetwo plants and one product), the union of these two chains should satisfy |�(⇤)| = |⇤|� 1 <|⇤|. The reverse is also true, |�(⇤)| < |⇤| only holds when the sub-structure (⇤,�(⇤)) is theunion of adjacent chains. In case (i), we characterize g(⇤) for an isolated chain. Now we seethe e↵ects of combining chain 1 and chain 2. When ⇤ increases from chain 1 to the unionof chains 1 and 2, g(⇤) raises by

2k

m|ij3 � i

j2 |� 2|ij3 � i

j2 |� 1 =k �m

m(2|i

j3 � ij2 |)� 1. (A.8)

If |ij3 � i

j2 | is no less than m

2(k�m)

, then g(⇤) increases, otherwise g(⇤) decreases.

Sum up cases (i) and (iii). The minimal g(⇤) in case (i) is commensurate to min{mod (i

j1 � ij2 ,m)}. Case (iii) suggests that the minimal g(⇤) obtained in case (i) can be

further decreased if the adjacent chain (i2j2�1

�! i2j3�1

�! i2j3 �! i

2j2 �! i2j2�1

) satisfies|ij3 � i

j2 | < m

2(k�m)

. Both cases advocate the implementation of Flexibility Design Rule (iii).

When Flexibility Design Rule (iii) is satisfied, the minimal g(⇤) in case (i) is maximized,and the potential drop of g(⇤) in case (iii) is minimized. Thus, Flexibility Design Rule (iii)accords with g-Measure.

It should be noted that Flexibility Design Rule (iii) is a necessary, but not su�cientcondition for the maximal g-Measure. When m

2(k�m)

is an integer, we can cut the circle evenly.

g-Measure is maximized when Flexibility Design Rule (iii) is satisfied. When m

2(k�m)

is not

an integer, Flexibility Design Rule (iii) leads to some segments contain b m

2(k�m)

c products

and others contain d m

2(k�m)

e products. Then, as bm/(k � m)c < m/(k � m), we do not

want to build adjacent chains with lengths bm/(k�m)c to prevent the drop of g(⇤) in case(iii). Instead, if possible, we build a chain with length dm/(k �m)e next to all chains withlengths bm/(k � m)c. In this way, g-Measure is maximized. This idea is not captured by


Flexibility Design Rule (iii), so in general Flexibility Design Rule (iii) does not characterizeall structures that attain the maximum g-Measure. If we cannot build a chain with lengthdm/(k � m)e) next to all chains with lengths bm/(k � m)c, Flexibility Design Rule (iii)alone characterizes all structures that attain the maximum g-Measure. In either cases, thestructures that attain the maximum g-Measure must satisfy Flexibility Design Rule (iii).

When m is an odd number, the above argument still holds. Instead of 2|ij1 � i

j2 | (seeFigure A.8), the total number of products in a chain is 2|i

j1 � ij2 | ± 1. ⇤

APL Metric’s Relationship with Flexibility Design Rules (ii) and (iii). Theexpected minimum distance between two randomly and uniformly chosen points on the righthand structure of Figure 3.13 is

p3

4+

(1� p)3

4+

p(1� p)

2.

Before unfolding the main analysis, two relevant results are listed below.

Lemma 19. The expected minimum distance between two randomly and uniformly chosenpoints on the circumference of a circle of perimeter a is a/4.

Lemma 20. The expected minimum distance between two randomly and uniformly chosenpoints on a line of length a is a/3.

Mathematically speaking, given two continuously uniform random variables X ⇠ U(0, a)and Y ⇠ U(0, a), Lemma 19 suggests E[min{|X � Y |, a � |X � Y |}] = a/4 and Lemma 20suggests E|X�Y | = a/3. The proofs of these two lemmas are straight-forward, and are thusomitted.

In the right hand structure of Figure 3.13, the line separates the circle into two segmentsof lengths p and 1�p. Given that both two points are on the segment of length p, Lemma 19suggests that the expected minimum distance between them is p/4. Similarly, given thatboth two points are on the segment of length 1�p, the expected minimum distance betweenthem is (1 � p)/4. When the two points lie on di↵erent segments, the expected minimumdistance between them is 1/4. From the telescope property,

Expected minimum distance

=p2 · Expected minimum distance given that both points lie on the segment of length p

+(1� p)2·Expected minimum distance given that both points lie on the segment of length 1� p

+2p(1� p)·Expected minimum distance given that the two points lie on di↵erent segments

=p3

4+

(1� p)3

4+

p(1� p)

2.

When p ! 0 or p ! 1, the expected minimum distance between two randomly and uniformlychosen points on the right hand structure of Figure 3.13 is 1/4, which accords with Lemma 19.


2/1p2/2p

2/3p

2/4p

2/mkp �

2/1��mkp

Figure A.10: APL and Flexibility Design Rules

2/1p2/2p

2/3p

2/4p

2/mkp �

2/1��mkp

AB

C

D

A'B'

C'

D'

2/1p

2/2p

2/3p

2/4p

AA'

BB'

C' C

DD'

2/mkp �

2/mkp �2/1p

2/2p

2/3p

2/4p

Figure A.11: An Equivalent Structure to Figure A.10

When p = 1

2

, the expected minimum distance between two randomly and uniformly chosenpoints on the right hand structure of Figure 3.13 is 3

16

, which accords with Lemma 22.In the remaining we prove that given that Flexibility Design Rules (i) and (ii) are satisfied,

APL-metric supports Flexibility Design Rule (iii).Provided that Flexibility Design Rules (i) and (ii) are satisfied, the circle is cut into

2(k�m) segments (see Figure A.10). We assume m is su�ciently large, the perimeter of thecircle is 1, k � m+2, and the lengths of the 2(k�m) circle segments are p

i

/2, 1 i k�m,where

P

k�m

i=1

pi

= 1.We suggest that the two structures in Figure A.11 have equivalent APL, and thus study

the right hand structure instead. In the left hand structure of Figure A.11, A and A’ areconnected by a line of length 0 and are regarded as one point. For the same reason, we think


2/1p

2/2p

2/3p

2/4p

A

B

C

D

2/mkp �

Figure A.12: APL Metric Calculation

of B and B’, C and C’, and so forth as one point. (See Figure A.11).The right hand structure in Figure A.11 is a chain of k�m small circles. If the two chosen

points are on the same circle of length pi

, Lemma 19 shows that their expected minimumdistance is p

i

/4 (this holds only when pi

1/2, otherwise the shortest path may not alwayslie on the small circle of length p

i

). Suppose the two chosen points lie on di↵erent circles,their expected minimum distance is the same as that of Figure A.12 provided that the twopoints are not on the same segment.

The expected minimum distance between two randomly and uniformly chosen points onFigure A.12 is 1/2 ·1/4 = 1/8. Meanwhile, from Lemma 20, the expected minimum distanceis p

i

/2·1/3 = pi

/6 if both points lie on segment i (this holds only when pi

1/2, otherwise theshortest path may not always lie on the segment of length p

i

/2). So the expected minimumdistance when the two points do not lie on the same segment is

1

8�

k�m

X

i=1

p2i

· (pi6) =

1

8� 1

6

k�m

X

i=1

p3i

, pi

� 1

2, 81 i k �m.

In summary, the APL of structures in Figure A.11 is

k�m

X

i=1

p2i

· pi4

| {z }

both lie on segment i

+1

8� 1

6

k�m

X

i=1

p3i

| {z }

lie on di↵erent segments

=1

8+

1

12

k�m

X

i=1

p3i

, pi

1

281 i k �m. (A.9)

Under the constraintP

k�m

i=1

pi

= 1, the above expression is minimized at p1

= p2

= . . . =pk�m

= 1/(k�m). Hence, Flexibility Design Rule (iii) is supported by the WS-APL methodwhen m is su�ciently large.

Another way to see (A.9) is by comparing the right hand structure in Figure A.11 andthe one in Figure A.12. When the two points do not lie on the same small segment, the


�

length�a�

length�b�

X�

Y�

aͲX�

bͲY�

Structure�(1)�

length�b�

length�b�

length�a�

Structure�(2)�

a<b

Figure A.13: Illustrations of Lemmas 21 and 22

shortest path has identical length in both structures. When the two points lie on the samesegment, the two structures yield di↵erent shortest pathes. Suppose, if both points lie onthe segment A-A’-B’-B, then by Lemma 19 the right hand structure in Figure A.11 givesthe expected minimum distance p

1

/4, while by Lemma 20 the expected minimum distanceis p

1

/2/3 = p1

/6 in the structure in Figure A.12. The di↵erence is p1

/4 � p1

/6 = p1

/12.Hence, the overall di↵erence in expected minimum distance is 1

12

P

k�m

i=1

p3i

, which explainsthe second term in (A.9).

We end the proof by explaining the last technical detail. The above analysis does notwork for the case when one of the p

i

is greater than 1/2. For example, 1

8

� 1

6

P

k�m

i=1

p3i

, whichis the expected minimum distance when the two points do not lie on the same segment ofFigure A.12, falls below 0 when p

i

goes toward 1. This is because, when p1

is close to 1, theexpected minimum distance between two points on segment A-B is no longer p

1

/6.The remaining proof is not complicated but tedious. Two relevant lemmas are listed

below.

Lemma 21. Consider a circle of perimeter a+ b. The circle is separated into two segmentsof lengths a and b, as illustrated in structure (1) of Figure A.13. Uniformly and randomlypick one point from each of the two segments, then the expected minimum distance betweenthe two points is

(�a2/12 + ab/2 + b2/4)/b.

Mathematically speaking, given two continuously uniform random variables X ⇠ U [0, a]and Y ⇠ U [0, b] (see structure (1) in Figure A.13),

E[min{X + Y, a+ b�X � Y }] = (�a2/12 + ab/2 + b2/4)/b.


Proof. Proof of Lemma 21.

E[min{X + Y, a+ b�X � Y }]

=

Z

a

0

[

Z

a+b2 �x

0

(x+ y)dy

b+

Z

b

a+b2 �x

(a+ b� x� y)dy

b]dx

a

=1

ab

Z

a

0

(ax� x2 � a2

4+

ab

2+

b2

4)dx

=(�a2/12 + ab/2 + b2/4)/b.

⇤In the extreme case of a = 0, Lemma 21 gives the correct result b/4.

Lemma 22. Consider structure (2) in Figure A.13. Uniformly and randomly choose twopoints from the structure, then the expected minimum distance between the two points is1

12

5b

3+3a

3+9a

2b+9ab

2

(a+b)

2 when b 3a and 1

8

2a

2+4ab+3b

2

a+b

otherwise.

Proof. Proof of Lemma 22. As illustrated in Figure A.14, there are four cases to consider.In case (1), both two points lie on exactly one of the segments of length b; in case (2), bothtwo points lie on the segment of length a; in case (3), one point lies on the segment of lengtha and the other point lies on the segment of length b; finally in case (4), each of the twosegments of length b contains one point.

In case (1), Lemmas 19and 20 suggest that the expected minimum distance between thetwo points is min{b/3, (a+b)/4}. The probability for case (1) to occur is b

2

2(a+b)

2 . Similarly, in

case (2), the expected minimum distance between the two points is min{a/3, (a+b)/4} = a/3.The probability for case (2) to occur is a

2

(a+b)

2 .

In case (3), Lemma 21 suggests that the minimum distance between the two points is(�a2/12 + ab/2 + b2/4)/b. The probability for case (3) to occur is 2ab

(a+b)

2 .

Finally, case (4) involves the comparison of routes 1 and 2 in Figure A.14. When b�a

2

X b+a

2

, route 1 is shorter than route 2; when X b�a

2

or b�X b�a

2

, route 2 is shorterthan route 1. When route 2 is shorter, the expected minimum distance is

b� a

4+

Z

b+a2

0

xdx

b+

Z

b

a+b2

(a+ b� x)dx

b

=b� a

4+ [

1

2(b+ a

2)2 + a

b� a

2+

1

2(b� a

2)2]

1

b

=2b+ a

4� a2

4b.

When route 1 is shorter, by Lemma 21 the expected minimum distance is (�a

2

12

+ ab

2

+ b

2

4

)/b+b�a

2

. To sum up, the expected minimum distance under case (4) is

b� a

b· (2b+ a

4� a2

4b) +

a

b· [(�a2

12+

ab

2+

b2

4) · 1

b+

b� a

2] =

a+ b

2� 1

2· a

2

b+

1

6· a

3

b2.


�

�

Case�(1)�

length�b�

length�b�

length�a�

Case�(4)�

a<blength�b�

length�b�

length�a�

Xroute�1�

route�2

Case�(2)�

length�b�

length�b�

length�a�

Case�(3)�

length�b�

length�b�

length�a�

��the�route�

the�route�

the�route�

Figure A.14: Illustrations of Lemma 22


�

p1/2�

p1/2�

A,�A’�

B,�B’�

p1/2�

p1/2�

(1Ͳp1)/2�A,�A’�B,�B’�

structure�(1)� structure�(2)�

C,�C’�

Figure A.15: p1

> 1

2

The probability for case (4) to occur is b

2

2(a+b)

2 .Lemma 22 holds as the weighted summation of all four cases equals

min{ b3,a+ b

4} · b2

2(a+ b)2+

a

3· a2

(a+ b)2+ (�a2/12 + ab/2 + b2/4)/b · 2ab

(a+ b)2

+[a+ b

2� 1

2· a

2

b+

1

6· a

3

b2] · b2

2(a+ b)2

=min{ 1

12(5b3 + 3a3 + 9a2b+ 9ab2)/(a+ b)2,

1

8

2a2 + 4ab+ 3b2

a+ b}

⇤When at least one of the p

i

is greater than 1

2

, without loss of generality, assume p1

> 1

2

.

AsP

k�m

i=1

pi

= 1, pi

< 1

2

8i � 2. See structure (1) in Figure A.15 for an illustration. ByLemma 22 and substituting a = (1 � p

1

)/2 and b = p1

/2, the expected minimum distancebetween two randomly and uniformly chosen points on the structure (2) in Figure A.15 is1/8 + 1/12 · p3

1

when 1

2

< p1

3

4

and 1/8 + 1/16 · p21

when p1

� 3

4

.There is only one more step to finish the proof - to show that the APL of structure (2) is

1

12

P

k�m

i=2

p3i

smaller than that of structure (1) in Figure A.15. Compare the two structures inFigure A.15. The only di↵erence is that some circles in structure (1) are replaced with linesegments in structure (2). When the two chosen points lie on di↵erent small circles, bothstructures yield the same minimum distance. When both the two points lie on exactly onesmall circle, structure (2) yields a shorted path, i.e., by Lemmas 19 and 20, the di↵erence isa/4� a/3/2 = a/12. So the overall downward bias in structure(2) is 1

12

P

k�m

i=2

p3i

.In summary, when p

1

> 1

2

, the expected minimum distance between two randomly and

uniformly chosen points on Figure A.10 is 1

8

+ 1

12

P

k�m

i=1

p3i

when pi

3

4

81 i k � m


and 1

8

+ 1

16

p21

+ 1

12

P

k�m

i=2

p3i

when p1

� 3

4

. It is straightforward to show the unique optionalsolution is p

1

= p2

= . . . = pk�m

. ⇤

Proof. Proof of Proposition 9. The dual problem of (4.1) is

W (C,µ,�) = min✓,⇢,�

"

✓ +X

j2V 0

µj

· ⇢j

+X

j2V 0

✓

(µj

)2 + (�j

)2◆

�j

#

s.t. ✓ +X

j2V 0

dj

· ⇢j

+X

j2V 0

(dj

)2 · �j

� R(d), 8 d 2 R|V 0|+

,

(A.10)

where ✓, ⇢ and � are dual decision variables. Since the constraints in (4.1) are equalities,the dual decision variables are unrestricted in sign. We observe that (A.10) is simpler than(4.1) because there is no integral of d in (A.10). The same trick is used in Mak, Rong, andZhang, 2012. Some recent distributionally-robust optimization models include Bertsimas,Natarajan, and Teo, 2004; Bertsimas, Natarajan, and Teo, 2006; Bertsimas et al., 2010, Gohand Sim, 2010, Delage and Ye, 2010, Natarajan, Teo, and Zheng, 2011 and Chen, He, andZhang. 2011.

The constraint in (A.10) can be further simplified. Because the constraint holds for all

d 2 R|V 0|+

, it is identical to

✓ � maxd2R|V 0|

+

R(d)�X

j2V 0

dj

· ⇢j

�X

j2V 0

(dj

)2 · �j

�

. (A.11)

The total shortfall R(d) is the optimal value of the flexibility design problem. We have

R(d) = max(x,y)2⇤

✓

X

j2V 0

dj

· yj

�X

i2V

Ci

· xi

◆

, (A.12)

where

⇤ =

⇢

(x,y) | yj

xi

, 8(i, j) 2 E; xi

� 0, 8i 2 V ; 0 yj

1, 8j 2 V 0.

�

Replacing R(d) with (A.12) in (A.11) yields,

✓ � maxd2R|V 0|

+

max(x,y)2⇤

X

j2V 0

dj

· yj

�X

i2V

Ci

· xi

�X

j2V 0

dj

· ⇢j

�X

j2V 0

(dj

)2 · �j

�

Exchanging the order of maxd2R|V 0|

+and max

(x,y)2⇤ yields

✓ � max(x,y)2⇤

maxd2R|V 0|

+

X

j2V 0

dj

· yj

�X

i2V

Ci

· xi

�X

j2V 0

dj

· ⇢j

�X

j2V 0

(dj

)2 · �j

�

(A.13)


Exchanging the order of maxd2R|V 0|

+and max

(x,y)2⇤ is an important step. Fixing xi

and

yj

, optimizing over all possible values of d is an easy task in (A.13). This is also why we takethe dual of the flexibility design problem. R(d) is the minimum value of an optimizationproblem. We purposely take the dual so that R(d) is expressed as the maximum valueof an optimization problem. No optimality is lost if we exchange the order of max

d2R|V 0|+

and max(x,y)2⇤. However, optimality would be compromised if we exchange the order of

maxd2R|V 0|

+and min

fij ,sj .

The inner optimization maxd2R|V 0|

+in (A.13) can be further simplified.

maxd2R|V 0|

+

X

j2V 0

dj

· yj

�X

i2V

Ci

· xi

�X

j2V 0

dj

· ⇢j

�X

j2V 0

(dj

)2 · �j

�

= maxd2R|V 0|

+

X

j2V 0

(yj

� ⇢j

) · dj

�X

j2V 0

�j

· (dj

)2�

�X

i2V

Ci

· xi

=X

j2V 0

(

maxdj�0

(yj

� ⇢j

) · dj

� �j

· (dj

)2�

)

�X

i2V

Ci

· xi

In summary,

W (C,µ,�) = min✓,⇢,�

"

✓ +X

j2V 0

µj

· ⇢j

+X

j2V 0

✓

(µj

)2 + (�j

)2◆

�j

#

s.t. ✓ � max(x,y)2⇤

(

X

j2V 0

maxdj�0

(yj

� ⇢j

) · dj

� �j

· (dj

)2�

�X

i2V

Ci

· xi

)

⇤

Proof. Proof of Proposition 10.Part (i). We exclude the following three cases in Figure A.16.

(1): �⇤j

< 0 for any j 2 V 0.

(2): �⇤j

= 0 for any j 2 V 0.

(3): �⇤j

> 0 and ⇢⇤j

� 1 for any j 2 V 0.

Case (1): we cannot have �⇤j

< 0 for any j 2 V 0. Suppose not, then there must existj0

2 V 0 such that �⇤j0

< 0. Then player 3’s objective value (P3) becomes infinite when d⇤j0

approaches positive infinity. Player 1’s objective value goes to positive infinity as well. Socase (1) cannot occur.


= 0 for any j 2 V 0. Suppose not, there exists j0

2 V 0 suchthat �⇤

j0= 0. When �⇤

j0= 0, (P3) becomes a linear function of d

j0 . To avoid the unbounded


Figure A.16: Three cases that cannot happen

𝜙∗

𝜌∗ (1): 𝜙∗ < 0

(2): 𝜙∗ = 0

𝜌∗ = 1

(3): 𝜙∗ > 0 and 𝜌∗ ≥ 1

case, we must have d⇤j0

= 0 when �⇤j0

= 0. We infer that players 2 and 3 always get zeroutility.

�⇤j0

= 0 leads to ⇢⇤j0

= 1. To see this, we first show that �⇤j0

= 0 leads to ⇢⇤j0

1.Whenever �⇤

j0= 0 and ⇢⇤

j0> 1, player 1 can slightly reduce the value of ⇢⇤

j0and improve his

objective function. Players 2 and 3 are una↵ected because they always have zero utility. Onthe other hand, �⇤

j0= 0 leads to ⇢⇤

j0� 1. Whenever ⇢⇤

j0< 1 and �⇤

j0= 0, the combination of

y⇤j0= 1 and d⇤

j0= 1 would make both (P2) and (P3) unbounded. Then, player 1’s objective

value goes to infinity as well. So �⇤j0= 0 leads to ⇢⇤

j0= 1.

In sum, we have d⇤j0

= 0 and ⇢⇤j0

= 1 when �⇤j0

= 0. Then, the term associated withproduct j is µ

j

in (P1). Because the term associated with product j0

is its worst expectedshortfall, we infer that the worst expected shortfall equals the average demand. This canonly happen if product j

0

is not connected to any plants. Following Assumption 7, we cansafely ignore case (2).


> 0 and ⇢⇤j

� 1 for any j 2 V 0. Suppose not, there existsj0

2 V 0 such that �⇤j0> 0 and ⇢⇤

j0� 1. On one hand, (P3) implies that �⇤

j0> 0 and ⇢⇤

j0� 1

leads to d⇤j0

= 0. On the other hand, whenever d⇤j0

= 0, players 2 and 3’s objective values(P2) and (P3) are una↵ected by �⇤

j0. Therefore, player 1 has incentive to reduce the value

of �⇤j0as long as �⇤

j0> 0 and ⇢⇤

j0� 1. �⇤

j0> 0 and ⇢⇤

j0� 1 leads to contradicting conclusions

on d⇤j0.

Part (ii). By definition, d⇤j

is the optimal solution to

maxdj�0

[(y⇤j

� ⇢⇤j

)dj

� �⇤j

(dj

)2].

Because the objective function is a second order polynomial of dj

, the optimal solution d⇤j

must satisfy part (ii).


Part (iii). Recall that x⇤i

is the optimal solution to

max(x,y)2⇤

(

X

j2V 0

(yj

� ⇢⇤j

) · d⇤j

� �⇤j

· (d⇤j

)2�

�X

i2V

Ci

· xi

)

. (P2)

The above objective function is a decreasing function of xi

. We would like to have xi

as smallas possible. The constraint (x,y) 2 ⇤ suggests that x

i

is bounded below by maxj:(i,j)2E y

j

.Therefore, the optimal (x⇤,y⇤) satisfies x⇤

i

= maxj:(i,j)2E y⇤

j

for all i 2 V . ⇤

Proof. Proof of Proposition 11.We prove by contradiction. Without loss of generality, we assume y⇤

1

= 1 holds for all theoptimal solutions of (P2). The main idea of the proof is to show the contradictions aboutplayer 1’s optimal decisions ⇢⇤ and �

⇤. The proof consists of three major steps.

Step 1. Transform (P2) into a linear programming problem.

Step 2. What is the implication when y⇤1

= 1 holds for all the optimal solutions of (P2)?

Step 3. Check the first order conditions on ⇢⇤1

and �⇤1

.

Step 1. Transform (P2) into a linear programming problem.Due to part (ii) of Proposition 10, (P2) can be simplified as

max(x,y)2⇤

X

j2V 0

(max{yj

� ⇢⇤j

, 0})2

4�⇤j

�X

i2V

Ci

· xi

�

, (A.14)

where ⇢

⇤ and �

⇤ are player 1’s optimal decisions. It is easy to verify that in (A.14), theobjective function is a convex function of x and y. Since (A.14) is maximizing a convexfunction, one of the optimal solutions must be an extreme point. Because the extremepoints of ⇤ are always integers, restricting the values of x and y to integers would notchange the optimal value. That is,

max(x,y)2⇤,

xi,yj2{0,1}

X

j2V 0

(max{yj

� ⇢⇤j

, 0})2

4�⇤j

�X

i2V

Ci

· xi

�

(A.15)

has the same optimal value as (A.14). As yj

is either 0 or 1, (A.15) is equivalent to

max(x,y)2⇤,

xi,yj2{0,1}

(

X

j2V 0

(max{�⇢⇤j

, 0})2

4�⇤j

· (1� yj

) +(max{1� ⇢⇤

j

, 0})2

4�⇤j

· yj

�

�X

i2V

Ci

· xi

)

.


Due to part (i) of Proposition 10, max{1 � ⇢⇤j

, 0} = 1 � ⇢⇤j

. The above optimization isequivalent to

max(x,y)2⇤,

xi,yj2{0,1}

(

X

j2V 0

(max{�⇢⇤j

, 0})2

4�⇤j

· (1� yj

) +(1� ⇢⇤

j

)2

4�⇤j

· yj

�

�X

i2V

Ci

· xi

)

. (A.16)

Because the objective function is a linear function of x and y in (A.16) and because (x,y) 2 ⇤is totally unimodular, dropping o↵ the binary constraints x

i

, yj

2 {0, 1} does not compromisethe optimality of (A.16). The optimal value of (A.16) equals the optimal value of

max(x,y)2⇤

(

X

j2V 0

(max{�⇢⇤j

, 0})2

4�⇤j

· (1� yj

) +(1� ⇢⇤

j

)2

4�⇤j

· yj

�

�X

i2V

Ci

· xi

)

. (A.17)

In sum, the optimal values of (P2) and (A.17) are the same. In fact, they also have thesame set of optimal solutions (it can be verified that optimal solutions of one problem mustmaximize the other problem). So we transform the non-linear optimization (P2) to the linearoptimization (A.17).

Step 2. What is the implication when y⇤1

= 1 holds for all the optimal solutions of (P2)?Note the close resemblance between (4.3) and (A.17). The only di↵erence is the coe�cientassociated with y. Some sensitivity analysis about the optimal solution of (4.3) is providedin the following lemma.

Lemma 23. Let e

1

2 R|V 0| be a column vector where the first element is 1 and all theother elements are 0. Let ✏ be a scaler. Consider the following family of optimizationsparameterized by ✏

max(x,y)2⇤

✓

(d+ ✏e1

)Ty �C

T

x

◆

. (A.18)

Let (X⇤(✏),Y ⇤(✏)) be the set of optimal solutions to (A.18). If y⇤1

= 1 holds for all y⇤ 2Y

⇤(0), then there exists ✏ > 0 such that (X⇤(✏),Y ⇤(✏)) is constant over 0 < |✏| < ✏.

The idea behind Lemma 23 is as follows. If all the optimal solutions of (4.3) satisfyy⇤1

= 1, then all the optimal solutions remain optimal and all the non-optimal solutionsremain non-optimal when d

1

changes by an infinitesimal amount ✏. Applying Lemma 23 to(A.17) we infer that if y⇤

1

= 1 holds for all the optimal solutions of (P2), an infinitesimalchange in ⇢⇤

1

and �⇤1

would not a↵ect x⇤ and y

⇤.

Proof. Proof of Lemma 23.By definition, for any (x⇤,y⇤) 2 (X⇤(0),Y ⇤(0)) and (x,y) 2 ⇤ � (X⇤(0),Y ⇤(0)), we

haved

T

y

⇤ �C

T

x

⇤ > d

T

y �C

T

x.

Define

✏ =1

2

8

<

:

d

T

y

⇤ �C

T

x

⇤ � max(x,y)2⇤

(x,y)/2(X⇤(0),Y

⇤(0))

(dT

y �C

T

x)

9

=

;

> 0.


Then for any (x⇤,y⇤) 2 (X⇤(0),Y ⇤(0)) and (x,y) 2 ⇤� (X⇤(0),Y ⇤(0)),

(d+ ✏e1

)Ty⇤ �C

T

x

⇤ ��

(d+ ✏e1

)Ty �C

T

x

�

=d

T

y

⇤ �C

T

x

⇤ + ✏y⇤1

� (dT

y �C

T

x)� ✏y1

>d

T

y

⇤ �C

T

x

⇤ � |✏|� (dT

y �C

T

x)� |✏y1

|=2✏� 2|✏| > 0

as long as |✏| < ✏. We infer that (x⇤,y⇤) 2 (X⇤(✏),Y ⇤(✏)) and (x,y) 2 ⇤� (X⇤(✏),Y ⇤(✏))when |✏| < ✏. So (X⇤(✏),Y ⇤(✏)) is constant when |✏| < ✏. ⇤

Step 3. Check the first order conditions on ⇢⇤1

and �⇤1

.Part (i) of Proposition 10 specifies an open feasible region of ⇢⇤

1

and �⇤1

, and thus the firstorder condition on ⇢⇤

1

and �⇤1

is expected to hold. In (P1), the terms associate with ⇢1

and�1

are

µ1

⇢1

+(µ2

1

+�2

1

)�1

+X

j2V 0

(max{�⇢⇤j

, 0})2

4�⇤j

·(1�y⇤j

(⇢,�))+(1� ⇢⇤

j

)2

4�⇤j

·y⇤j

(⇢,�)

�

�X

i2V

Ci

·x⇤i

(⇢,�).

(A.19)Because y⇤

1

= 1 for all the optimal solution x

⇤ and y

⇤, Lemma 23 implies that x

⇤ and y

⇤

remain the same when ⇢⇤1

and �⇤1

changes by an infinitesimal amount. So we can ignore theterms associated with x⇤

i

(⇢⇤,�⇤) and y⇤i

(⇢⇤,�⇤) when taking first order derivative of (A.19)with respect to ⇢

1

and �1

at ⇢⇤1

and �⇤1

. The first order conditions of (A.19) with respect to⇢1

and �1

at ⇢⇤1

and �⇤1

are

µ1

� 1� ⇢⇤1

2�⇤1

= 0,

(µ1

)2 + (�1

)2 � (1� ⇢⇤1

)2

4(�⇤1

)2= 0.

The above two equalities cannot both be satisfied as long as �1

> 0 (see Assumption 8).In summary, player 1’s optimal decision (⇢⇤,�⇤) would never be such that y⇤

1

= 1 holdsfor all the optimal solutions y⇤. It can be argued similarly that we would never have y⇤

1

= 0holds for all the optimal solutions y⇤. The main reason is that the following two equalitiescannot be satisfied simultaneously,

µ1

� max{⇢⇤1

, 0}2�⇤

1

= 0,

(µ1

)2 + (�1

)2 � (max{⇢⇤1

, 0})2

4(�⇤1

)2= 0.

Though we focus only on product 1’s decision y⇤1

, the analysis can be applied to any arbitraryproduct j 2 V 0. ⇤


Proof. Proof of Proposition 12. For any product set � ✓ V 0, let �(�) be the set of plantsthat are directly connected to product set �, i.e., �(�) = {i 2 V |9j 2 � s.t. (i, j) 2 E}.Consider the following optimization, which has the same optimal value as (4.3).

R(d) = max�✓V

0

⇢

X

j2�

dj

�X

i2�(�)

Ci

�

(A.20)

Lemma 24. If �⇤1

and �⇤2

are two distinct optimal solutions to (A.20), d � 0 and C � 0,then

(i) �⇤1

[�⇤2

is an optimal solution to (A.20),

(ii) �⇤1

\�⇤2

is an optimal solution to (A.20),

Proof. Proof of Lemma 24. For any two product sets �1

and �2

, by definition we have

�(�1

[�2

) = �(�1

) [ �(�2

)

and�(�

1

\�2

) ✓ �(�1

) \ �(�2

).

Given any two distinct optimal solutions �⇤1

and �⇤2

,X

j2�⇤1[�⇤

2

Dj

�X

i2�(�⇤1[�⇤

2)

Ci

=X

j2�⇤1[�⇤

2

Dj

�X

i2�(�⇤1)[�(�⇤

2)

Ci

=X

j2�⇤1

Dj

+X

j2�⇤2

Dj

�X

j2�⇤1\�⇤

2

Dj

�✓

X

i2�(�⇤1)

Ci

+X

i2�(�⇤2)

Ci

�X

i2�(�⇤1)\�(�⇤

2)

Ci

◆

=

✓

X

j2�⇤1

Dj

�X

i2�(�⇤1)

Ci

◆

+

✓

X

j2�⇤2

Dj

�X

i2�(�⇤2)

Ci

◆

�✓

X

j2�⇤1\�⇤

2

Dj

�X

i2�(�⇤1)\�(�⇤

2)

Ci

◆

=

✓

X

j2�⇤1

Dj

�X

i2�(�⇤1)

Ci

◆

+

✓

X

j2�⇤2

Dj

�X

i2�(�⇤2)

Ci

◆

�✓

X

j2�⇤1\�⇤

2

Dj

�X

i2�(�⇤1\�⇤

2)

Ci

◆

| {z }

�0 as �

⇤2 is maximizing (A.20)

�✓

X

i2�(�⇤1\�⇤

2)

Ci

�X

i2�(�⇤1)\�(�⇤

2)

Ci

◆

| {z }

0

.

(A.21)

So (A.21) suggestsX

j2�⇤1[�⇤

2

Dj

�X

i2�(�⇤1[�⇤

2)

Ci

�X

j2�⇤1

Dj

�X

i2�(�⇤1)

Ci


Meanwhile, �⇤1

is an optimal solution to (A.20) by definition, i.e.,

X

j2�⇤1[�⇤

2

Dj

�X

i2�(�⇤1[�⇤

2)

Ci

X

j2�⇤1

Dj

�X

i2�(�⇤1)

Ci

.

So we must haveX

j2�⇤1[�⇤

2

Dj

�X

i2�(�⇤1[�⇤

2)

Ci

=X

j2�⇤1

Dj

�X

i2�(�⇤1)

Ci

.

In addition, (A.21) holds at equality only if

X

j2�⇤2

Dj

�X

i2�(�⇤2)

Ci

=X

j2�⇤1\�⇤

2

Dj

�X

i2�(�⇤1\�⇤

2)

Ci

and �(�⇤1

) \ �(�⇤2

) = �(�⇤1

\ �⇤2

). We infer that both �⇤1

[ �⇤2

and �⇤1

\ �⇤2

are optimalsolutions to (A.20). ⇤

Recall that player 2’s optimization problem (P2) is equivalent to (A.17), which can berewritten as

X

j2V 0

(max{�⇢⇤j

, 0})2

4�⇤j

+ max�✓V

0

(

X

j2�

(1� ⇢⇤j

)2

4�⇤j

�(max{�⇢⇤

j

, 0})2

4�⇤j

�

�X

i2�(�)

Ci

)

. (A.22)

Proposition 11 suggests that for any product j, there exists two optimal solutions �⇤1

and�⇤

2

of (A.22) where j 2 �⇤1

but j /2 �⇤2

. Then the union of all the optimal solutions of (A.22)is V 0 and the intersection of all the optimal solutions of (A.22) is the empty set ;. Because(A.22) is of the form of (A.20), Lemma 24 can be applied. Parts (i) and (ii) of Lemma 24suggest that both V 0 and ; are optimal solutions of (A.22). ⇤

Proof. Proof of Corollary 13.Because Proposition 12 suggests that (x⇤,y⇤) = (0,0) is an optimal solution to

max(x,y)2⇤

(

X

j2V 0

(max{�⇢⇤j

, 0})2

4�⇤j

· (1� yj

) +(1� ⇢⇤

j

)2

4�⇤j

· yj

�

�X

i2V

Ci

· xi

)

.

=X

j2V 0

(max{�⇢⇤j

, 0})2

4�⇤j

+ max(x,y)2⇤

(

X

j2V 0

(1� ⇢⇤j

)2

4�⇤j

�(max{�⇢⇤

j

, 0})2

4�⇤j

�

yj

�X

i2V

Ci

· xi

)

,

(x⇤,y⇤) = (0,0) is also an optimal solution to

max(x,y)2⇤

(

X

j2V 0

(1� ⇢⇤j

)2

4�⇤j

�(max{�⇢⇤

j

, 0})2

4�⇤j

�

yj

�X

i2V

Ci

· xi

)

(A.23)


We infer that the optimal value of (A.23) is zero. Strong duality implies that the optimalvalue of the dual of (A.23) is also zero,

0 = minfij ,sj

X

j2V 0

sj

s.t.X

j:(i,j)2E

fi,j

Ci

, 8i 2 V,

X

i:(i,j)2E

fi,j

+ sj

�(1� ⇢⇤

j

)2

4�⇤j

�(max{�⇢⇤

j

, 0})2

4�⇤j

, 8j 2 V 0,

9fi,j

� 0, 8(i, j) 2 E; 9sj

� 0, 8(j) 2 V 0.

So there must exist fij

� 0 for all (i, j) 2 E such that

X

j:(i,j)2E

fi,j

Ci

, 8i 2 V,

X

i:(i,j)2E

fi,j

�(1� ⇢⇤

j

)2

4�⇤j

�(max{�⇢⇤

j

, 0})2

4�⇤j

, 8j 2 V 0, (A.24)

Because (x⇤,y⇤) = (1,1) is also an optimal solution to (A.23),

X

j2V 0

(1� ⇢⇤j

)2

4�⇤j

�(max{�⇢⇤

j

, 0})2

4�⇤j

�

1�X

i2V

Ci

· 1 = 0.

We infer that the inequalities in (A.24) can be replaced with equalities. ⇤

Proof. Proof of Theorem 14.In the previous step, we show that W (C,µ,�) is the optimal value of minimizing

X

j2V 0

[µj

⇢j

+ (µ2

j

+ �2

j

)�j

] +X

j2V 0

[(max{�⇢

j

, 0})2

4�j

(1� y⇤j

) +(1� ⇢

j

)2

4�j

y⇤j

]�X

i

Ci

x⇤i

with respect to ⇢ and �. Because of Proposition 12, we can get rid of (x,y) by replacing itwith (0,0). The objective function becomes

X

j2V 0

[µj

⇢j

+ (µ2

j

+ �2

j

)�j

+(max{�⇢

j

, 0})2

4�j

].


Corollary 13 summarizes all the constraints on ⇢ and �. (P1) becomes

W (C,µ,�) = minfij ,⇢j ,�j

X

j2V 0

⇢

µj

· ⇢j

+ ((µj

)2 + (�j

)2) · �j

+(max{�⇢

j

, 0})2

4�j

�

X

j:(i,j)2E

fi,j

= Ci

, 8i 2 V,

X

i:(i,j)2E

fi,j

=(1� ⇢

j

)2

4�j

� (max{�⇢j

, 0})2

4�j

, 8j 2 V 0.

fi,j

� 0, 8(i, j) 2 E.

⇢j

< 1,�j

> 0, 8j 2 V 0.

The above optimization problem is a minimization problem over fi,j

, ⇢j

and �j

. To simplify,we transform the above optimization to a two-stage optimization problem. We first minimizeover ⇢

j

and �j

, and then minimize over fi,j

.

W (C,µ,�) = minfij ,

¯

Dj

X

j2V 0

rj

(Dj

)

�

.

s.t.X

j:(i,j)2E

fi,j

= Ci

, 8i 2 V,

X

i:(i,j)2E

fi,j

= Dj

, 8j 2 V 0,

fi,j

� 0, 8(i, j) 2 E.

where

rj

(Dj

) = min⇢j ,�j

⇢

µj

· ⇢j

+ ((µj

)2 + (�j

)2) · �j

+(max{�⇢

j

, 0})2

4�j

�

s.t.(1� ⇢

j

)2

4�j

� (max{�⇢j

, 0})2

4�j

= Dj

.

⇢j

< 1,�j

> 0, 8j 2 V 0.

(A.25)

So the proof boils down to solving the above optimization problem for rj

(Dj

). We proceedaccording to the following two cases (i) ⇢⇤

j

2 [0, 1) and �⇤j

> 0, and (ii) ⇢⇤j

< 0 and �⇤j

> 0.We find the optimal value and optimal solution for each case, and compare these two casesto find the global optimal value.

Case (i): ⇢⇤j

2 [0, 1) and �⇤j

> 0. As max{�⇢j

, 0} = 0 when 0 ⇢j

< 1, (A.25) can be


simplified as

min⇢j ,�j

⇢

µj

· ⇢j

+ ((µj

)2 + (�j

)2) · �j

�

s.t.(1� ⇢

j

)2

4�j

= Dj

.

0 ⇢j

< 1,�j

> 0.

(A.26)

Figure A.17(a) depicts the optimization problem (A.26). The optimal solution of (A.26)would be such that the objective function is a tangent line to the curve that represents theconstraint. The tangent of the line that represents the objective function is

d�j

d⇢j

= � µj

µ2

j

+ �2

j

.

The tangent of the curve that represents the constraint is

d�j

d⇢j

= � 1

2Dj

(1� ⇢j

).

Setting the two tangents equal, we have

⇢⇤j

= 1� 2Dj

µj

µ2

j

+ �2

j

.

As (A.26) requires that 0 ⇢j

< 1,

⇢⇤j

= 1� 2Dj

µj

µ2

j

+ �2

j

is optimal only when Dj

µ

2j+�

2j

2µj. When D

j

>µ

2j+�

2j

2µj, the tangent of the line that represents

the objective function is always smaller than that of the line that represents the constraint.

We have ⇢⇤j

= 0 when Dj

>µ

2j+�

2j

2µj.

In sum, when Dj

µ

2j+�

2j

2µj, the optimal value and optimal solution of (A.26) are µ

j

�µ

2j

µ

2j+�

2jD

j

and (⇢⇤j

,�⇤j

) = (1 � 2µj

µ

2j+�

2jD

j

,µ

2j

(µ

2j+�

2j )

2 Dj

), respectively. When Dj

>µ

2j+�

2j

2µj, the

optimal value and optimal solution of (A.26) areµ

2j+�

2j

4

¯

Djand (⇢⇤

j

,�⇤j

) = (0, 1

4

¯

Dj), respectively.

Case (ii): ⇢⇤j

< 0 and �⇤j

> 0. We solve

min⇢j ,�j

⇢

µj

· ⇢j

+ ((µj

)2 + (�j

)2) · �j

+(max{�⇢

j

, 0})2

4�j

�

s.t.(1� ⇢

j

)2

4�j

� (max{�⇢j

, 0})2

4�j

= Dj

.

⇢j

< 0,�j

> 0.


Figure A.17: Illustration of problems (A.26) and (A.27)

(a) Case (i): 0 ⇢⇤j < 1 and �⇤j > 0

𝜇 𝜌 + 𝜇 + 𝜎 𝜙

𝜌

𝜙

1 − 𝜌4𝜙 = 𝐷

optimal solution

0

(b) Case (ii): ⇢⇤j < 0 and �⇤j > 0

1 − 2𝜌4𝜙 = 𝐷

optimal solution

𝜌

𝜙

0

As max{�⇢j

, 0} = �⇢j

when ⇢j

< 0, (A.25) can be simplified as

min⇢j ,�j

⇢

µj

· ⇢j

+ ((µj

)2 + (�j

)2) · �j

+(⇢

j

)2

4�j

�

s.t.1� 2⇢

j

4�j

= Dj

.

0 ⇢j

< 1,�j

> 0.

(A.27)

Figure A.17(b) depicts the optimization problem (A.27). The optimal solution to (A.27)would be such that the constraint is a tangent line to the curve that represents the objectivefunction. The tangent of the line that represents the constraint is

d�j

d⇢j

= � 1

2Dj

.

The contour function of the objective function of (A.27) is

µj

· ⇢j

+ ((µj

)2 + (�j

)2) · �j

+(⇢

j

)2

4�j

= constant.

Taking derivative with respect to ⇢j

on both sides of the above equation yields

µj

+ ((µj

)2 + (�j

)2)d�

j

d⇢j

+2⇢

j

4�j

�⇢2j

4�2

j

d�j

d⇢j

= 0.


After simplification, we infer that the tangent of the curve that represents the objectivefunction is

d�j

d⇢j

= �µj

� (� ⇢j

2�j)

µ2

j

+ �2

j

� (� ⇢j

2�j)2.

Setting the two tangents equal, we have

1

2Dj

=µj

� (� ⇢j

2�j)

µ2

j

+ �2

j

� (� ⇢j

2�j)2.

Solving the above equation and 1� 2⇢j

= 4�j

Dj

yields

⇢⇤j

= �Dj

�p

(Dj

� µj

)2 + (�j

)2

2p

(Dj

� µj

)2 + (�j

)2.

and

�⇤j

=1

4p

(Dj

� µj

)2 + (�j

)2.

The optimal value to (A.27) is

�1

2(D

j

� µj

) +1

2

q

(Dj

� µj

)2 + (�j

)2.

Comparing Cases (i) and (ii). When Dj

>µ

2j+�

2j

2µj, the following inequalities show that

� 1

2(D

j

� µj

) +1

2

q

(Dj

� µj

)2 + (�j

)2 µ2

j

+ �2

j

4Dj

()q

(Dj

� µj

)2 + (�j

)2 µ2

j

+ �2

j

2Dj

+ (Dj

� µj

)

()(Dj

� µj

)2 + (�j

)2 (µ2

j

+ �2

j

)2

4(Dj

)2+ (D

j

� µj

)2 +(µ2

j

+ �2

j

)(Dj

� µj

)

Dj

()0 µ2

j

+(µ2

j

+ �2

j

)2

4(Dj

)2�

µj

(µ2

j

+ �2

j

)

Dj

()(µ2

j

+ �2

j

2Dj

� µj

)2 � 0.

We infer that case (ii) leads to a lower optimal value than case (i) when Dj

> (µ2

j

+�2

j

)/(2µj

).Similarly, it can be verified straight-forwardly that

�1

2(D

j

� µj

) +1

2

q

(Dj

� µj

)2 + (�j

)2 � µj

� Dj

µ2

j

µ2

j

+ �2

j

.

So case (i) has a lower optimal value than case (ii) when Dj

(µ2

j

+ �2

j

)/(2µj

). In sum, weget the functional form of r

j

(Dj

) specified in Theorem (14). ⇤


Proof. Proof of Result 1. Part (a). Taking first order derivative of rj

(Dj

) with respect toD

j

yields

drj

(Dj

)

dDj

=

8

<

:

� (µj)2

(µj)2+(�j)

2 0, when Dj

(µj)2+(�j)

2

2µj,

�1

2

+ 1

2

¯

Dj�µjp(

¯

Dj�µj)2+(�j)

2 0, when D

j

>(µj)

2+(�j)

2

2µj.

Similarly, we have

drj

(Dj

)

dµj

=

8

<

:

1� Dj

2µj(�j)2

((µj)2+(�j)

2)

2 � 0, when Dj

(µj)2+(�j)

2

2µj,

1

2

� 1

2

¯

Dj�µjp(

¯

Dj�µj)2+(�j)

2� 0, when D

j

>(µj)

2+(�j)

2

2µj.

and

drj

(Dj

)

d�j

=

8

<

:

Dj

2�j(µj)2

((µj)2+(�j)

2)

2 � 0, when Dj

(µj)2+(�j)

2

2µj,

1

2

�jp(

¯

Dj�µj)2+(�j)

2� 0, when D

j

>(µj)

2+(�j)

2

2µj.

It can be verified that drj(¯

Dj)

d

¯

Dj 0, drj(

¯

Dj)

dµj� 0 and drj(

¯

Dj)

d�j� 0. (4.6) tell that W (C,µ,�)

and rj

(Dj

) are positively correlated. An increase in either µ or � increases rj

(Dj

), whichin turn increases W (C,µ,�). An increase in C increases D which in turn decreases r

j

(Dj

)and W (C,µ,�).

Part (b). rj

(Dj

) is convex in Dj

because

d2rj

(Dj

)

d(Dj

)2=

8

<

:

0, when Dj

(µj)2+(�j)

2

2µj,

1

2

�

2j

[(

¯

Dj�µj)2+(�j)

2]

32, when D

j

>(µj)

2+(�j)

2

2µj

is non-negative.We examine the dual form of (4.6). Let �, � and ⌧ be the dual decision variables that

correspond to the three constraints of (4.6). Define

L(�,�, ⌧ )

= inffij ,

¯

Dj

8

<

:

X

j2V 0

rj

(Dj

) +X

i2V

�i

(Ci

�X

j:(i,j)2E

fij

) +X

j2V 0

�j

(Dj

�X

i:(i,j)2E

fij

)�X

(i,j)2E

fij

⌧ij

9

=

;

=X

i2V

�i

Ci

+ inffij ,

¯

Dj

8

<

:

X

j2V 0

[rj

(Dj

) + �j

Dj

]�X

(i,j)2E

(�i

+ �j

+ ⌧ij

)fij

9

=

;

=X

i2V

�i

Ci

+X

j2V 0

⇢

inf¯

Dj

[rj

(Dj

) + �j

Dj

]

�

�X

(i,j)2E

(

supfij

(�i

+ �j

+ ⌧ij

)fij

)

For all values of �, �, and ⌧ � 0, L(�,�, ⌧ ) is a lower bound ofW (C,µ,�). We are interestedin the tightest lower bound, that is, the maximum of L(�,�, ⌧ ). Because L(�,�, ⌧ ) goes to


minus infinity when �i

+�j

+ ⌧ij

6= 0, we can focus on cases where �i

+�j

+ ⌧ij

= 0. Similarly,we focus on �

j

� 0 because L(�,�, ⌧ ) goes to minus infinity when there exists j 2 V 0 suchthat �

j

< 0. We have

W (C,µ,�) = max�,�,⌧

(

X

i2V

�i

Ci

+X

j2V 0

min¯

Dj

{rj

(Dj

) + �j

Dj

})

s.t. �j

� 0, 8j 2 V 0;

�i

+ �j

+ ⌧ij

= 0, 8(i, j) 2 E;

⌧ij

� 0, 8(i, j) 2 E.

We can get rid of ⌧ij

,

W (C,µ,�) = max�,�

(

X

i2V

�i

Ci

+X

j2V 0

min¯

Dj

{rj

(Dj

) + �j

Dj

})

s.t. �j

� 0, 8j 2 V 0;

�i

��j

, 8(i, j) 2 E.

The maximum attainable value of �i

is maxj:(i,j)2E{��

j

}. We can further get rid of �i

,

W (C,µ,�) = max�

(

X

i2V

Ci

maxj:(i,j)2E

{��j

}+X

j2V 0

min¯

Dj

{rj

(Dj

) + �j

Dj

})

s.t. 0 �j

µ2

j

µ+

j

�2

j

, 8j 2 V 0.

(A.28)

DefineL1

(�) =X

i2V

Ci

maxj:(i,j)2E

{��j

}

andL2

(�) =X

j2V 0

min¯

Dj

{rj

(Dj

) + �j

Dj

}.

Let D⇤be the optimal solution of the optimization in L

2

(�). In (A.28), we focus on �j

µ

2j

µ

2j+�

2jbecause L

2

(�) does not increase and L1

(�) decreases when �j

increases beyondµ

2j

µ

2j+�

2j.

By first order condition D

⇤satisfies

�1

2+

1

2

D⇤j

� µj

q

(D⇤j

� µj

)2 + (�j

)2= ��

j

(A.29)

It can be verified that L1

(�) is a piecewise-linear and decreasing function of �, and thatL2

(�) is a concave increasing function of �j

for each j 2 V 0. By the Envelop Theorem,


the first order derivative of L2

(�) with respect to �j

is D⇤j

. Because D⇤j

is nonnegative anddecreasing in �, L

2

(�) is increasing and concave in �j

. Let �

⇤ be the optimal solution to(A.28). �⇤ balances the tradeo↵ between L

1

(�) and L2

(�). We argue that �⇤ in increasingin each C

i

. The reason is as follows. The optimal �⇤j

is such that the decreasing slope ofL1

(�) equals the increasing slope of L2

(�). The argument also holds for � at which L1

(�)is not di↵erentiable. Therefore, an increase in C

i

raises the decreasing slope of the first part,and thus increases �⇤

j

for all j 2 V 0. By the Envelop theorem, the first order e↵ects of Ci

isgiven by

dW (C,µ,�)

dCi

= maxj:(i,j)2E

��⇤j

, (A.30)

where �⇤j

is the optimal solution to the dual problem (A.28). To show that W (C,µ,�) isconvex in C

i

, we need only to show that maxj:(i,j)2E ��⇤

j

is increasing in Ci

, which is alreadyshown. ⇤

Part (c). It can be verified that

@2rj

(Dj

)

@Dj

@µj

=

8

<

:

� 2µ(�j)2

[(µj)2+(�j)

2]

2 , when Dj

(µj)2+(�j)

2

2µj,

�1

2

�

2j

[(

¯

Dj�µj)2+(�j)

2]

32, when D

j

>(µj)

2+(�j)

2

2µj.

Because @

2rj(

¯

Dj)

@

¯

Dj@µj< 0, �dr

j

(Dj

)/dDj

is increasing in µj

for each j 2 V 0. Figure 4.1(b)

provides examples where �drj

(Dj

)/dDj

is neither increasing nor decreasing in �j

.Because of (A.30), we need only to show that �⇤ is increasing in µ

j

for each j 2 V 0. Recallthat �⇤ is the optimal solution to (A.28) and that �⇤ balances the tradeo↵ between L

1

(�)and L

2

(�). Because L1

(�) is una↵ected by µj

, we need to only examine the e↵ects of µj

onL2

(�). Recall that the first order derivative of L2


is D⇤j

(by the Enveloptheorem), we need only to examine the e↵ects of µ

j

on D⇤j

. (A.29) suggests that D⇤j

increasesin µ

j

. Therefore, we conclude that (1) the first order derivative of L2


isdecreasing in µ

j

; that (2) �⇤ is increasing in µj

; and that (3) dW (C,µ,�)/dCi

is decreasingin µ

j

. ⇤Part (d). The first part can be easily verified. The second part follows directly from

(A.30). We require that �j

µ

2j

µ

2j+�

2jin the dual form in the proof of part (b). ⇤

Proof. Proof of Result 2. We first find out the optimal Dj

for the total-flexibility configura-tion. For the total-flexibility configuration, (4.6) becomes

R(C,µ,�) = min¯

Dj

X

j2V 0

rj

(Dj

).

s.t.X

j2V 0

Dj

=X

i2V

Ci

.


Because rj

(Dj

) is decreasing convex in Dj

, the above optimization is easy to solve. Theoptimal solution is given by

D⇤j

= µj

+ �j

·P

i2V Ci

�P

j2V 0 µj

P

j2V 0 �j

.

We infer that a flexibility configuration E achieves the same optimal value if (4.7) can besatisfied. ⇤

89

Appendix B

Numerical examples for Figure 1.1

In this electronic companion, we provide the full set of configurations tested when plottingFigure 1. Table B.1 reports the simulation results when k = 13. Column �(13) denotes thetwo plants that are directly connected to product 13. We assume that the first 12 productsform a chain with all 12 plants. That is, plant i is directly connected to products i andi + 1 for 1 i < 12 and plant 12 is directly connected to products 1 and 12. Column Exp.S lists the expected total shortfall of each structure. �p

n

is the standard deviation of thesample mean, where � is the sample standard deviation and n = 10, 000 is the sample size.Structures that have the maximal or minimal shortfall are highlighted in bold.

Table B.1: k = 13

�(13) Exp. S �pn

1,2 5.3637 0.08511,3 4.6664 0.08331,4 4.1666 0.08111,5 3.8279 0.07941,6 3.6312 0.07831,7 3.5439 0.0779

The following 5 tables report the simulation results for 14 k 18. For each k, wetested 100 structures that satisfy the Chaining Guideline. This allows us to estimate

maxEi2JG(k)

R(Ei

)� minEi2JG(k)

R(Ei

)

minEi2JG(k)

R(Ei

)

numerically.

APPENDIX B. NUMERICAL EXAMPLES FOR FIGURE 1.1 90

Table B.2: k = 14

�(13) �(14) Exp. S �pn

1,2 3,4 7.0585 0.09301,3 2,4 6.7921 0.09352,3 1,4 6.8693 0.09331,7 4,10 2.6567 0.06712,12 1,10 5.4088 0.087210,11 2,5 3.9855 0.07364,10 6,11 3.2408 0.07213,4 2,11 4.8311 0.08117,2 11,8 3.2871 0.07215,7 1,3 4.0575 0.07592,3 6,1 4.6893 0.080811,12 6,5 4.7623 0.076611,5 2,10 3.1484 0.07145,3 2,12 4.6069 0.080312,7 1,3 3.5388 0.07315,10 1,3 3.4514 0.07218,9 6,7 7.0744 0.09274,9 3,5 3.9743 0.07728,10 1,12 4.9358 0.081210,6 4,7 3.6799 0.07597,10 8,5 4.3191 0.080910,7 5,12 3.2014 0.070911,7 8,3 3.0078 0.07034,6 3,11 3.5888 0.07303,6 4,12 3.6033 0.07476,3 11,12 3.9427 0.07296,2 4,5 5.5278 0.08648,4 9,3 3.8161 0.07652,4 6,7 4.8954 0.08102,4 10,1 3.9378 0.076112,9 6,7 4.2715 0.07623,6 12,7 3.2503 0.07137,3 6,8 4.4771 0.08089,5 12,1 3.7461 0.071811,10 2,4 4.4852 0.07705,9 2,8 3.1243 0.07146,10 9,11 4.5141 0.0812

Continued on next page


Table B.2 – continued from previous page�(13) �(14) Exp. S �p

n

11,5 9,3 2.9118 0.06941,9 7,6 3.9415 0.073911,8 10,7 5.3177 0.08703,11 1,6 3.0768 0.07063,12 9,7 3.4533 0.07086,1 9,7 3.5541 0.07312,10 9,8 4.4359 0.07867,12 8,10 4.3795 0.08006,10 2,3 3.7943 0.07265,10 1,7 2.7647 0.06826,8 4,1 3.5084 0.071912,3 2,5 4.3034 0.08003,6 5,12 3.5752 0.074612,1 9,4 3.8166 0.07316,7 12,4 3.8837 0.07329,8 7,3 4.3085 0.07762,12 3,1 6.7944 0.09417,11 9,3 2.9415 0.06975,6 12,2 4.4097 0.076511,8 5,3 3.4098 0.07066,2 8,3 3.5525 0.07515,7 4,8 5.3492 0.08664,10 12,9 3.3671 0.07265,8 2,11 3.0968 0.069011,10 4,8 3.9921 0.07431,6 4,2 4.3422 0.08003,6 2,8 3.6508 0.07536,9 8,1 3.2781 0.07211,4 7,8 3.9446 0.07345,10 9,12 3.3614 0.07257,4 2,8 3.6413 0.075310,6 2,4 3.3727 0.07142,4 6,7 4.8954 0.08106,11 7,12 3.7572 0.07598,12 3,9 3.1690 0.07154,9 1,3 3.5863 0.07359,11 5,10 4.0506 0.07769,1 8,5 3.2011 0.0706




n

11,1 6,10 3.6054 0.07314,10 6,1 2.7656 0.06813,9 6,2 3.1344 0.07165,8 3,9 3.6303 0.07513,12 4,10 3.6873 0.07543,4 2,7 4.7004 0.08109,7 6,8 6.7862 0.09318,9 12,3 3.9885 0.07379,3 2,8 3.6595 0.07566,8 10,5 4.3700 0.08088,5 11,10 4.2691 0.07674,8 7,11 3.1599 0.07134,2 12,8 3.3374 0.07066,8 7,9 6.7831 0.09317,12 3,2 3.8727 0.07362,1 5,6 5.2053 0.08125,10 8,12 3.0695 0.070312,3 2,9 3.6943 0.07612,7 11,6 2.9709 0.06995,9 7,2 3.0417 0.07108,4 1,10 3.0791 0.06933,6 9,5 3.6406 0.07549,5 6,1 2.9729 0.06994,6 3,10 3.5590 0.07356,11 5,10 3.8045 0.0767

Table B.3: k = 15

�(13) �(14) �(15) Exp. S �pn

1,7 3,9 5,11 2.3244 0.06351,2 3,4 5,6 7.1573 0.095310,4 7,2 9,6 2.8679 0.06913,6 2,1 11,7 3.4577 0.071412,9 7,10 11,1 4.8069 0.084511,8 12,7 6,10 4.6098 0.08423,6 11,7 9,8 3.7743 0.0744



Table B.3 – continued from previous page�(13) �(14) �(15) Exp. S �p

n

3,8 2,9 11,12 3.8522 0.075710,7 12,1 2,11 4.8401 0.08076,11 3,7 8,1 2.6510 0.06658,5 1,6 3,2 3.8050 0.07603,2 1,8 4,7 4.0153 0.07729,6 7,2 11,4 2.5315 0.06533,7 8,6 12,1 3.5647 0.07162,8 7,1 12,9 3.7228 0.07609,1 11,12 10,7 4.8896 0.08453,5 2,1 12,4 6.3181 0.09424,6 8,1 11,7 2.8921 0.067811,5 6,1 3,8 2.5050 0.06524,11 2,12 7,9 3.0816 0.068612,4 5,6 10,2 3.2286 0.06963,5 1,7 11,9 2.8808 0.06656,11 2,9 7,3 2.6016 0.06628,4 2,3 11,1 3.8967 0.07453,1 6,11 2,4 4.6220 0.07996,2 12,4 1,7 3.4964 0.074810,8 2,1 11,7 4.1965 0.07722,10 5,4 9,1 3.5198 0.07201,9 8,7 10,4 3.1882 0.06959,7 5,1 10,8 4.3013 0.07749,2 7,6 11,10 3.8660 0.07309,1 2,10 12,8 4.5923 0.08384,8 9,7 3,12 3.1539 0.069911,2 5,9 8,10 3.1921 0.07035,3 2,10 7,8 3.4201 0.07116,2 10,4 3,7 3.1231 0.07092,5 10,4 8,12 2.5309 0.06526,9 10,8 2,12 3.6382 0.07273,4 10,6 5,1 3.7117 0.07319,6 8,1 4,10 2.7169 0.06739,2 1,6 8,7 3.6087 0.07372,8 3,4 11,9 3.5054 0.07217,3 1,11 9,4 2.7734 0.06652,8 12,3 4,5 3.4252 0.07161,9 5,12 8,2 2.8054 0.0678




n

5,2 10,11 9,4 3.3096 0.07107,10 8,5 4,6 4.8262 0.08576,5 7,9 2,1 4.1886 0.07464,8 12,6 3,10 2.5043 0.065310,9 2,6 3,8 3.3190 0.070911,7 9,2 12,1 3.7956 0.075210,9 2,7 4,5 3.6389 0.07155,3 4,1 12,8 3.6013 0.073412,2 10,7 6,4 2.9856 0.067410,3 1,9 8,6 2.8802 0.06775,10 4,11 7,8 3.8535 0.076712,6 1,11 8,5 3.1250 0.070112,3 8,5 2,1 4.4630 0.07766,3 9,5 11,7 2.9497 0.06973,12 4,5 2,8 3.4820 0.07203,1 9,5 8,11 2.8192 0.066910,9 5,8 7,4 5.0118 0.08573,6 10,12 1,7 2.9387 0.06752,10 12,1 9,7 4.1608 0.076911,8 2,3 1,12 5.0252 0.08125,12 9,10 8,3 2.9499 0.06754,9 7,5 8,10 4.7055 0.08518,7 1,9 12,5 3.2339 0.070112,5 11,6 3,2 3.8360 0.07594,9 10,1 11,12 4.7979 0.081010,1 5,9 3,4 3.4053 0.07159,6 8,3 10,12 2.9510 0.06812,3 6,11 1,7 3.3078 0.070510,4 3,5 7,11 3.0970 0.07048,2 5,1 7,6 3.8678 0.076512,10 6,11 2,5 3.1738 0.070112,9 8,5 2,10 2.9595 0.06845,9 11,4 7,12 2.8321 0.06897,4 8,5 10,6 4.7833 0.08579,12 4,11 1,5 3.0347 0.07018,4 3,9 1,5 3.0338 0.07066,3 9,11 4,2 3.5325 0.071511,2 8,5 10,7 2.9568 0.0685




n

6,11 5,12 1,3 3.6988 0.07559,8 5,10 1,2 3.5929 0.070612,8 3,5 2,4 4.3268 0.07734,2 5,11 12,1 4.9225 0.08445,9 10,7 11,1 3.2152 0.07044,1 3,9 11,7 2.5306 0.06521,12 10,4 7,9 3.2627 0.069211,6 7,4 9,8 3.9142 0.07724,2 6,5 10,9 4.2902 0.07542,1 7,4 12,10 3.7404 0.072011,8 12,7 9,1 4.5897 0.08406,8 7,5 12,10 4.7902 0.080411,5 8,12 9,3 2.6250 0.06678,1 5,9 12,10 3.2122 0.07096,10 12,11 5,3 3.5001 0.071210,11 7,8 2,6 3.9808 0.07373,11 10,4 9,8 3.7493 0.0754

Table B.4: k = 16

�(13) �(14) �(15) �(16) Exp. S �pn

1,7 2,8 4,10 5,11 2.0153 0.05821,2 3,4 5,6 7,8 5.4834 0.084410,11 2,8 4,7 12,6 2.2481 0.059610,2 6,11 12,8 1,9 2.9928 0.069110,9 5,8 3,1 4,2 3.2322 0.067210,9 4,12 1,6 5,3 2.8141 0.06506,8 9,10 4,2 12,5 2.5694 0.06148,3 10,4 7,9 11,12 2.7273 0.06437,2 4,11 10,3 12,5 2.3424 0.06203,4 8,6 5,10 7,12 2.5967 0.06264,10 5,7 1,12 2,6 2.5478 0.06281,5 2,10 4,7 8,9 2.4002 0.06119,6 2,3 11,10 7,12 2.8765 0.06471,6 2,12 10,11 5,4 3.4511 0.071110,6 11,3 4,2 7,8 2.6056 0.0623



Table B.4 – continued from previous page�(13) �(14) �(15) �(16) Exp. S �p

n

5,7 1,3 2,6 11,12 3.3832 0.07106,5 11,2 10,3 12,7 2.4611 0.06201,3 5,10 8,9 6,7 3.6806 0.07074,9 3,5 8,10 1,12 2.6645 0.063110,6 4,7 8,5 12,11 3.2146 0.06927,8 3,4 6,11 12,2 2.8365 0.06404,5 8,9 3,2 6,7 5.4419 0.08412,4 10,1 12,9 6,7 2.6236 0.06213,6 12,7 8,9 5,1 2.5642 0.062811,10 2,4 5,9 8,6 2.9804 0.065610,9 11,5 3,1 7,6 2.8427 0.063611,8 10,7 3,1 6,12 2.7219 0.06459,7 6,1 2,10 8,12 2.8277 0.06748,10 6,2 3,5 1,7 2.8270 0.06526,8 4,1 12,3 2,5 3.7628 0.07463,6 5,12 1,9 4,7 2.4632 0.062412,6 4,9 8,7 3,2 2.6004 0.062912,3 1,7 11,9 5,6 2.4125 0.060612,2 11,8 5,3 6,7 2.7366 0.06384,8 10,12 9,5 2,11 2.5351 0.062711,10 4,8 1,6 2,3 2.5201 0.06136,2 8,9 1,4 7,5 3.2724 0.070110,9 12,7 4,2 8,6 2.7523 0.06332,4 6,7 11,12 8,3 2.9181 0.06509,4 1,3 11,5 10,8 2.4692 0.06235,11 1,6 10,4 3,9 2.2950 0.06116,2 5,8 3,9 12,4 2.3329 0.062010,3 4,2 7,9 6,8 3.2977 0.06938,9 12,3 2,6 10,5 2.2323 0.05928,5 11,10 4,7 2,12 2.6850 0.06258,6 7,9 12,3 2,1 3.5369 0.06815,6 10,8 12,3 2,9 2.4060 0.06032,7 11,6 5,9 8,4 2.3755 0.06181,10 3,6 9,5 4,11 2.2893 0.06125,10 3,12 4,9 6,11 2.5680 0.064010,3 11,12 7,8 2,5 2.5231 0.06139,10 4,7 2,6 3,1 2.7869 0.064711,7 12,9 10,1 8,6 4.0038 0.0781




n

10,3 6,11 7,9 8,2 2.4844 0.06318,9 11,12 10,7 1,2 4.9795 0.081811,6 3,7 8,1 5,2 2.3499 0.06193,2 1,8 4,7 9,6 2.9727 0.06827,6 2,11 4,3 8,12 2.7104 0.06301,2 8,7 12,9 11,10 5.0510 0.08307,3 5,2 1,12 4,6 4.5723 0.08248,1 11,7 5,6 3,4 2.9617 0.064411,2 12,7 9,4 5,6 2.2354 0.059310,2 3,5 1,7 11,9 2.2847 0.06026,11 2,9 7,3 8,4 2.3558 0.06172,3 11,1 6,4 12,7 3.2233 0.069810,8 2,1 11,7 5,4 2.8720 0.06349,1 8,7 10,4 5,2 2.3722 0.06122,7 6,11 10,9 1,12 3.1289 0.06849,2 8,4 7,3 12,11 2.6040 0.06312,5 9,8 10,3 7,6 3.4135 0.06962,10 4,3 7,5 8,12 2.4369 0.06066,9 10,8 2,12 3,4 2.6967 0.062110,6 5,4 1,9 8,2 2.3764 0.06092,1 6,8 9,7 3,4 4.0232 0.07274,3 11,9 7,1 2,8 2.6271 0.063012,3 4,5 1,9 8,2 2.7365 0.06515,2 10,11 9,4 7,8 2.7566 0.06345,4 6,7 9,2 1,8 3.4358 0.070212,6 3,10 9,2 8,11 2.3278 0.06157,9 2,12 1,10 4,5 2.6615 0.06265,3 4,1 12,8 2,10 2.5488 0.06327,6 4,10 3,1 9,8 3.1726 0.06606,5 10,4 11,7 8,12 2.9801 0.06766,1 11,8 5,12 3,2 2.7568 0.06481,6 3,9 5,11 7,12 2.1051 0.05904,12 3,5 2,8 1,9 2.5531 0.06435,8 1,11 10,9 7,4 2.9758 0.06523,6 10,12 1,7 2,9 2.3860 0.061710,7 11,8 2,3 1,12 3.6907 0.07155,12 9,10 8,3 4,7 2.5062 0.06225,8 10,7 1,9 12,11 3.1810 0.0689




n

6,5 3,2 4,9 10,1 3.0292 0.066311,12 10,1 5,9 3,4 3.5878 0.07159,6 8,3 10,12 2,11 2.4811 0.06242,1 7,10 4,3 5,11 3.2193 0.06728,2 5,1 7,6 12,10 2.5525 0.06306,11 2,5 12,9 8,10 2.6985 0.06425,9 11,4 7,12 8,10 2.8678 0.06725,6 9,12 4,11 1,8 2.4319 0.06144,3 9,8 1,5 6,11 2.6696 0.0619

Table B.5: k = 17

�(13) �(14) �(15) �(16) �(17) Exp. S �pn

1,7 2,8 3,9 4,10 5,11 1.8384 0.05491,2 3,4 5,6 7,8 9,10 3.5242 0.06709,10 11,6 1,3 7,2 5,8 1.9750 0.05546,9 8,3 2,1 11,10 4,5 2.2751 0.05777,10 12,8 3,5 11,2 6,1 1.9198 0.05566,7 11,1 10,12 5,3 8,2 2.3403 0.05805,8 3,11 2,1 7,9 6,10 2.3356 0.05857,8 2,11 6,3 9,1 5,4 2.4111 0.05854,10 5,12 9,3 8,11 7,6 2.2229 0.05911,7 8,10 9,4 3,6 5,2 2.2547 0.05946,12 1,5 8,3 4,9 7,2 2.1711 0.05894,9 7,2 12,11 5,1 10,3 1.9545 0.05573,1 12,9 11,2 5,6 10,8 2.3658 0.05934,11 12,2 10,1 7,8 9,5 2.0930 0.056512,2 3,11 9,6 10,8 1,7 2.2248 0.05885,4 11,1 10,7 6,8 2,9 2.1318 0.05682,11 7,4 6,8 1,5 9,10 2.0097 0.055410,8 4,7 5,3 1,12 2,9 2.1066 0.05661,10 11,9 12,7 5,4 8,2 2.3007 0.058812,4 3,9 10,11 8,7 5,6 2.5651 0.060812,9 4,10 6,7 2,3 5,1 2.1191 0.05658,5 10,4 11,6 1,9 7,12 2.1383 0.05869,7 2,11 12,10 4,5 1,8 2.3314 0.0591



Table B.5 – continued from previous page�(13) �(14) �(15) �(16) �(17) Exp. S �p

n

7,6 4,10 8,12 11,9 1,2 2.2331 0.05768,7 3,1 11,5 10,9 4,6 2.4961 0.05985,8 7,4 9,3 12,11 10,1 2.2212 0.056912,6 8,4 11,9 2,10 7,5 1.9337 0.05562,12 8,10 11,7 1,4 6,9 2.2552 0.05787,10 9,5 8,11 4,6 12,2 2.2560 0.058711,7 2,12 6,8 10,5 1,4 1.9838 0.055711,2 6,5 4,9 1,12 10,8 2.3023 0.05772,8 7,10 12,11 3,9 1,4 2.2443 0.05932,9 5,10 4,3 6,12 1,8 1.8797 0.05505,6 4,8 7,10 9,2 11,12 2.1415 0.05684,10 2,5 6,8 3,7 12,1 2.0852 0.056710,1 5,7 8,3 12,6 4,2 1.9491 0.05596,1 9,12 10,7 11,5 2,3 1.9609 0.05549,6 3,2 5,4 1,8 10,11 2.2609 0.05717,10 6,12 5,8 9,1 4,2 1.9012 0.05538,6 3,4 2,10 12,11 1,7 2.1182 0.056511,5 8,9 4,1 7,3 10,12 1.9615 0.05526,11 4,12 2,7 10,5 8,9 1.9892 0.05605,8 4,6 10,12 7,2 3,11 1.9604 0.05586,4 12,8 11,7 3,9 5,1 1.8455 0.05474,7 10,1 3,2 6,5 12,8 2.4491 0.05899,12 11,5 6,7 4,10 1,3 2.0537 0.05631,11 2,3 7,12 4,5 6,9 2.2779 0.05741,4 5,2 12,3 6,10 9,7 2.2388 0.05804,7 1,5 10,12 3,2 8,11 2.0670 0.05649,3 11,7 10,2 12,8 6,4 1.9055 0.05559,7 10,12 2,6 3,1 8,5 1.9991 0.05573,9 4,11 6,1 7,8 10,5 1.9457 0.05546,1 11,7 10,5 12,3 4,2 2.2335 0.05948,4 1,7 12,9 5,6 11,10 2.4786 0.06016,2 11,7 10,12 3,5 9,4 1.9136 0.05526,1 4,8 5,3 7,2 12,11 2.5149 0.061811,8 10,9 6,5 1,4 2,7 2.2721 0.056912,9 11,5 2,6 10,1 3,4 2.2613 0.059311,4 7,5 6,2 10,3 12,1 2.2770 0.059512,10 3,5 7,4 2,1 8,9 2.3334 0.05799,12 2,11 6,5 8,4 10,3 2.0529 0.0565




n

8,4 2,10 9,5 11,1 3,12 2.4249 0.06113,9 10,1 7,11 12,2 8,6 2.2110 0.05874,11 10,5 2,3 9,12 7,8 2.1175 0.05691,3 2,9 12,5 10,11 7,8 2.1432 0.05685,9 10,2 4,12 6,11 8,7 2.0168 0.05616,5 10,8 2,4 3,11 12,1 2.6167 0.061312,8 11,4 1,2 5,6 7,3 2.3822 0.06011,9 7,11 5,3 2,12 4,6 2.1459 0.056610,12 7,3 1,11 2,5 9,8 2.4037 0.059010,1 8,3 9,12 11,7 6,5 2.0763 0.05638,10 12,2 1,7 3,6 9,11 2.5596 0.06147,8 12,5 4,1 6,2 9,11 2.0777 0.05627,12 1,11 6,5 3,2 9,10 2.3832 0.05835,2 8,9 11,4 7,10 3,1 2.1987 0.05681,7 2,11 9,8 4,10 5,3 1.9310 0.05545,1 4,7 9,2 12,6 3,8 2.1778 0.05908,6 10,1 12,7 11,3 5,2 1.8451 0.05484,1 7,5 11,8 2,6 10,9 2.2182 0.05718,9 5,6 3,10 1,7 12,2 2.0964 0.05639,12 3,10 6,2 4,7 11,1 1.9262 0.05539,1 4,8 12,11 2,3 10,6 2.0993 0.05639,12 10,8 1,7 2,11 4,5 2.2876 0.05859,3 12,1 4,10 5,6 8,7 2.5137 0.05971,10 9,8 7,4 6,5 3,12 2.3946 0.05814,8 1,6 10,12 7,2 11,3 1.8405 0.05494,9 3,1 6,7 12,10 8,2 2.0748 0.05669,10 3,12 6,11 1,7 8,5 1.9894 0.05577,10 5,8 3,2 11,9 6,1 2.0532 0.05608,6 10,3 9,11 5,12 7,2 1.8484 0.054911,7 9,5 12,1 4,10 3,6 1.9561 0.05563,7 9,11 4,6 5,1 8,2 2.2763 0.05908,5 7,1 11,10 2,3 12,9 2.3102 0.05775,3 2,10 12,11 7,4 1,6 2.3280 0.05965,4 9,1 6,8 10,7 3,11 2.0406 0.05604,7 6,12 8,5 3,11 9,10 2.2662 0.05943,10 5,9 7,8 2,11 12,1 2.4930 0.059112,4 3,9 10,11 8,7 5,6 2.5651 0.060812,9 4,10 6,7 2,3 5,1 2.1191 0.0565




n

8,5 10,4 11,6 1,9 7,12 2.1383 0.0586

Table B.6: k = 18

�(13) �(14) �(15) �(16) �(17) �(18) Exp. S �pn

1,7 2,8 3,9 4,10 5,11 6,12 1.5576 0.05091,2 3,4 5,6 7,8 9,10 11,12 2.4482 0.05691,4 9,2 3,12 8,11 10,6 7,5 1.5917 0.051112,1 5,8 11,7 2,9 4,6 10,3 1.6469 0.05151,2 7,10 11,9 4,3 6,8 12,5 1.9131 0.05331,4 7,5 9,2 3,8 12,11 6,10 1.6134 0.05133,1 5,2 11,10 12,4 8,7 6,9 1.8922 0.05319,4 11,8 5,7 6,12 1,3 10,2 1.5755 0.05113,7 6,2 10,1 8,4 11,12 9,5 1.7105 0.05193,12 8,1 2,11 10,5 7,6 4,9 1.6276 0.051511,6 5,8 2,7 3,9 10,4 1,12 1.6025 0.05144,2 5,8 7,3 1,12 11,10 9,6 1.7860 0.05238,9 4,12 1,11 7,6 5,10 2,3 1.9126 0.05359,3 1,11 12,6 2,10 7,8 5,4 1.6869 0.051812,4 6,1 5,8 7,3 10,9 2,11 1.6064 0.05136,2 8,3 1,12 5,4 7,10 9,11 1.6951 0.051910,8 11,3 4,7 6,5 1,2 9,12 1.7917 0.05278,3 12,11 1,2 7,4 5,9 6,10 1.7485 0.05226,1 2,11 4,3 10,7 12,5 9,8 1.7692 0.05248,3 12,10 1,2 4,11 5,9 6,7 1.6811 0.05185,8 6,4 9,12 2,1 10,7 11,3 1.6212 0.05136,4 9,5 3,1 7,10 11,8 2,12 1.7166 0.05193,2 8,4 9,11 1,6 5,7 10,12 1.7488 0.05211,5 7,9 11,3 4,8 2,6 10,12 1.5730 0.05108,9 11,1 3,6 4,7 2,10 5,12 1.6351 0.05159,3 6,5 11,2 1,12 7,4 10,8 1.8454 0.05291,9 5,6 2,4 10,8 3,7 12,11 1.8017 0.05272,7 4,10 6,5 12,9 8,1 11,3 1.6038 0.05132,10 4,7 9,1 11,3 12,6 8,5 1.5733 0.05106,3 1,7 8,2 12,10 11,5 4,9 1.5619 0.05103,4 6,5 7,9 2,11 12,10 8,1 1.7779 0.0525



Table B.6 – continued from previous page�(13) �(14) �(15) �(16) �(17) �(18) Exp. S �p

n

2,5 4,3 6,12 11,10 7,1 8,9 1.9074 0.053511,1 4,5 10,6 3,7 9,12 2,8 1.6370 0.051512,5 10,1 3,7 11,9 8,4 2,6 1.5770 0.05102,8 10,9 3,5 6,1 12,11 4,7 1.7731 0.05239,7 2,11 12,10 4,1 8,3 5,6 1.6536 0.05165,10 6,12 8,4 9,3 11,1 7,2 1.5605 0.05102,1 11,9 7,6 4,5 12,3 8,10 2.1081 0.05492,9 3,11 12,4 5,1 8,6 10,7 1.5754 0.05101,4 11,10 5,7 2,9 8,12 6,3 1.6509 0.051510,1 11,12 7,2 5,8 4,3 6,9 1.7958 0.05256,4 5,2 10,3 7,9 11,1 12,8 1.5928 0.051112,5 11,6 8,7 4,1 2,9 3,10 1.6115 0.051411,12 9,6 8,1 4,5 7,10 3,2 1.8262 0.05294,5 8,6 11,10 9,3 2,1 12,7 1.7264 0.052111,1 7,8 10,12 9,4 2,3 6,5 2.0314 0.05442,5 8,1 12,4 9,7 6,11 3,10 1.5658 0.05109,11 5,6 2,12 7,1 4,3 8,10 1.8663 0.05318,9 11,6 4,5 10,1 3,12 7,2 1.6333 0.051511,2 7,5 12,10 4,8 6,9 1,3 1.7632 0.052412,6 2,3 8,11 10,5 4,7 1,9 1.6015 0.05145,9 6,10 8,2 7,4 3,11 12,1 1.6138 0.05149,11 4,3 10,1 2,5 6,12 7,8 1.7846 0.052411,5 4,12 7,6 8,10 3,2 1,9 1.6968 0.05224,11 2,6 9,10 5,12 1,3 7,8 1.7579 0.05212,12 4,3 7,9 1,6 10,11 8,5 1.7348 0.05212,10 6,11 9,7 4,8 12,3 5,1 1.5581 0.05095,6 1,9 8,11 10,7 4,2 3,12 1.6683 0.05182,6 3,11 4,7 12,1 10,5 8,9 1.6636 0.05174,6 10,5 9,11 12,2 3,8 7,1 1.5694 0.05102,11 8,5 4,3 12,1 9,10 6,7 2.1281 0.05492,10 12,6 9,1 4,5 7,11 3,8 1.6045 0.05147,10 2,9 12,1 3,6 5,4 11,8 1.7676 0.05248,7 10,3 6,5 11,9 12,1 2,4 1.8471 0.052812,9 1,8 7,2 3,6 4,10 11,5 1.5628 0.05106,1 7,5 3,2 11,10 9,12 4,8 1.7850 0.052610,7 8,5 6,3 2,4 12,1 9,11 1.6659 0.05174,6 2,5 8,1 11,12 10,3 9,7 1.6278 0.051412,2 11,10 5,7 6,4 9,3 8,1 1.7681 0.0523




n

5,10 9,12 2,1 8,6 11,7 4,3 1.7556 0.05226,12 1,2 3,7 5,9 10,8 11,4 1.6094 0.05142,10 8,7 9,4 6,5 12,11 3,1 1.9293 0.05369,8 1,11 3,12 10,6 5,7 2,4 1.7794 0.05256,11 1,10 9,12 5,3 4,8 7,2 1.5797 0.05112,7 11,6 1,5 3,4 8,12 10,9 1.6800 0.051911,2 5,1 4,3 10,12 7,6 9,8 1.8401 0.05291,3 4,11 5,10 8,7 12,2 9,6 1.9472 0.05376,10 7,1 11,4 12,2 9,8 3,5 1.6358 0.05154,5 7,1 6,9 2,3 12,8 11,10 1.8065 0.05278,11 6,12 5,10 1,9 7,4 3,2 1.5930 0.051311,1 5,8 2,6 3,10 7,4 9,12 1.6088 0.05133,6 5,4 10,12 8,7 1,11 2,9 1.9450 0.053511,5 4,3 2,9 7,8 6,10 1,12 1.7349 0.05212,5 11,10 6,1 7,8 4,9 3,12 1.6858 0.05183,8 12,4 7,6 2,10 11,1 9,5 1.6386 0.05172,5 1,8 11,3 7,6 12,4 10,9 1.6769 0.051811,4 8,2 12,7 5,1 10,9 6,3 1.6097 0.05142,3 8,6 5,1 12,10 9,4 7,11 1.6343 0.051511,12 1,2 3,10 9,4 8,6 7,5 1.9783 0.05374,3 1,9 12,7 5,11 2,8 10,6 1.6019 0.05139,3 7,11 10,5 4,12 1,2 8,6 1.6317 0.05151,2 12,8 6,7 3,11 4,9 10,5 1.6893 0.05194,5 12,9 2,6 7,11 1,8 3,10 1.6163 0.05149,6 2,3 11,10 7,12 1,5 4,8 1.6577 0.05175,7 1,3 2,6 11,12 10,8 9,4 1.6538 0.05169,3 5,8 10,1 12,6 4,7 11,2 1.5885 0.05124,5 8,9 3,2 6,7 10,1 12,11 2.2173 0.055511,10 2,4 5,9 8,6 3,1 7,12 1.7707 0.05249,7 6,1 2,10 8,12 3,5 4,11 1.5634 0.05109,3 5,6 12,2 11,8 7,4 10,1 1.7196 0.0521

When 19 k 24, we not only need to evaluate

maxEi2JG(k)

R(Ei

)� minEi2JG(k)

R(Ei

)

minEi2JG(k)

R(Ei

),


but also need to examine

maxEi2JGFD(k)

R(Ei

)� minEi2JGFD(k)

R(Ei

)

minEi2JGFD(k)

R(Ei

).

Based on the simulation for 13 k 18, we infer that the following structure performs verypoorly: �(1) = {1, 2}, �(2) = {2, 3}, �(3) = {3, 4}, �(4) = {4, 5}, �(5) = {5, 6}, �(6) ={6, 7}, �(7) = {7, 8}, �(8) = {8, 9}, �(9) = {9, 10}, �(10) = {10, 11}, �(11) = {11, 12},�(12) = {12, 1}, �(13) = {1, 2}, �(14) = {3, 4}, �(15) = {5, 6}, and etc. Therefore, we usethis structure to obtain the worst-case performance of the Chaining Guideline. Table B.7reports the expected shortfall of this structure as a function of number of products k.

Table B.7: 19 k 24

k Exp. S �pn

19 3.4646 0.061320 4.5472 0.067821 4.6827 0.069822 3.9172 0.064323 2.7303 0.053624 1.9839 0.0453

All the structures in Tables B.8-B.13 satisfy the Chaining Guideline and the FlexibilityDesign Guideline. For each tested structure, the links for the first 18 products are �(1) ={1, 2}, �(2) = {2, 3}, �(3) = {3, 4}, �(4) = {4, 5}, �(5) = {5, 6}, �(6) = {6, 7}, �(7) ={7, 8}, �(8) = {8, 9}, �(9) = {9, 10}, �(10) = {10, 11}, �(11) = {11, 12}, �(12) = {12, 1},�(13) = {1, 7}, �(14) = {2, 8}, �(15) = {3, 9}, �(16) = {4, 10}, �(17) = {5, 11}, and�(18) = {6, 12}.

Table B.8: k = 19

�(19) Exp. S �pn

1,3 1.445769 0.0475061,4 1.431933 0.047431,5 1.446049 0.0474891,6 1.476208 0.047699


Table B.9: k = 20

�(19) �(20) Exp. S �pn

1,4 2,5 1.2523 0.04347,2 6,4 1.2665 0.043510,7 1,3 1.2758 0.04358,10 5,3 1.2654 0.04354,12 1,5 1.2777 0.043812,10 4,7 1.2745 0.04373,5 4,1 1.2527 0.04343,6 7,11 1.2575 0.04355,12 2,9 1.2622 0.04353,10 8,11 1.2746 0.04359,1 10,2 1.2878 0.04379,6 5,12 1.2851 0.04391,3 9,6 1.2777 0.04361,10 9,12 1.2545 0.04351,9 11,7 1.2602 0.04353,11 7,2 1.2621 0.043410,5 1,11 1.2780 0.04375,10 11,9 1.3602 0.04444,2 12,5 1.2609 0.04358,6 2,10 1.2558 0.04357,5 1,9 1.2552 0.043511,1 2,7 1.2924 0.04384,9 8,10 1.3758 0.04451,5 10,12 1.2738 0.043710,12 3,5 1.2725 0.04366,8 4,7 1.2604 0.04356,3 12,10 1.2736 0.043712,7 4,2 1.2579 0.04341,5 12,2 1.2787 0.04376,3 4,12 1.2786 0.043711,4 12,9 1.2814 0.04389,1 12,4 1.2561 0.043512,3 1,11 1.2590 0.04355,12 11,8 1.2829 0.04388,3 4,2 1.3730 0.04447,2 11,1 1.2924 0.04387,12 4,8 1.2600 0.0435




n

7,10 8,6 1.2572 0.04357,9 2,4 1.2818 0.043612,8 7,4 1.2558 0.043511,7 9,2 1.2610 0.043512,7 6,8 1.3792 0.04459,12 10,7 1.2542 0.04354,8 7,2 1.2937 0.04374,2 11,9 1.2689 0.04367,12 8,11 1.2903 0.04386,1 3,7 1.2894 0.043712,10 4,11 1.3763 0.04467,4 3,1 1.2829 0.04363,12 2,11 1.2525 0.04352,11 8,3 1.2778 0.043610,2 3,8 1.3656 0.04441,6 12,8 1.3746 0.04458,4 6,1 1.2615 0.043510,8 11,3 1.2707 0.04358,1 12,9 1.2931 0.04385,12 1,9 1.2601 0.04357,4 2,9 1.2848 0.04366,10 4,11 1.3751 0.04479,4 7,11 1.2584 0.04353,12 6,9 1.2905 0.04397,4 1,8 1.2898 0.04378,1 3,10 1.2908 0.04378,1 7,12 1.3962 0.04471,9 5,10 1.2569 0.04351,8 7,9 1.3991 0.04472,11 10,5 1.2686 0.04366,8 4,2 1.2576 0.043511,8 6,10 1.2542 0.043512,9 3,11 1.2765 0.04372,9 1,8 1.4084 0.04484,9 3,12 1.2767 0.04371,6 8,5 1.2833 0.04381,8 3,5 1.2581 0.04348,3 5,10 1.2774 0.0435




n

2,12 7,3 1.2820 0.04368,4 7,2 1.2937 0.04374,7 9,6 1.2548 0.04348,10 9,2 1.3752 0.04457,12 10,3 1.2592 0.04341,6 3,8 1.2874 0.04373,7 5,12 1.2606 0.04356,4 10,2 1.2549 0.04351,6 9,7 1.2940 0.043812,4 11,7 1.2799 0.04377,11 10,1 1.2753 0.04378,3 2,10 1.3671 0.04448,5 1,6 1.2864 0.04385,3 12,9 1.2704 0.043612,8 5,1 1.2796 0.04372,6 8,10 1.2555 0.04352,9 10,7 1.2871 0.04361,5 12,2 1.2787 0.043710,5 4,7 1.2796 0.04375,12 2,4 1.2581 0.04354,2 10,8 1.3553 0.04436,8 4,9 1.2644 0.04359,6 8,12 1.2825 0.04381,8 5,2 1.2829 0.0437

Table B.10: k = 21

�(19) �(20) �(21) Exp. S �pn

1,4 2,5 3,6 1.1893 0.042212,2 6,10 11,8 1.1700 0.04191,11 12,9 10,5 1.1876 0.04218,3 9,1 4,2 1.2882 0.043111,4 10,3 12,5 1.3019 0.04321,5 2,10 4,7 1.1788 0.04205,7 1,3 2,6 1.1952 0.04225,10 7,12 11,8 1.1788 0.0419




n

4,6 7,2 10,1 1.1802 0.04212,4 5,9 8,6 1.1776 0.04209,7 6,11 8,10 1.1758 0.04199,7 6,1 2,10 1.1756 0.04208,10 6,2 3,5 1.1699 0.041910,1 5,7 6,8 1.1747 0.04208,4 6,1 12,3 1.1717 0.04197,9 8,3 2,12 1.2625 0.04283,1 7,11 9,5 1.1656 0.04194,8 10,12 9,5 1.1738 0.04198,10 6,2 4,7 1.1700 0.04206,11 7,12 8,3 1.1993 0.04219,7 6,8 12,3 1.2368 0.04257,11 4,2 12,8 1.1713 0.04196,8 7,9 12,3 1.2365 0.04258,4 1,10 3,6 1.1678 0.04194,9 6,11 10,3 1.3341 0.04332,9 6,3 1,11 1.1696 0.041911,8 12,7 6,10 1.1807 0.04206,11 3,7 8,1 1.1903 0.04209,6 7,2 11,4 1.1671 0.04184,6 8,1 11,7 1.1702 0.04194,11 2,12 7,9 1.1666 0.04186,11 2,9 7,3 1.1846 0.04203,1 6,11 2,4 1.1808 0.04219,7 5,1 10,8 1.1713 0.04199,1 2,10 12,8 1.1859 0.04204,8 9,7 3,12 1.1717 0.041811,2 5,9 8,10 1.2333 0.04266,9 10,8 2,12 1.1720 0.04197,3 1,11 9,4 1.1716 0.04197,10 8,5 4,6 1.1779 0.04205,3 4,1 12,8 1.1684 0.041912,2 10,7 6,4 1.1734 0.042010,3 1,9 8,6 1.1741 0.04196,3 9,5 11,7 1.1712 0.04193,1 9,5 8,11 1.1686 0.04194,9 7,5 8,10 1.2010 0.0421




n

9,6 8,3 10,12 1.1734 0.041912,10 6,11 2,5 1.1957 0.042212,9 8,5 2,10 1.1645 0.04185,9 11,4 7,12 1.1984 0.04217,4 8,5 10,6 1.1780 0.04209,12 4,11 1,5 1.1833 0.04206,3 9,11 4,2 1.1742 0.042011,2 8,5 10,7 1.1677 0.04196,11 5,12 1,3 1.2835 0.042812,8 3,5 2,4 1.1764 0.04195,9 10,7 11,1 1.1735 0.042011,8 12,7 9,1 1.1815 0.04196,8 7,5 12,10 1.1767 0.04208,1 5,9 12,10 1.1677 0.041812,7 4,6 1,11 1.2693 0.04291,6 10,5 8,11 1.1741 0.04191,3 11,9 7,2 1.1994 0.04215,3 7,11 2,6 1.1772 0.04203,10 1,5 2,4 1.2116 0.04235,7 4,1 6,3 1.1900 0.04217,3 10,5 11,4 1.2834 0.04272,7 5,3 9,6 1.1847 0.042012,4 8,3 5,2 1.1747 0.04193,11 12,7 4,9 1.1927 0.042011,9 3,7 10,5 1.1897 0.042012,2 4,7 1,9 1.1743 0.04197,10 11,4 6,2 1.1702 0.04202,4 7,12 9,11 1.1648 0.041811,4 8,12 1,5 1.1763 0.04196,10 5,1 8,11 1.1684 0.04193,6 9,1 12,10 1.1787 0.04205,12 6,3 9,7 1.1765 0.041910,12 7,2 1,4 1.1804 0.04207,11 6,9 1,10 1.1757 0.04205,9 12,7 4,1 1.1719 0.04191,3 6,2 9,5 1.1774 0.04201,8 2,5 11,7 1.2224 0.04241,9 12,4 2,11 1.1756 0.0420




n

6,10 9,2 8,12 1.1732 0.041911,7 1,5 9,2 1.1845 0.04205,7 12,10 3,1 1.1712 0.04197,3 4,2 9,5 1.1801 0.04205,1 3,12 9,6 1.1792 0.04208,12 3,10 9,6 1.1779 0.04195,8 10,2 11,6 1.1702 0.04197,11 4,8 12,2 1.1692 0.04187,2 11,6 10,8 1.1691 0.04194,6 11,3 5,10 1.3034 0.04326,8 11,3 5,12 1.1776 0.04192,4 3,8 1,11 1.2014 0.04221,6 4,8 3,7 1.1827 0.04203,6 12,10 5,1 1.1790 0.04219,7 3,6 10,5 1.1809 0.0420

Table B.11: k = 22

�(19) �(20) �(21) �(22) Exp. S �pn

1,4 2,5 3,6 7,10 1.1333 0.040810,2 6,11 12,8 1,9 1.1482 0.04107,2 4,11 10,3 12,5 1.1472 0.04099,7 6,1 2,10 8,12 1.1700 0.04126,8 4,1 12,3 2,5 1.1350 0.04083,6 5,12 1,9 4,7 1.1273 0.04074,8 10,12 9,5 2,11 1.1428 0.040910,3 4,2 7,9 6,8 1.1637 0.04122,7 11,6 5,9 8,4 1.1249 0.04071,10 3,6 9,5 4,11 1.1455 0.04095,10 3,12 4,9 6,11 1.2860 0.042611,7 12,9 10,1 8,6 1.1544 0.041111,6 3,7 8,1 5,2 1.1469 0.04106,11 2,9 7,3 8,4 1.1526 0.04105,3 4,1 12,8 2,10 1.1305 0.04089,6 8,3 10,12 2,11 1.1407 0.04106,11 2,5 12,9 8,10 1.1383 0.0409




n

5,9 11,4 7,12 8,10 1.1505 0.04094,9 2,11 8,5 10,7 1.1404 0.04096,8 7,5 12,10 11,9 1.1490 0.04093,8 1,5 9,12 10,6 1.1222 0.04087,4 6,9 11,1 2,10 1.1211 0.04079,6 12,10 2,5 8,1 1.1339 0.04093,11 1,9 7,2 8,6 1.1640 0.041312,8 6,11 7,9 5,3 1.1345 0.040910,1 5,3 2,4 8,11 1.1434 0.041010,2 4,7 12,9 11,8 1.1347 0.040910,5 7,9 11,4 8,12 1.1584 0.04099,6 10,5 1,8 11,3 1.1360 0.04101,4 7,11 6,9 10,2 1.1239 0.04076,4 5,8 3,11 12,10 1.1439 0.04097,4 1,3 2,6 11,8 1.1279 0.04081,3 6,2 9,5 10,8 1.1325 0.040912,2 9,7 6,11 5,3 1.1300 0.04099,12 4,2 11,8 3,10 1.1707 0.04136,10 9,2 8,12 11,7 1.1460 0.04107,11 1,5 9,2 8,4 1.1363 0.040810,1 5,9 11,3 2,4 1.1581 0.04121,3 5,12 9,6 8,10 1.1246 0.04088,12 3,10 9,6 5,2 1.1302 0.04092,4 3,8 1,11 6,10 1.1290 0.04088,10 2,12 11,1 6,4 1.1326 0.04081,9 7,3 6,10 5,12 1.1351 0.04093,7 9,12 6,2 1,8 1.3101 0.043012,4 1,8 3,10 7,2 1.1948 0.041411,1 2,10 12,7 5,3 1.1337 0.040910,8 1,11 7,5 2,9 1.1464 0.04109,6 5,12 2,4 11,3 1.1410 0.04096,8 11,9 7,4 5,12 1.1329 0.040711,6 1,8 3,7 9,12 1.1558 0.041212,9 7,4 11,6 10,2 1.1234 0.04078,12 1,4 10,6 9,5 1.1247 0.040711,8 6,2 1,9 4,7 1.1246 0.040712,3 10,8 5,7 6,9 1.1245 0.04081,9 8,3 5,7 2,6 1.1603 0.0412




n

11,4 9,5 12,2 10,7 1.1264 0.040712,10 9,6 4,7 3,11 1.1398 0.04097,12 3,5 9,1 6,10 1.1240 0.04084,2 6,8 10,5 11,9 1.1423 0.04084,9 8,3 12,7 11,6 1.1355 0.04086,4 1,10 2,5 11,9 1.1273 0.04073,12 8,4 11,1 9,2 1.1538 0.04123,7 8,4 10,5 1,9 1.1545 0.041012,2 11,6 10,3 1,8 1.1503 0.041111,9 3,7 12,8 10,1 1.1482 0.04116,1 12,3 5,10 9,7 1.1321 0.04099,7 5,1 3,10 2,11 1.1297 0.04099,12 5,3 7,11 8,10 1.1283 0.040810,6 4,9 12,8 11,1 1.1264 0.040812,4 2,5 9,1 11,8 1.1313 0.040812,5 2,10 11,7 8,1 1.1529 0.041010,8 6,4 3,1 9,11 1.1253 0.040810,3 12,2 8,5 7,9 1.1243 0.04089,5 11,1 2,6 12,10 1.1384 0.040912,3 10,8 9,5 4,1 1.1319 0.04086,8 4,7 3,11 5,12 1.1284 0.040711,8 9,7 2,5 10,3 1.1429 0.04104,12 10,3 1,6 5,2 1.1396 0.04097,11 12,8 2,6 1,9 1.1722 0.041310,7 11,1 6,2 4,12 1.1437 0.040911,3 8,12 2,9 1,6 1.1616 0.04136,11 2,10 12,4 3,7 1.1345 0.040812,8 2,4 10,6 3,11 1.1322 0.04094,11 3,10 1,8 7,5 1.1334 0.04083,10 2,6 1,11 9,12 1.1419 0.04107,3 9,1 11,2 5,12 1.1353 0.04094,12 5,3 8,1 2,11 1.1373 0.040911,1 5,10 8,6 9,2 1.1347 0.04098,10 12,2 5,9 11,4 1.1466 0.04092,9 3,11 10,6 1,4 1.1345 0.041011,8 4,6 12,7 9,1 1.1252 0.040710,8 9,4 7,11 1,6 1.1293 0.040711,1 7,9 6,4 8,12 1.1253 0.0407




n

5,9 3,12 6,8 7,2 1.1467 0.04114,6 10,8 7,5 9,1 1.1276 0.04073,6 11,4 8,1 2,10 1.1284 0.04085,9 1,6 12,4 2,7 1.1255 0.04077,11 3,1 8,4 10,12 1.1347 0.040911,7 9,12 2,10 3,6 1.1312 0.0409

Table B.12: k = 23

�(19) �(20) �(21) �(22) �(23) Exp. S �pn

1,4 2,5 3,6 7,10 8,11 1.0401 0.03919,11 12,7 2,4 10,3 5,8 1.0720 0.03959,7 6,11 8,10 3,1 12,2 1.0590 0.03935,7 6,8 4,1 12,3 2,9 1.0586 0.03944,8 10,12 9,5 2,11 1,6 1.0470 0.03914,6 8,1 11,7 5,3 2,12 1.0563 0.03933,1 6,11 2,4 12,7 10,8 1.0493 0.03927,2 10,12 1,9 11,8 3,5 1.0440 0.03919,12 4,11 1,5 8,3 6,2 1.0442 0.03918,1 5,9 12,10 6,11 3,7 1.0421 0.039112,7 4,6 1,11 10,5 8,3 1.0658 0.03943,11 1,9 7,2 8,6 10,12 1.0590 0.039312,7 4,2 8,10 1,3 6,11 1.0496 0.03927,9 5,3 11,2 6,4 10,1 1.0429 0.03913,6 9,1 12,10 7,11 5,2 1.0424 0.03911,4 7,11 6,9 10,2 5,8 1.0427 0.03917,4 1,3 2,6 11,8 9,5 1.0386 0.039112,2 6,11 1,9 4,8 3,10 1.0517 0.03927,11 1,5 9,2 8,4 3,10 1.0860 0.03958,12 3,10 9,6 5,2 11,7 1.0436 0.039210,8 2,12 11,1 6,4 3,7 1.0464 0.03925,12 4,7 1,8 3,10 2,11 1.0474 0.03914,7 11,1 2,10 12,5 3,8 1.0493 0.039212,4 10,8 1,6 9,7 11,2 1.0482 0.039211,3 7,2 1,6 5,8 4,9 1.0455 0.03924,11 10,12 3,7 8,5 9,2 1.0467 0.0392




n

3,12 7,4 11,6 1,10 2,5 1.0561 0.039210,5 1,9 12,2 11,6 3,8 1.0453 0.03912,4 9,7 12,5 3,11 8,10 1.0510 0.039210,6 4,9 12,8 11,1 2,7 1.0508 0.03929,12 7,5 8,6 11,3 4,1 1.0377 0.03914,2 5,9 1,11 8,10 12,3 1.0613 0.03934,12 7,9 11,3 2,6 10,1 1.0472 0.03927,4 6,9 3,8 11,2 12,10 1.0467 0.039310,8 6,4 3,1 9,11 12,2 1.0488 0.03925,7 10,1 9,6 3,12 4,2 1.0477 0.03923,8 9,5 10,6 11,7 12,2 1.0432 0.03925,3 10,2 6,1 11,9 12,8 1.0437 0.03913,1 11,2 5,7 12,4 8,10 1.0455 0.039111,1 5,10 8,6 9,2 12,4 1.0475 0.03922,9 3,11 10,6 1,4 8,12 1.0478 0.03924,7 1,3 12,10 9,2 11,8 1.0610 0.03948,12 7,9 2,5 4,11 6,10 1.0461 0.039212,4 7,5 1,10 3,8 9,11 1.0444 0.03916,9 1,8 7,10 11,2 3,5 1.0402 0.039112,4 5,2 6,9 7,10 11,8 1.0461 0.03923,7 8,1 6,11 2,4 12,9 1.0500 0.03926,4 9,2 10,5 8,12 3,7 1.0513 0.03934,6 12,3 1,11 9,7 2,5 1.0416 0.039110,3 1,8 9,2 11,6 7,12 1.0636 0.03937,3 6,9 8,11 5,2 1,10 1.0405 0.039110,7 8,3 11,4 9,5 6,2 1.0580 0.03941,8 2,9 7,10 4,11 12,3 1.0616 0.03943,1 10,5 4,8 11,9 12,2 1.0613 0.03937,10 8,12 1,6 4,2 5,3 1.0475 0.03929,4 6,1 5,10 8,11 2,7 1.0461 0.03927,3 10,6 9,1 8,11 4,12 1.0441 0.03921,5 9,2 4,8 6,3 7,10 1.0580 0.03946,9 5,1 8,12 2,7 11,4 1.0443 0.03918,4 11,2 9,6 12,5 10,1 1.0449 0.039110,12 4,9 8,5 6,1 7,11 1.0570 0.03934,12 5,8 1,3 9,7 11,6 1.0396 0.03917,4 6,9 11,2 8,1 5,3 1.0392 0.03917,11 3,8 4,1 9,12 10,6 1.0449 0.0392




n

10,5 8,12 7,3 9,11 6,4 1.0526 0.03937,12 2,6 4,8 1,3 5,9 1.0618 0.03948,3 1,5 6,11 2,12 4,7 1.0573 0.03938,6 7,11 5,3 4,1 2,12 1.0575 0.03938,6 11,4 10,12 3,7 1,5 1.0438 0.039111,6 7,4 12,8 9,5 10,1 1.0537 0.03933,8 9,11 1,6 5,10 7,4 1.0441 0.039211,6 7,10 8,3 2,9 4,1 1.0805 0.03969,12 3,7 10,8 1,6 5,2 1.0479 0.03925,7 3,12 4,9 8,10 11,1 1.0454 0.039112,2 1,3 5,10 6,4 11,7 1.0584 0.03924,6 3,12 8,1 11,9 10,7 1.0441 0.03923,8 2,6 7,11 12,9 1,5 1.0547 0.039312,10 1,8 4,11 6,2 3,5 1.0462 0.03919,11 7,12 6,4 2,10 8,5 1.0456 0.03927,4 9,5 12,2 6,11 8,3 1.0412 0.03928,3 2,11 6,4 10,12 9,5 1.0588 0.039411,2 1,5 8,12 4,7 10,6 1.0537 0.03925,3 10,7 9,2 4,12 8,11 1.0471 0.03923,10 9,11 8,6 7,12 5,2 1.0485 0.03924,2 1,10 7,12 9,6 8,3 1.0685 0.03952,9 5,7 8,10 4,1 11,6 1.0486 0.03922,12 6,9 4,1 8,5 11,7 1.0427 0.039111,9 2,12 1,6 8,3 7,10 1.0592 0.03937,9 1,3 4,11 8,12 2,10 1.0629 0.03949,7 3,6 10,5 8,1 11,2 1.0431 0.03915,8 12,2 9,1 10,6 11,4 1.0467 0.03914,8 9,12 10,5 3,1 6,11 1.0461 0.03921,8 7,11 4,2 9,6 5,3 1.0397 0.03913,1 9,7 10,5 2,6 12,8 1.0740 0.03957,11 3,10 5,12 9,1 2,6 1.0457 0.03911,4 6,9 11,7 12,8 3,10 1.0440 0.03926,8 4,2 1,3 7,11 12,9 1.0453 0.03922,10 12,8 4,11 1,9 6,3 1.0490 0.03927,3 10,1 12,2 11,8 5,9 1.0426 0.0391


Table B.13: k = 24

�(19) �(20) �(21) �(22) �(23) �(24) Exp. S �pn

1,4 2,5 3,6 7,10 8,11 9,12 0.9725 0.03729,7 6,1 2,10 8,12 3,5 4,11 0.9736 0.03727,12 8,3 9,4 1,11 5,10 6,2 0.9744 0.03728,1 5,9 12,10 6,11 3,7 2,4 0.9736 0.03721,4 7,11 6,9 10,2 5,8 3,12 0.9740 0.03722,6 7,10 9,11 1,3 12,5 8,4 0.9724 0.037210,5 4,12 9,2 8,1 6,11 7,3 0.9836 0.03726,1 11,8 3,7 9,5 12,4 10,2 0.9734 0.037210,5 1,9 12,2 11,6 3,8 7,4 0.9739 0.03723,8 7,10 2,5 11,4 1,6 9,12 0.9735 0.03722,4 9,7 12,5 3,11 8,10 6,1 0.9737 0.03727,4 6,9 3,8 11,2 12,10 5,1 0.9728 0.03725,7 10,1 9,6 3,12 4,2 11,8 0.9730 0.03723,8 9,5 10,6 11,7 12,2 1,4 0.9738 0.03725,9 3,12 6,8 7,2 4,11 10,1 0.9734 0.03723,7 8,1 6,11 2,4 12,9 10,5 0.9746 0.03728,6 2,10 9,4 12,3 5,7 1,11 0.9730 0.03729,7 2,12 5,8 1,3 11,4 10,6 0.9733 0.037211,8 1,3 5,10 7,4 9,12 2,6 0.9748 0.03723,12 1,9 11,7 2,4 5,8 6,10 0.9737 0.037211,2 12,4 10,8 6,3 9,1 7,5 0.9734 0.03725,12 1,10 8,4 11,2 9,6 7,3 0.9738 0.03724,12 5,8 1,3 9,7 11,6 10,2 0.9735 0.03727,11 3,8 4,1 9,12 10,6 2,5 0.9739 0.037212,2 8,5 9,7 4,6 11,3 1,10 0.9745 0.037210,7 8,11 1,3 5,9 2,6 12,4 0.9729 0.03729,1 7,4 5,12 10,8 2,11 6,3 0.9735 0.03723,5 7,9 2,4 6,11 8,12 10,1 0.9722 0.03727,3 6,9 12,10 1,8 4,11 2,5 0.9733 0.03722,7 1,6 12,9 11,4 3,10 5,8 0.9743 0.03723,12 1,4 8,5 2,10 9,6 7,11 0.9728 0.03729,6 8,1 3,12 7,4 11,2 10,5 0.9739 0.03727,9 3,8 1,11 10,5 2,12 6,4 0.9738 0.03724,8 9,12 10,5 3,1 6,11 2,7 0.9754 0.03724,6 5,1 2,9 3,7 10,12 8,11 0.9733 0.03724,8 5,2 12,7 3,11 9,6 10,1 0.9728 0.03723,8 9,2 10,12 4,11 1,6 5,7 0.9824 0.0373




n

8,10 3,7 5,12 11,6 2,4 9,1 0.9776 0.03727,12 3,8 9,5 6,2 11,4 10,1 0.9730 0.03721,3 6,4 9,12 2,7 10,5 11,8 0.9731 0.037210,6 5,9 2,7 1,11 3,12 4,8 0.9728 0.037211,1 12,9 6,3 10,7 8,5 4,2 0.9730 0.03729,11 4,1 10,6 3,7 12,8 5,2 0.9740 0.03729,2 11,7 1,4 6,8 12,5 10,3 0.9740 0.03724,12 10,6 8,1 11,9 3,7 5,2 0.9737 0.03725,3 2,6 4,7 1,10 9,12 8,11 0.9723 0.03724,11 10,7 2,9 6,3 8,1 12,5 0.9738 0.03725,10 6,2 9,4 12,8 7,3 11,1 0.9749 0.03726,8 2,12 1,4 3,10 11,7 9,5 0.9737 0.03728,5 4,6 2,11 3,1 7,10 9,12 0.9732 0.03729,1 10,8 7,4 3,12 2,5 11,6 0.9736 0.03728,4 2,5 3,7 1,6 9,11 10,12 0.9726 0.03723,7 5,12 2,11 10,1 4,9 8,6 0.9728 0.03723,7 9,12 1,10 11,6 5,8 4,2 0.9729 0.03727,11 1,4 9,6 8,3 12,10 2,5 0.9740 0.037212,3 4,7 8,5 9,2 1,11 10,6 0.9736 0.03729,1 12,4 6,11 7,3 2,5 10,8 0.9731 0.037211,7 1,10 12,3 6,2 5,9 8,4 0.9728 0.037212,8 10,5 2,6 11,9 3,7 4,1 0.9731 0.03729,5 8,4 11,7 1,10 3,6 2,12 0.9745 0.03722,4 12,3 7,11 1,9 5,8 10,6 0.9737 0.037212,2 3,10 1,6 8,5 9,11 4,7 0.9729 0.03726,8 1,11 7,3 2,10 9,4 5,12 0.9742 0.03726,10 8,1 3,5 4,12 11,9 2,7 0.9785 0.03729,5 8,6 12,2 3,7 10,1 4,11 0.9740 0.03727,12 3,10 11,8 4,1 2,6 9,5 0.9727 0.03727,10 2,11 4,9 6,8 3,12 1,5 0.9731 0.037211,3 7,12 8,1 5,9 10,2 6,4 0.9729 0.03729,2 11,7 3,1 4,12 10,5 8,6 0.9741 0.03726,9 7,4 11,2 8,5 12,10 1,3 0.9726 0.037210,8 7,4 9,11 1,3 12,5 2,6 0.9740 0.03721,10 3,7 8,6 12,5 9,4 11,2 0.9731 0.03727,4 2,11 3,6 5,10 1,8 12,9 0.9733 0.03724,12 3,1 2,5 11,8 9,7 6,10 0.9739 0.03723,1 12,5 6,8 11,4 9,7 2,10 0.9738 0.0372




n

1,9 7,11 12,10 4,2 6,8 5,3 0.9745 0.037211,6 4,7 2,9 3,10 1,5 8,12 0.9741 0.03723,6 8,4 7,5 12,9 1,10 11,2 0.9740 0.03729,5 8,1 2,12 11,6 4,7 10,3 0.9739 0.037212,8 9,11 5,1 3,6 10,7 2,4 0.9739 0.037211,2 6,10 3,8 4,9 12,7 5,1 0.9745 0.03724,1 11,2 7,12 3,10 8,6 5,9 0.9734 0.03722,11 6,4 7,12 8,1 9,5 3,10 0.9738 0.03729,4 7,3 5,2 1,11 8,12 6,10 0.9728 0.03726,1 3,11 8,10 5,9 2,7 12,4 0.9730 0.037212,8 11,4 5,3 1,9 2,7 10,6 0.9747 0.03727,10 1,3 6,11 4,2 9,12 5,8 0.9737 0.037211,2 7,5 4,9 12,10 8,6 3,1 0.9738 0.03724,9 6,11 3,10 8,12 5,1 2,7 0.9820 0.03734,7 6,9 2,12 5,10 1,8 3,11 0.9739 0.03729,12 11,2 5,3 4,1 8,6 10,7 0.9726 0.03722,7 5,12 6,10 9,11 4,1 8,3 0.9739 0.03725,2 3,10 1,8 4,7 12,9 11,6 0.9742 0.03721,10 11,3 6,2 5,8 7,4 9,12 0.9730 0.03725,1 6,3 9,4 10,12 8,11 2,7 0.9730 0.03724,7 2,12 5,3 1,11 8,10 9,6 0.9729 0.037212,7 9,4 5,1 8,10 11,3 6,2 0.9752 0.03724,7 3,8 1,6 2,5 11,9 12,10 0.9730 0.03725,7 6,11 2,9 1,4 12,10 3,8 0.9813 0.0373

The expected total shortfall of a total-flexibility structure is provided in Table B.14 as abenchmark.

Table B.14: Expected Total Shortfall of the Total-Flexibility Structure as a Function of k

k Expected Total Shortfall13 2.424814 2.194315 1.993716 1.817817 1.662618 1.5248



Table B.14 – continued from previous pagek Expected Total Shortfall19 1.40220 1.29221 1.193122 1.103923 1.023124 0.9498

120

Bibliography

Aksin, O. and F. Karaesmen (2007). “Characterizing the performance of process flexibilitystructures”. In: Operations Research Letters 35.4, pp. 477–484.

Bassamboo, A., L.Y. Chu, and R.S. Randhawa (2011). Note: Designing Flexible SystemsUsing a New Notion of Submodularity. Working paper.

Bassamboo, A., R.S. Randhawa, and J. A. Van Mieghem (2010). “Optimal Flexibility Config-urations in Newsvendor Networks: Going beyond Chaining and Pairing”. In:ManagementSci. 56.8, pp. 1285–1303.

Bertsimas, D., K. Natarajan, and C. Teo (2004). “Probabilistic combinatorial optimization:Moments, semidefinite programming, and asymptotic bounds”. In: SIAM Journal onOptimization 15.185.

— (2006). “Persistence in discrete optimization under data uncertainty”. In: Mathematicalprogramming 108.2.

Bertsimas, D. et al. (2010). “Models for minimax stochastic linear optimization problemswith risk aversion”. In: Mathematics of Operations Research 35.3.

Bish, E., A. Muriel, and S. Biller (2005). “Managing flexible capacity in a make-to-orderenvironment”. In: Management Sci. 51.2, pp. 167–180.

Bish, E. and Q. Wang (2004). “Optimal Investment Strategies for Flexible Resources, Con-sidering Pricing and Correlated Demand”. In: Management Sci. 52.6, pp. 954–964.

Chen, L., S. He, and S. Zhang. (2011). “Tight bounds for some risk measures, with applica-tions to robust portfolio selection”. In: Oper. Res. 59.4, pp. 847–865.

Chou, M.C., G. A. Chua, and C.P. Teo (2010). “On range and response: Dimensions ofprocess flexibility”. In: Eur. J. Oper. Res. 207.2, pp. 711–724.

Chou, M.C., C.P. Teo, and H. Zheng (2008). “Process Flexibility: Design, Evaluation andApplications”. In: Flexible Services and Manufacturing Journal 20.1-2, pp. 59–94.

Chou, M.C. et al. (2010). “Design for process flexibility: e�ciency of the long chain andsparse structure”. In: Oper. Res. 58.1, pp. 43–58.

Chou, M.C. et al. (2011). Process Flexibility Revisited: The Graph Expander and Its Appli-cations. Oper. Res. forthcoming.

CVX Research, Inc. (Sept. 2012). CVX: Matlab Software for Disciplined Convex Program-ming, version 2.0 beta. http://cvxr.com/cvx.

Delage, E. and Y. Ye (2010). “Distributionally robust optimization under moment uncer-tainty with application to data-driven problems”. In: Oper. Res. 58.3, pp. 595–612.

BIBLIOGRAPHY 121

Deng, T. and Z.-J. Shen (2012). Process Flexibility Design in Unbalanced Networks. Manu-facturing Service Oper. Management, forthcoming.

Fine, C. and R. Freund (1990). “Optimal Investment in Product-Flexible ManufacturingCapacity”. In: Management Sci. 36.4, pp. 449–466.

Goh, J. and M. Sim (2010). “Distributionally robust optimization and its tractable approx-imations”. In: Oper. Res. 58.4, pp. 902–917.

Grant, M. and S. Boyd (2008). “Graph implementations for nonsmooth convex programs”. In:Recent Advances in Learning and Control. Ed. by V. Blondel, S. Boyd, and H. Kimura.Lecture Notes in Control and Information Sciences. http://stanford.edu/

~

boyd/

graph_dcp.html. Springer-Verlag Limited, pp. 95–110.Graves, S.C. and B.T. Tomlin (2003). “Process Flexibility in Supply Chains”. In: Manage-

ment Sci. 49.7, pp. 907–919.Hopp, W.J., S.M.R. Iravani, and W.L. Xu (2010). “Vertical Flexibility in Supply Chains”.

In: Management Sci. 56.3, pp. 495–502.Hopp, W.J., E. Tekin, and M. P. Van Oyen (2004). “Benefits of skill chaining in serial

production lines with cross-trained workers”. In: Management Sci. 50.1, pp. 83–98.Iravani, S.M.R., B. Kolfal, and M.P. Van Oyen (2007). “Call-Center Labor Cross-Training:

It’s a Small World After All”. In: Management Sci. 53.7, pp. 1102–1112.Iravani, S.M.R., M.P. Van Oyen, and K.T. Sims (2005). “Structural Flexibility: A New

Perspective on the Design of Manufacturing and Service Operations”. In: ManagementSci. 51.2, pp. 155–166.

Jordan, W.C. and S.C. Graves (1995). “Principles on the Benefits of Manufacturing ProcessFlexibility”. In: Management Sci. 41.4, pp. 577–594.

Kent, D. (2009). Alabama’s auto plants use flexibility to deal with industry downturn. TheBirmingham News.

Kong, Q. et al. (2012). Scheduling Arrivals to a Stochastic Service Delivery System usingCopositive Cones. workinig paper, National University of Singapore, Singapore.

LeBeau, P. (2008). Toyota Shows Flexibility With Prius/Highlander/Tundra Moves. CNBCNews.

— (2010). Ford’s Flexibility Is the Story to Watch. CNBC News.Lim, M. et al. (2011). Flexibility and Fragility: Supply Chain Network Design with Disruption

Risks. Working paper, UIUC.Mak, H., Y. Rong, and J. Zhang (2012). Appointment Scheduling with Limited Distributional

Information. Working paper, Hong Kong University of Science and Technology, HongKong.

Meilijson, I. and A. Nadas (1979). “Convex Majorization with An Application to the Lengthof Critical Paths”. In: Journal of Applied Probability 16.3, pp. 671–693.

Natarajan, K., D. Shi, and K. Toh (2012). A Probabilistic Model for Minimax Regret inCombinatorial Optimization. Working paper, Singapore University of Technology andDesign.

Natarajan, K., M. Song, and C. Teo (2009). “Persistency Model and Its Applications inChoice Modeling”. In: Management Sci. 55.3, pp. 453–469.

BIBLIOGRAPHY 122

Natarajan, K., C.P. Teo, and Z. Zheng (2011). “Mixed 0-1 Linear Programs Under ObjectiveUncertainty: A Completely Positive Representation”. In: Oper. Res. 59.3, pp. 713–728.

Sheikhzadeh, M., S. Benjaafar, and D. Gupta (1998). “Machine sharing in manufacturingsystems: Total Flexibility versus chaining”. In: Internat. J. Flexible Manufacturing Sys-tems 10.4, pp. 351–378.

Simchi-Levi, David and Yehua Wei (2011). Understanding the performance of the long chainand sparse designs in process flexibility. Working Paper, Operations Research Center,Massachusetts Institute of Technology.

van Mieghem, J.A. (1998). “Investment Strategies for flexible resources”. In: ManagementSci. 44.8, pp. 1071–1078.

process flexibility design in unbalanced and...

Documents