Essays on Inventory Management, Capacity Management, and Resource-Sharing Systems Ph.D

ESSAYS ON INVENTORY MANAGEMENT, CAPACITY MANAGEMENT,

AND RESOURCE-SHARING SYSTEMS

Yang Bo

APPROVED BY SUPERVISORY COMMITTEE:

Milind Dawande, Co-Chair

Ganesh Janakiraman, Co-Chair

Alp Muharremoglu

Anyan Qi Copyright c 2017

Yang Bo

AND RESOURCE-SHARING SYSTEMS

YANG BO, BS

DISSERTATION

Presented to the Faculty of

The University of Texas at Dallas

in Partial Fulﬁllment

of the Requirements

for the Degree of

DOCTOR OF PHILOSOPHY IN

MANAGEMENT SCIENCE

THE UNIVERSITY OF TEXAS AT DALLAS

August 2017 ACKNOWLEDGMENTS

I am deeply thankful to my advisors, Drs. Milind Dawande and Ganesh Janakiraman, for patiently guiding me in my research endeavors, and for providing me with constructive advice and feedback. My frequent beneﬁcial interactions with them have sharpened my research ability and have enabled me to become an independent researcher.

I owe much to my parents and my wife; their unconditional love and support continues to give me courage and make me stronger.

I would also like to express my gratitude to the following professors in my department, from whom I have beneﬁted a lot: Alain Bensoussan, Metin Cakanyildirim, Dorothee Honhon, Elena Katok, Alp Muharremoglu, Shun-Chen Niu, Ozalp Ozer, Anyan Qi, Suresh P. Sethi, Kathryn Stecke, Serdar Simek, Shouqiang Wang, John J. Wiorkowski, and Shengqi Ye.

Last but not the least, I would like to thank my fellow PhD students: Ying Cao, Bahriye Cesaret, Jiayu Chen, Shaokuan Chen, Wei Chen, Ilhan Emre Ertan, Zhichao Feng, Blair Flicker, Shivam Gupta, Harish Guda, Bharadwaj Kadiyala, Ismail Kirci, Chungseung Lee, Jingyun Li, Ting Luo, Sandun Perera, Xi Shan, Sina Shokoohyar, Yulia Vorotyntseva, Xiao Zhang, and many others. Many thanks for making my stay at UTD joyful.

April 2017

v ESSAYS ON INVENTORY MANAGEMENT, CAPACITY MANAGEMENT,

AND RESOURCE-SHARING SYSTEMS

Yang Bo, PhD The University of Texas at Dallas, 2017

Supervising Professors: Milind Dawande, Co-Chair Ganesh Janakiraman, Co-Chair

This dissertation consists of three essays, each focusing on one of the following three important topics in operations management: inventory management, capacity management, and management of resource-sharing systems. These topics are each summarized below.

In the first essay, we investigate the following integrality question for inventory control on distribution systems: Given integral demands, does an integral optimal policy exist? We show that integrality holds under deterministic demand, but fails to hold under stochastic demand. In distribution systems with stochastic demand, we identify three factors that influence the gap between integral and real optimal policies: shipping cost variation across time, holding cost difference across stages, and economies of scale. We then obtain a tight worst-case bound for the gap, which captures the impact of all these factors.

In the second essay, we investigate the computational complexity of determining the capacity of a process, a fundamental concept in Operations Management. We show that it is hard to calculate process capacity exactly and, furthermore, also hard to eﬃciently approximate it to within a reasonable factor; e.g., within any constant factor. These results are based on a novel characterization, which we establish, of process capacity that relates it to the

vi fractional chromatic number of an associated graph. We also show that capacity can be eﬃciently computed for processes for which the collaboration graph is a perfect graph.

The third essay addresses an important problem in resource-sharing systems. We study the minimum-scrip rule in such a system: whenever a service request arises, among those who volunteer and are able to provide service, the one with the least number of scrips (also known as coupons) is selected to provide service. Under mild assumptions, we show that everybody in the system being always willing to provide service is a Nash Equilibrium under the minimum-scrip rule. This suggests that the minimum scrip rule can lead to a high level of social welfare.

vii TABLE OF CONTENTS

ACKNOWLEDGMENTS ...... v ABSTRACT ...... vi LIST OF FIGURES ...... xi CHAPTER 1 INTRODUCTION ...... 1 CHAPTER 2 ON INTEGRAL POLICIES IN DETERMINISTIC AND STOCHASTIC DISTRIBUTION SYSTEMS ...... 4 2.1 Introduction ...... 5 2.2 Preliminaries ...... 8 2.2.1 Network Representation and Notation ...... 8 2.2.2 Economies of Scale Structure ...... 10 2.3 Distribution Systems with Deterministic Demand ...... 10 2.4 Distribution Systems with Stochastic Demand ...... 13 2.4.1 Failure of Integrality When Assumption S Does Not Hold ...... 14 2.4.2 Failure of Integrality When Assumption H Does Not Hold ...... 16 2.4.3 Bounding the Integrality Gap ...... 17 2.5 Managerial Implications of Results ...... 22 2.5.1 Unpredictable Demand: Variation in Shipping Costs Across Time . . 22 2.5.2 Unpredictable Demand: Holding Cost Variation Across Stages . . . . 23 2.5.3 Unpredictable Demand: Economies of Scale ...... 24 2.5.4 Predictable Demand ...... 24 2.6 Proofs of Results ...... 24 2.6.1 Proof of Theorem 2.3.1 ...... 24 2.6.2 Proof of Proposition 2.4.1 ...... 26 2.6.3 Proof of Proposition 2.4.2 ...... 29 2.6.4 Proof of Theorem 2.4.1 ...... 31 CHAPTER 3 ON FINDING PROCESS CAPACITY ...... 40 3.1 Introduction ...... 40 3.1.1 A Brief Overview of the Related Literature ...... 43

viii 3.2 Preliminary Results ...... 44 3.2.1 Formal Deﬁnition of Process Capacity ...... 45 3.2.2 Suﬃciency of Cyclic Policies and Redundance of Precedence and Non- Preemption Constraints ...... 47 3.2.3 Static Planning Problem for Collaboration ...... 50 3.3 Characterizing Process Rates for Subclasses of Cyclic Policies ...... 52 3.3.1 The Maximum Process Rate Over 1-Unit Cyclic Policies ...... 52 3.3.2 The Maximum Process Rate Over k-Unit Cyclic Policies ...... 54 3.3.3 The Maximum Process Rate Over Cyclic Policies (Process Capacity) 55 3.4 Hardness and Inapproximability Results ...... 57 3.4.1 Inapproximability of Determining Process Capacity ...... 57 3.5 Good News ...... 58 3.6 Extension: Relaxation of Assumption (3.2.14) ...... 61 3.7 Proofs of Some Results ...... 63 3.7.1 Proof of Lemma 3.2.1 ...... 63 3.7.2 Proof of Lemma 3.3.2 ...... 65 3.7.3 Proof of Theorem 3.4.1 ...... 66 3.7.4 Proof of Theorem 3.5.3 ...... 66 3.7.5 Proof of Lemma 3.6.2 ...... 71 CHAPTER 4 ANALYSIS OF SCRIP SYSTEMS ...... 76 4.1 Introduction ...... 76 4.2 Model ...... 78 4.3 Analysis and Results ...... 79 4.3.1 The Main Result ...... 80 4.3.2 Intermediate Results ...... 82 4.3.3 Proof of Theorem 4.3.1 ...... 83 4.4 Proofs of the Intermediate Results – Lemmas 4.3.2 – 4.3.4 ...... 84 4.4.1 Proof of Lemma 4.3.2 ...... 84 4.4.2 Proof of Lemma 4.3.3 ...... 86

ix 4.4.3 Proof of Lemma 4.3.4 ...... 87 APPENDIX PROOFS OF LEMMA 3.2.2 ...... 95 REFERENCES ...... 104 BIOGRAPHICAL SKETCH ...... 108 CURRICULUM VITAE

x LIST OF FIGURES

2.1 (a) A Distribution System and (b) An Assembly System...... 9 2.2 The Distribution Network in the Proofs of Propositions 2.4.1 and 2.4.2...... 15

A.1 Policy π1 in Example 1 ...... 97 A.2 Policyπ ˆ in Example 1 ...... 98

A.3 Policy π0 in Example 2 ...... 101 A.4 Policyπ ˇ in Example 2 ...... 102

xi CHAPTER 1

INTRODUCTION

This dissertation includes three essays. The first two essays focus on inventory and capacity management, and the third essay addresses an important problem in resource-sharing systems. We now briefly describe the problems analyzed in these essays and summarize our contributions. In the first essay, we study the “integrality” question for dynamic optimization models of inventory control: Given integral initial inventory levels, capacity constraints, and demand realizations, does there exist an integral optimal policy? One practical implication of this question lies in whether or not full-truckload (FTL) shipping is optimal if customer demand is in integral number of truckloads. In this essay, we investigate the integrality question in single-product, multi-echelon distribution systems and show that integrality holds under deterministic demand, but fails to hold under stochastic demand. In distribution systems with stochastic demand, Less-Than-Truckload (LTL) shipping can be significantly cheaper than the cost of the optimal FTL shipping policy, even in the presence of economies of scale. For instance, this occurs in settings where shipping costs are expected to increase in the future and/or inventories are more expensive to hold upstream than downstream. In such situations, our results highlight the importance of strategically positioning inventory: LTL shipments can offer a more balanced allocation of inventory across the distribution network, leading to benefits that can exceed the savings from FTL shipments due to economies of scale. However, when the cost parameters are fairly constant across time and inventory holding costs are not significantly higher upstream than downstream, then the difference between the costs of optimal FTL and optimal LTL shipping is provably marginal. The second essay investigates the analytical tractability of process capacity. The concept of the capacity of a process and the associated managerial insights on the investment and management of capacity are of fundamental importance in Operations Management. Most

1 OM textbooks use the following simple approximation to determine process capacity: the

capacity of each resource is ﬁrst calculated by examining that resource in isolation; process

capacity is then deﬁned as the smallest (bottleneck) among the capacities of the resources.

In a recent paper, (Gurvich and Van Mieghem, 2015) show that this “bottleneck formula”

can be signiﬁcantly inaccurate, and obtain a necessary and suﬃcient condition under which

it correctly determines process capacity. We provide further clarity on the intractability

of process capacity by showing that it is hard to calculate process capacity exactly and,

furthermore, also hard to eﬃciently approximate it to within a reasonable factor; e.g., within

any constant factor. These results are based on a novel characterization, which we establish,

of process capacity that relates it to the fractional chromatic number of an associated graph.

An important implication is that it is unlikely that we can replace the bottleneck formula

with a simple but close approximation of process capacity. From a practical viewpoint, our

analysis results in a natural hierarchy of subclasses of policies that require an increasing

amount of sophistication in implementation and management: While process capacity is the

maximum long-term process rate achievable over all feasible policies, we provide a precise

expression for the maximum process rate over policies in each subclass in this hierarchy, thus

highlighting the tradeoﬀ between operational diﬃculty and the achievable process rate.

The third essay addresses an important problem in resource-sharing systems. In a recent

paper, (Johnson et al., 2014) use an inﬁnitely-repeated game with discounting, among a

set of homogeneous players, to model a scrip system. In each period, a randomly-chosen

player requests service; all the other players have a choice of whether or not to volunteer

to provide service. Among the players who volunteer, the service provider is chosen using

the minimum-scrip rule: A player with the minimum number of scrips is chosen as the service provider. The authors study the always-trade strategy for a player, i.e., the strategy

of a player always willing to provide service, regardless of the distribution of scrips among

the players. A key result of their work is that, under the minimum-scrip rule, for any

2 number of players and for every discount factor close enough to one, there exists a Nash equilibrium in which each player plays the always-trade strategy. This result, however, is established under an assumption of severe punishment: If a player selected to provide service refuses to do so even once, then that player will be forever banned from participating in the system, thus losing all potential future beneﬁt. (Johnson et al., 2014) explain that this assumption is unsatisfactory, particularly for large scrip systems, due to diﬃculties in detecting players who refuse to provide service and verifying the reason(s) for their refusal, and the consequent possibility of players being unfairly banned. Motivated by this concern, the authors suggest an important direction of future research – investigate whether or not the always-trade strategy is an equilibrium without the punishment assumption. In this essay, we address the question posed in (Johnson et al., 2014) for a generalization of their model. Our generalization allows for the possibility that players might genuinely not be available to provide service, say due to sickness. Without using the assumption of punishment, we show that when the number of players is large enough and the discount factor is close enough to one, there exists an -Nash equilibrium in which each player plays the always-trade strategy.

3 CHAPTER 2

ON INTEGRAL POLICIES IN DETERMINISTIC AND STOCHASTIC

DISTRIBUTION SYSTEMS

Authors – Yang Bo, Milind Dawande, Ganesh Janakiraman, and S. Thomas McCormick

Naveen Jindal School of Management, SM 30

The University of Texas at Dallas

800 West Campbell Road

Richardson, Texas 75080-3021

Portions of this chapter are reprinted by permission, Yang Bo, Milind Dawande, Ganesh Janakiraman, S. Thomas McCormick, “On Integral Policies in Deterministic and Stochastic Distribution Systems”, Operations Research, Published online in Articles in Advance 04 Apr 2017. Copyright (2017), the Institute for Operations Research and the Management Sciences, 5521 Research Park Drive, Suite 200, Catonsville, Maryland 21228 USA.

4 2.1 Introduction

In a recent paper, (Chen et al., 2014) pose the following fundamental “integrality” question

for dynamic optimization models of inventory control: When the starting inventory levels are

integer-valued and all demands are integer-valued random variables, for what kind of inven-

tory problems do there exist integral optimal ordering quantities, even if we allow them to be

real valued? For a single-product, multi-echelon assembly network with stochastic or deter-

ministic demand, (Chen et al., 2014) prove that there always exists an integral optimal policy.

Our main goal in this paper is to examine the integrality question for both deterministic-

and stochastic-demand single-product multi-echelon distribution networks, and contrast the

answers with those for assembly networks.

Why do the notion of integrality and our results matter? To answer this, consider the

following practical setting for an arbitrary distribution network:

Suppose that customer demand is in full truckloads (FTL) and that shipments to cus-

tomers from the retail locations are also required to be in full truckloads. We deﬁne a unit

to be a truckload of shipment. In any period, the ordering/shipping cost incurred in trans-

porting a quantity q from a node of the network to its immediate successor in the network is:

C(q) = cfullbqc + f q − bqc ,

where cfull is the unit cost of shipping a full truckload, f(x) is the cost of shipping a fraction x

of a full truckload for x ∈ (0, 1), f(0) = 0, and f(1) = cfull. Assume that f(x) is concave non-

decreasing in x; thus, we have that f(x) ≥ cfullx for x ∈ [0, 1]. Notice that this “economy

of scale” cost structure favors FTL shipping. We also note that limx→0+ f(x) exists; this limit captures the ﬁxed cost for ordering/shipping any strictly positive quantity up to a full

truckload.

5 For this setting, the integrality question becomes: Given that ordering/shipping capacities are integral and demands are integer-valued random variables, is FTL shipping (within the network) optimal? The main message from the results in this paper is as follows:

1. In distribution systems with deterministic demand, FTL shipping is optimal.

2. In distribution systems with stochastic demand:

◦ A Less-Than-Truckload (LTL) shipping policy can be signiﬁcantly cheaper than an optimal FTL shipping policy.

◦ If (i) the extent of the variation in the cost parameters, i.e., shipping, holding, and back-ordering costs, is limited across time and (ii) inventory holding costs are not signiﬁcantly higher upstream than downstream, then the diﬀerence between the costs of LTL shipping and FTL shipping is provably marginal.

◦ In particular, when the cost parameters are stationary and inventory holding costs increase as we go downstream, there always exists an optimal policy which only uses FTL shipments.

◦ Moreover, when the ﬁxed ordering costs (i.e., economies of scale) are large enough and inventory holding costs increase as we go downstream, there always exists an optimal policy which only uses FTL shipments, even when the cost parameters vary with time.

Together, the above statements for distribution systems represent a fundamental departure from the optimality of FTL shipping in both deterministic and stochastic assembly systems, a result that was established in (Chen et al., 2014). Also, we note that the conclusions in the above statements still apply, with suitable units, if the units are not full truckloads. Fractional optimal solutions to deterministic optimization problems (with integral data) are common. In the case of distribution systems, however, it is surprising that there always

6 exists an integral optimal solution under deterministic demand but a fractional solution may be signiﬁcantly cheaper than all integral solutions under stochastic demand. Thus, demand uncertainty and the consequent need to strategically position inventory at diﬀerent nodes of the distribution network may result in fractional (i.e., LTL) shipments being substantially cheaper.

We also investigate the integrality question for assembly-distribution networks, in which assembly is followed by distribution. The results for such networks are similar to those for distribution networks.

While the examination of the integrality question is a relatively new area of investiga- tion, distribution systems have been fundamental to the study of supply chain management.

Indeed, inventory theorists and practitioners have studied stochastic distribution systems from at least as early as 1960. In terms of identifying the optimal policy itself, the only result we have to date is the optimality of echelon-order-up-to policies for the special case of serial inventory systems; this is due to the seminal paper of (Clark and Scarf, 1960). The diﬃculty in analyzing the optimal policy, in general, was also discussed in that same paper.

Consequently, research on these systems has since focussed on proposing easy-to-compute policies, and studying their cost performance analytically (e.g., in asymptotic regimes) or numerically. (Eppen and Schrage, 1981) is an early and inﬂuential paper along these lines.

A recent contribution to this line of research, which presents progress on the asymptotic performance of certain simple policies, is due to (Gallego et al., 2007). For a more extensive review of research on stochastic distribution systems, see (Federgruen, 1993), (Zipkin, 2000), or (Axsater, 2003).

The remainder of the paper is organized as follows: In Section 2.2, we present some preliminaries that will be useful in our subsequent analysis. In Section 2.3, we ﬁrst describe precisely the dynamic optimization model in a single-product, multi-echelon distribution system under deterministic demand, and then establish the existence of an integral optimal

7 policy for this problem. In Section 2.4, we investigate the integrality question for the same

problem in a stochastic-demand distribution system. Section 2.5 discusses the managerial

implications of our results. The proofs of our results are in Section 2.6.

2.2 Preliminaries

In this section, we present some preliminaries that will be useful in our subsequent analysis.

Section 2.2.1 explains the network representation and notation for the inventory systems of

interest. Section 2.2.2 explains the economies of scale structure in ordering/shipping that

will be considered throughout the paper.

2.2.1 Network Representation and Notation

A distribution system can be represented by an out-tree, which is a connected, directed

acyclic graph, with one node designated as the root node and all edges directed away from

the root node. There exists only one directed path from the root node to any other node;

the number of edges on this path is the height of that node. In particular, the height of the root node is 0. Each path from the root node ends at a leaf node. The maximum height over

all the nodes is the height of the tree, which we denote by Λ.

Consider1 any (directed) path/sub-path x1 → x2 → · · · → xk, k ∈ N. For every pair of nodes (xi, xj), 1 ≤ i < j ≤ k, on this path, node xi is a predecessor of node xj and node xj is a successor of node xi. For every pair of consecutive nodes (xj, xj+1), 1 ≤ j ≤ k − 1, on this path, node xj is the immediate predecessor of node xj+1 and node xj+1 is the immediate

successor of node xj. For every node x, let Succ(x) denote the set of immediate successors of node x, and let Pred(x) denote the unique immediate predecessor2 of node x.

1Throughout the paper, we use N (resp., Z+) to denote the set of all positive (resp., non-negative) integers.

2For convenience, we let the immediate predecessor of the root node be the external supplier.

8 To facilitate our formulation in Section 2.3, we label the nodes of the tree using the standard breadth-ﬁrst-search traversal; see, e.g., (Diestel, 2010). For a distribution network, let N denote the index set of all the nodes. For λ ∈ {0, 1, ··· , Λ}, let Nλ denote the index set of all the nodes with height λ; in particular, N∗ is the index set of all the leaf nodes. S These sets Nλ are mutually exclusive and satisfy N = 0≤λ≤∗ Nλ.

External Supplier External Suppliers

Factory 0 -4 -5 -6

Distribution 1 Center 2 -2 Subassembly -3

Retail Retail 3 Locations 4 Locations 5 -1 Assembly Final Product

External Demand External Demand External Demand External Demand (a) (b)

Figure 2.1. (a) A Distribution System and (b) An Assembly System.

The distribution system in Figure 2.1(a) helps illustrate the notions above. This out- tree is labeled using the breadth-ﬁrst search from the root node all the way down to the leaf nodes. Nodes 1, 2, 3, 4, and 5, constitute all the successors of node 0. We also have

Succ(0) = 1, 2 and Pred(3) = Pred(4) = 1. Node 0 is the root node, and its height is 0. The height of both nodes 1 and 2 is 1. Nodes 3, 4, and 5, are the leaf nodes and their height is 2. The maximum height across all the nodes is 2; and therefore, the height of this out-tree is 2. An assembly system (see, e.g., Figure 2.1(b)) can also be represented in a similar manner by an in-tree; i.e., each edge points towards the root node. Similar notions introduced above for distribution systems apply here, with some minor modiﬁcation.

9 2.2.2 Economies of Scale Structure

Throughout our paper, we assume the following economies of scale (EOS) structure in ordering/shipping within the distribution network: Let customer demand be in integer multiples of any given unit. Without loss of generality, we assume this unit to be a truckload of shipment, and use this terminology throughout the paper. Thus, customer demand is in full truckloads (FTL). For every j ∈ N , the cost of

j ordering/shipping qt units in period t to node j from its unique immediate predecessor is

j j j j j j j Ct (qt ) = ct bqt c + ft qt − bqt c , (2.1)

j j where ct is the unit cost of shipping a full truckload and ft (x) is the cost of shipping a fraction j j j j x of a full truckload for x ∈ [0, 1], where ft (0) = 0, and ft (1) = ct . We assume that ft (x) j is concave non-decreasing in x. It follows that limx→0+ ft (x) exists; this limit captures the ﬁxed cost for ordering/shipping any strictly positive quantity up to a full truckload. Then,

j limx→0+ ft (x) 0 j (here, we adopt the convention that 0 = 1 for convenience) is the ratio of the ft (1) j limx→0+ ft (x) ﬁxed cost to the cost of a full truckload at node j in period t. Let Ψ = minj∈N ,t∈T j . ft (1) Thus, Ψ characterizes the extent of economies of scale in ordering/shipping.

2.3 Distribution Systems with Deterministic Demand

In this section, we ﬁrst describe precisely the dynamic optimization model in a single-product, multi-echelon distribution system under deterministic demand. We then establish the existence of an integral optimal policy for this problem. Let us now start with the description of the model. Throughout the paper, we assume that all lead times in ordering/shipping are zero purely for expositional convenience (we will comment on how the analysis will change for the extension with positive lead times in Section 2.4). Consider a distribution network with height Λ. The planning horizon is ﬁnite and consists of T periods. Let T = {∞, ∈, ··· , T}; then T

10 j is the index set of the time periods over the horizon. Let It denote the inventory level at

node j at the beginning of period t, for j ∈ N \NΛ and t ∈ T ∪ {T + ∞} (we introduce

j period T + 1 simply for convenience). For j ∈ N and t ∈ T , let qt denote the quantity j delivered to node j from its unique immediate predecessor in period t. The quantity qt is j non-negative and bounded from above by its corresponding capacity Ut . That is,

j j 0 ≤ qt ≤ Ut for j ∈ N and t ∈ T . (2.2)

Notice that these shipping quantities should satisfy the material availability constraints. That is, at any time, the total amount shipped out of any non-leaf node is bounded from

j above by the total inventory on hand at that node and that moment. Let dt denote the j j demand at leaf node j in period t, for j ∈ NΛ and t ∈ T . Let St and Bt denote, respectively, the inventory on hand and the amount of back-ordered demand at leaf node j at the beginning

j j j j of period t, for j ∈ NΛ and t ∈ T ∪ {T + ∞}. The input data, i.e., I1 , dt , S1, and B1, is non-negative and integer valued. The sequence of events in any period t, t ∈ T , is as follows:

1. For λ = Λ − 1, Λ − 2, ··· , 0, −1, the following delivery process happens sequentially3.

j Every node j with height λ delivers to its immediate successors. For j ∈ Nλ, let rt denote the residual inventory at node j immediately after these deliveries. Then we have

j j X i j rt = It − qt and rt ≥ 0, for j ∈ Nλ (2.3) i∈Succ(j) by the conservation of inventory ﬂow and material availability constraints at node j. Also by the conservation of inventory ﬂow, we have

j j j It+1 = qt + rt for j ∈ Nλ. (2.4)

3The node with height −1 refers to the external supplier and is introduced simply for ease of exposition. Note that the external supplier does not receive delivery.

11 By (2.2)-(2.4), we have

j It+1 ≥ 0 for j ∈ N \N∗. (2.5)

j 2. The demands arise at the leaf nodes. Each leaf node j decides a selling quantity zt j (variables zt are standard auxiliary variables). These ﬁnal shipping quantities from the retail locations to the customers are non-negative and constrained to be integer-valued, i.e.,

j + zt ∈ Z for j ∈ N∗. (2.6)

j Since the ﬁnal shipping quantities zt from retail locations are also bounded from above j j j j by the supply St + qt and by Bt + dt , we have

j j j j j j j j j j St+1 = St + qt − zt ,St+1 ≥ 0,Bt+1 = Bt + dt − zt , and Bt+1 ≥ 0 for j ∈ N∗. (2.7)

j We now consider the cost incurred over the horizon. For j ∈ N and t ∈ T , let ct , j j ft (·), and ht , represent, respectively, the unit ordering/shipping cost, the EOS structure in ordering/shipping, and the unit holding cost at each node j in period t. For j ∈ N∗ j and t ∈ T , let bt represent the unit back-ordering cost at each leaf node j in period t. Then, j j j j j in period t, the ordering/shipping cost at each node j is ct bqt c + ft (qt − bqt c), the holding j j cost at each non-leaf node j is ht It+1, and the holding and back-ordering costs at each leaf j j j j node j is ht St+1 + bt Bt+1. j j j Let π denote a feasible policy, i.e., π speciﬁes It+1 and qt for j ∈ N and t ∈ T , rt j j j for j ∈ N \N∗ and t ∈ T , and zt , St+1, and Bt+1, for j ∈ N∗ and t ∈ T . Thus, π satisﬁes the constraints (2.2)-(2.7). Let α, 0 ≤ α ≤ 1, denote the discount factor. The objective is to minimize the sum of discounted costs incurred at all the nodes over the horizon, where the total cost incurred in period t is discounted by a factor, αt−1, for t ∈ T . We denote this problem by P-DD. Note that in P-DD, all the cost parameters are allowed to be non-stationary, i.e., vary with time. We have

12 Theorem 2.3.1. There exists an integral optimal policy to P-DD.

The proof of Theorem 2.3.1 is in Section 2.6.1. We outline this proof below.

• We ﬁrst construct a mixed integer linear program (MILP) with the constraints (2.2)- (2.7) and a lower bounding objective with the following properties: (a) for every feasible policy π, the cost incurred by this policy is no more for this MILP than that for P-DD, and (b) for every integral feasible policy πint, the cost incurred by this policy is the same for both problems.

• By identifying and exploiting the TU property of the constraint matrix of the MILP, we then show that there exists an integral optimal policy π∗,int to this problem. This result, together with the two properties mentioned above, implies Theorem 2.3.1.

2.4 Distribution Systems with Stochastic Demand

This section investigates the integrality question in stochastic-demand distribution systems. The stochastic-demand version is identical to problem P-DD, with the following exception- s: (a) the demands are stochastic, (b) all the ordering/shipping decisions are made before the demands in that period are realised at the leaf nodes, and (c) the objective is to minimize the expected discounted costs incurred over the horizon. We denote the problem under stochastic demand by P-SD. Let us consider two assumptions:

• Assumption S: The cost parameters, i.e., the ordering/shipping and holding costs at each node and the back-ordering cost at each leaf node, are stationary across time.

• Assumption H: The holding costs at the nodes monotonically increase as we move downstream in the distribution system. That is, for every pair of nodes i, j such that node i supplies node j, the holding cost at node i is less than or equal to that at node j.

13 In Sections 2.4.1 and 2.4.2, we present examples to demonstrate that integral policies can

be signiﬁcantly sub-optimal in the absence of Assumptions S and H. Section 2.4.3 is devoted

to the analysis of the integrality gap4: We ﬁrst relax Assumptions S and H and establish

a general upper bound on the gap; this bound helps us analyze the impact of economies of

scale, non-stationary of the cost parameters across time, and the decrement in the holding

costs as we move downstream in the distribution system. This result also allows us to bound

the gap when Assumptions S and H hold.

2.4.1 Failure of Integrality When Assumption S Does Not Hold

Proposition 2.4.1. If Assumption H holds but Assumption S does not hold, then integrality

does not hold in general. Moreover, the integrality gap can be arbitrarily large.

To see this, we construct a 2-period example. Consider the distribution network in

1 Figure 2.2. Let N = 2k, k ∈ N. Let δ = N . Node 0 represents the distribution center, which ( k ) faces the external supplier. Each node n represents a retail location for n ∈ {1, 2, ··· ,N}, and faces external demand. The discount factor is 1. The initial on-hand inventory is N − 2

0 n at node 0, and 1 at node n, n ∈ {1, 2, ··· ,N}; that is, I1 = N − 2 and I1 = 1. All lead times for ordering/shipping are 0. In each period t, the ordering capacity is 1 at node 0,

0 n and inﬁnite at node n, 1 ≤ n ≤ N; that is, Ut = 1 and Ut = +∞. The cost parameters

n are as follows: The unit holding cost is 0 at each node in each period; that is, ht = 0 for n ∈ {0, 1, ··· ,N} and t ∈ {1, 2}. The unit back-ordering cost at every retail location n,

n 1 n ∈ {1, 2, ··· ,N}, is bt = δ2 in each period t. At node 0, the unit ordering/shipping cost 0 is 0 in each period t (i.e., ct = 0) while at every retail location n, n ∈ {1, 2, ··· ,N}, the

n n unit ordering/shipping cost is c1 = 0 in period 1 and c2 = 1 in period 2. The EOS structure

4Note that “integrality gap” for a minimization problem refers to the supremum of the ratios of the optimal values for an integer program and its LP relaxation across all instances; for details, see (Williamson and Shmoys, 2011).

14 n of the ordering/shipping quantity at each retail location n in each period t, ft (x), satisﬁes

n limx→0+ ft (x) = 0.

External Supplier

Distribution Center 0

Retail Retail Locations Locations 1 2 N

External Demand External Demand External Demand

Figure 2.2. The Distribution Network in the Proofs of Propositions 2.4.1 and 2.4.2.

We now specify the demand. In period 1, a set of k retail locations is randomly chosen;

every subset of k retail locations is equally likely to be chosen. Each of these k retail locations experiences a demand of 1 while the remaining N −k = k retail locations experience zero demand. Mathematically, the joint probability distribution Prob(·) of the demand in period 1 at all the retail locations is as follows:

n1 n2 nk 1, n Prob d1 = d1 = ··· = d1 = 1, d1 = 0 for n ∈ {1, 2, ··· ,N}\{n1, n2, ··· , nk} = δ

for 1 ≤ n1 < n2 < ··· < nk ≤ N. In period 2, the demand at each retail location is deterministically 1.

The detailed analysis of this example is in Section 2.6.2. The main steps of the analysis are as follows:

1 • We ﬁrst ﬁnd a feasible policy and show that its cost is kf( k ).

15 N−1 • We then show that the minimum cost attained by an integral policy is δ(k+1)· k−1 = k+1 2 .

• Therefore, the ratio of the cost of the best integral policy and the optimal cost is at least

k+1 2 k + 1 1 1 = · 1 . kf( k ) 2k f( k )

Notice that limx→0+ f(x) = 0. Thus, the ratio can be made arbitrarily large.

We now consider a special case of this example with N = 20 (i.e., 20 retail locations). Then k = 10, and the ratio of the costs of the best FTL shipping policy and a feasible LTL

11 1 shipping policy is 20 · f(0.1) . This ratio is greater than 1 if f(0.1) < 0.55. This example shows that, even when shipping costs are subject to substantial EOS (e.g., shipping 10% of a full truckload incurs 50% of the cost of shipping a full truckload), LTL shipping can still be cheaper than FTL shipping. If the purpose is only to show that integrality may fail for stochastic distribution systems (without making a statement about the integrality gap), then a much simpler example suﬃces. We note this in the remark below.

Remark 2.4.1. Integrality fails even for the simplest non-trivial distribution system, with one distribution center and two retail locations. To see this, consider the above example with the following modiﬁcation: (a) N = 2, (b) the initial on-hand inventory is 1 at each node,

1 and (c) the unit ordering cost is 2 at node 0 in each period. Then, it is easy to verify that 0 1 2 the unique period-1 optimal solution is q1 = q1 = q1 = 0.5.

2.4.2 Failure of Integrality When Assumption H Does Not Hold

Proposition 2.4.2. If Assumption S holds but Assumption H does not hold, then integrality does not hold in general. Moreover, the integrality gap can be arbitrarily large.

16 We establish this by analyzing the following 2-period example.

Consider the same example as in Section 2.4.1 with the following modiﬁcation: (a) the

0 unit holding cost is 1 at node 0 (the distribution center) in each period (i.e., ht = 1), (b) the unit ordering/shipping cost is 0 at retail location n, 1 ≤ n ≤ N, in each period (i.e.,

n ct = 0), (c) the initial inventory on hand is N −1 at node 0 and 1 at each retail location (i.e.,

0 n I1 = N − 1 and I1 = 1), and (d) in period 1, for n ∈ {1, 2, ··· ,N}, the scenario that one unit of demand arises at retail location n and no demand arises at the other retail locations

1 occurs with probability δ = N . Note that Assumption S holds but Assumption H fails for this example.

The detailed analysis of this example is in Section 2.6.3. The main steps of our analysis

1 are as follows: (a) we ﬁrst ﬁnd a feasible policy and show that its cost is N , (b) we then show that the cost incurred by any integral feasible policy is at least 1, and therefore, (c)

the ratio of the cost of the best integral policy and the optimal cost is at least N, which can

be made as large as desired by letting N go to inﬁnity.

Remark 2.4.2. Notice that the examples explained in Sections 2.4.1 and 2.4.2 use correlated

demand at the leaf nodes (the retail locations). Counterexamples to integrality can also be

constructed with independent demands.

2.4.3 Bounding the Integrality Gap

Propositions 2.4.1 and 2.4.2 imply that the integrality gap can be arbitrarily large when

Assumption S and Assumption H fail. Thus, a natural question arises: Do we have a bounded

integrality gap when Assumption S and Assumption H fail only to a limited extent?

We now introduce the following two assumptions:

17 5 • For each cost parameter ct, 1 ≤ t ≤ T , deﬁne the “relative range” of this parameter as max1≤t≤T ct . We have min1≤t≤T ct

◦ Assumption LN (Limited Non-Stationarity in the Cost Parameters): The rela-

tive range of each cost parameter is bounded from above by β for some constant

β ≥ 1.

• For every pair of nodes, (i, j), such that node j is a (not necessarily immediate) succes-

sor of node i, deﬁne the “relative decrement” of the holding costs in period t, 1 ≤ t ≤ T ,

j ht to be i . We have ht

◦ Assumption LH (Limited Decrement in Holding Costs Going Downstream):

Every relative decrement is bounded from below by γ, γ ∈ [0, 1].

Thus, β characterizes the extent to which Assumption S fails, and γ characterizes the extent

j limx→0+ ft (x) to which Assumption H fails. Also recall that Ψ = minj∈N ,t∈T j . Then, we have ft (1)

Theorem 2.4.1. If Assumptions LN and LH hold, then the minimum expected cost over the

1 β horizon of an integral policy is at most max{ γ , 1+βΨ } times the optimal cost.

The proof of Theorem 2.4.1 is in Section 2.6.4. The proof strategy is as follows: (a) we ﬁrst construct a feasible integral policyπ ˜ by a rounding scheme applied on π, (b) we then show in sequence that (i) the back-ordering cost underπ ˜ is equal to that under π, (ii) the holding

1 cost underπ ˜ is at most γ times the holding cost under π, and (iii) the ordering/shipping β cost underπ ˜ is at most max{1, 1+βΨ } times the ordering/shipping cost under π. Notice that the above theorem holds for any values of γ, β, and Ψ satisfying γ ∈ [0, 1],

β ≥ 1, and Ψ ∈ [0, 1]. As a consequence, we have

5Note that there is a slight abuse of notation here: the cost parameter c can stand for any cost parameter, i.e., the ordering/shipping cost at any node, the holding cost at any node, and the back-ordering cost at any leaf node.

18 Corollary 2.4.1. If Assumptions S and H hold, then there always exists an integral optimal

policy.

1 Corollary 2.4.2. If Assumptions LN and H hold, and if Ψ ≥ 1 − β , then there exists an integral optimal policy.

Remark 2.4.3. Throughout our discussion in this paper, we assumed that lead times are zero, and unsatisﬁed demand is backlogged. It is easy to verify that all our results continue to hold in the presence of positive lead times and/or lost sales instead of backlogging. We explain below.

Positive Lead Times. We assume that for each j, the lead time Lj in ordering/shipping at node j stays the same over the horizon; however, lead times can be diﬀerent at diﬀerent nodes. We now comment on how the analysis will change when there are positive lead times.

• For deterministic-demand distribution systems, the math programming formulation in

the positive-lead-time setting diﬀers from that in the zero-lead-time setting in the fol-

lowing aspects:

(a) There is a shift in the subscripts for all the ordering/shipping quantities in con-

straints (2.6.14) and (2.6.15); that is, qj becomes qj for every node j and every t t−Lj period t (for convenience, qj = 0 when t − L ≤ 0). t−Lj j

(b) The holding cost incurred by pipeline inventories is now incorporated in the total

holding cost, i.e., the expression HC (π) = P hjIj + P hjSj now t j∈N \N∗ t t+1 j∈N∗ t t+1 becomes

X X X HC (π) = hj Ij + (qi + qi + ··· + qi ) + hjSj t t t+1 t t−1 t−Li+1 t t+1 j∈N \N∗ i∈Succ(j) j∈N∗

for every period t.

19 The proof strategy remains the same as that in the zero-lead-time setting; that is, we exploit the TU property of the constraint matrix.

• For stochastic-demand distribution systems, the math programming formulation in the positive-lead-time setting diﬀers from that in the zero-lead-time setting in the following aspects:

j Pt−Lj j j P Pt i (a) Constraint (2.6.34) becomes It+1,ω = τ=1 qτ,ω + I1 − i∈Succ(j) τ=1 qτ,ω.

j j Pt−Lj j Pt j (b) Constraint (2.6.35) now becomes St+1,ω = S1 + τ=1 qτ,ω − τ=1 zτ,ω.

(c) The holding cost incurred by pipeline inventories is now incorporated in the total holding cost; that is, H (π; ω) = P hjIj + P hjSj becomes t j∈N \N∗ t t+1,ω j∈N∗ t t+1,ω X X H (π; ω) = hj Ij + (qi + qi + ··· + qi ) t t t+1,ω t,ω t−1,ω t−Li+1,ω j∈N \N∗ i∈Succ(j)

X j j + ht St+1,ω j∈N∗ for every period t.

For the proof of Theorem 2.4.1 in the positive-lead-time setting, the following steps are similar to that in the zero-lead-time setting: the construction of π˜ given π, the feasibility of π˜ given that of π, the comparison of total back-ordering costs under π˜ and π, and the comparison of total ordering/shipping costs under π˜ and π. We now comment on how the analysis will change for the comparison of holding costs under π˜ and π in the positive-lead-time setting:

j,P (a) For policy π, for j ∈ N \N∗, ω ∈ Ω, and t ∈ T , let It+1,ω denote the sum of the inventory on hand at node j and the total pipeline inventory out of node j at the beginning of period t + 1 under a sample path ω. That is,

X Ij,P = Ij + (qi + qi + ··· + qi ). t+1,ω t+1,ω t,ω t−1,ω t−Li+1,ω i∈Succ(j)

20 By substitution, we have

t−Lj t X X X X Ij,P = qj + Ij − qi + (qi + ··· + qi ) t+1,ω τ,ω 1 τ,ω t,ω t−Li+1,ω τ=1 i∈Succ(j) τ=1 i∈Succ(j)

t−Lj t−Li X j j X X i = qτ,ω + I1 − qτ,ω. τ=1 i∈Succ(j) τ=1

˜j,P j,P For policy π˜, the deﬁnition of It+1,ω is similar to that of It+1,ω.

(b) The same proof as that for the comparison of holding costs under π˜ and π in

the zero-lead-time setting carries over to the positive-lead-time setting with the j j,P ˜j ˜j,P following changes: It+1,ω → It+1,ω and It+1,ω → It+1,ω for j ∈ N \N∗ and t ∈ T ,

Pt j Pt−Lj j Pt j Pt−Lj j and τ=1 qτ,ω → τ=1 qτ,ω and τ=1 q˜τ,ω → τ=1 q˜τ,ω for j ∈ N and t ∈ T .

Lost Sales. We consider three cases.

j j • For deterministic-demand distribution systems with back-ordering, recall that zt , dt , j and bt , denote, respectively, the selling quantity, the demand, and the unit back-ordering j cost, at retail location j in period t. Also recall that Bt represents the back-ordering quantity at node j at the beginning of period t.

The lost-sales setting diﬀers from the back-ordering setting in the following aspects: (a)

j Instead of the unit back-ordering cost, we let bt denote the unit lost-sales cost at retail j location j in period t; (b) we drop all variables Bt and all the corresponding constraints, j j and instead add the constraints zt ≤ dt at retail location j in period t; and (c) in the objective function, we drop the total back-ordering cost while adding all the lost-sales

j j j costs, i.e., bt (dt − zt ) at node j in period t.

The proof strategy of the problem in deterministic-demand distribution systems in the

lost-sales setting is the same as in the back-ordering setting. That is, we exploit the

TU property of the constraint matrix.

21 • For stochastic-demand distribution systems with back-ordering, notice that Theorem 2 is the most general positive result, which subsumes all the other results. Thus, we only comment on the proof strategy of this theorem. In the lost-sales setting, the same result

1 β holds; in particular, the bound, i.e., max{ γ , 1+βΨ }, remains the same. Similar to the back-ordering setting, the main steps of the proof of this result for the lost-sales setting are as follows: (a) given any policy π, we ﬁnd a feasible integral policy π˜ by a rounding scheme applied on π, and (b) we then show in sequence that under each sample path (i) the lost-sales cost under π˜ is equal to that under π, (ii) the holding cost under π˜ is at

1 most γ times the holding cost under π, and (iii) the ordering/shipping cost under π˜ is β at most max{1, 1+βΨ } times the ordering/shipping cost under π.

• As far as the negative results are concerned (counterexamples to integrality), the examples that we use currently and their analysis hold for both the back-ordering and lost-sales settings. This is because, all of our examples are 2-period problems and shortages do not occur in the ﬁrst period under any policy. Thus, there is no diﬀerence between the back-ordering and lost-sales settings as far as these 2-period examples are concerned.

It follows from the above discussion that our results remain valid even for a combination of positive lead times and lost sales.

2.5 Managerial Implications of Results

We now explain the main intuition behind the technical results in the paper, which in turn provides managers with useful insights on when to consider LTL shipping.

2.5.1 Unpredictable Demand: Variation in Shipping Costs Across Time

Consider a situation in which shipping costs are expected to increase in the future. Then naturally there is an economic pressure to ship inventory downstream early. However, if

22 upstream inventory is currently limited, then it may not be possible to ship enough inventory to all downstream locations. Then, if only FTL shipments are used, some retail locations will not be allocated any stock. If, subsequently, demand occurs at these retail locations, then we would be forced to ship (newly available) inventory to these locations at a higher shipping cost. On the other hand, if LTL shipments are considered, then inventory can be more uniformly distributed across the retail locations, resulting in reduced shipping costs in the future. This is because retail locations that experience a higher demand than their allocated inventory need to be supplemented by shipments of a smaller size than those in the FTL scenario discussed above. Thus, if shipping costs are increasing and demands are unpredictable, then it may be worthwhile to explore LTL shipments.

More surprising and important to note is that an LTL shipping policy can be signiﬁcantly cheaper than the best FTL shipping policy even when shipping costs are subject to substantial EOS. This is the practical insight that stems from Proposition 2.4.1 (Section 2.4.1).

2.5.2 Unpredictable Demand: Holding Cost Variation Across Stages

Next, consider a situation in which inventory is more expensive to hold upstream than downstream. Here again, we face an economic pressure to ship downstream early. The use of FTL shipments will again leave some retail locations without any stock. Then, any subsequent demand at these locations would require a higher amount of inventory to ﬂow through the upstream locations, where inventory is more expensive to hold. On the other hand, LTL shipments oﬀer a more balanced positioning of inventory across the retail locations, resulting in less holding of stock upstream compared to that under FTL shipments. This is the practical insight that stems from Proposition 2.4.2 (Section 2.4.2).

Thus, the main import of our work is that if there are significant shipping costs differentials across time or significant holding cost differentials (in the above-mentioned ways), then

LTL shipping can result in substantial cost savings, even in the presence of substantial EOS.

23 2.5.3 Unpredictable Demand: Economies of Scale

For a given amount of volatility in the shipping costs, the greater the EOS, the more cost

eﬀective FTL shipping is. Conversely, for a given EOS structure, the greater the increase

in the shipping cost, the more is the potential of savings from LTL shipping. This is the

insight from Corollary 2.4.2 (Section 2.4.3). Similarly, for given holding cost diﬀerentials

across echelons, the savings from LTL shipments become less signiﬁcant with an increase in

the EOS.

2.5.4 Predictable Demand

If demands are highly predictable, the potential beneﬁt of LTL shipping in the strategic

positioning of inventory at the diﬀerent nodes of the distribution network no longer exists,

regardless of how signiﬁcant the cost diﬀerentials are. That is, there is no need to hedge

against uncertainty by using LTL shipments; therefore, FTL shipping emerges as the most

cost eﬀective policy. This is the main insight that stems from Theorem 2.3.1 (Section 2.3).

2.6 Proofs of Results

2.6.1 Proof of Theorem 2.3.1

We ﬁrst formulate problem P-DD as a math program. Recall from Section 2.3 that π denotes a generic feasible policy. For t ∈ T , let SCt(π), HCt(π) and BCt(π) denote, respectively, the sum of ordering/shipping costs at all the nodes in period t, the sum of holding costs at all the nodes in period t, and the sum of back-ordering costs at all the leaf nodes in period t. h i That is, SC (π) = P cjbqjc+f j(qj −bqjc) , HC (π) = P hjIj +P hjSj , t j∈N t t t t t t j∈N \N∗ t t+1 j∈N∗ t t+1 and BC (π) = P bjBj . Let TC(π) denote the sum of discounted costs incurred over t j∈N∗ t t+1 PT t−1 the entire horizon, i.e., TC(π) = t=1 α SCt(π)+HCt(π)+BCt(π) . The following math

24 program characterizes our problem:

minimize TC(π) (2.6.8)

j j subject to It+1, rt ≥ 0, j ∈ N \N∗, t ∈ T (2.6.9)

j j St+1,Bt+1 ≥ 0, j ∈ N∗, t ∈ T , (2.6.10)

j qt ≥ 0, j ∈ N , t ∈ T , (2.6.11)

j j −qt ≥ −Ut , j ∈ N , t ∈ T , (2.6.12)

j P i j It − i∈Succ(j) qt − rt = 0, j ∈ N \N∗, t ∈ T , (2.6.13)

j j j rt + qt − It+1 = 0, j ∈ N \N∗, t ∈ T , (2.6.14)

j j j j St + qt − zt − St+1 = 0, j ∈ N∗, t ∈ T , (2.6.15)

j j j j zt + Bt+1 − Bt − dt = 0, j ∈ N∗, t ∈ T , (2.6.16)

j + zt ∈ Z , j ∈ N∗, t ∈ T . (2.6.17)

We now establish Theorem 2.3.1 using the following steps:

• Consider an MILP which is the same as the math program above, with the only modiﬁcation that the ordering/shipping cost in period t is

X j j SCf t(π) = ct qt j∈N

for t ∈ T . Let TCg(π) denote the objective of this “modiﬁed” MILP. Notice that the two problems have the same policy space. Recall from Section 2.2.2 that for each j and

j j j j each t, ft (·) is concave non-decreasing in [0, 1], ft (0) = 0, and ft (1) = ct . It follows immediately that TC(π) ≥ TCg(π) for every feasible policy π and TC(πint) = TCg(πint) for every integral feasible policy πint.

• It suﬃces to show the existence of an integral policy π∗,int such that TCg(π) ≥ TCg(π∗,int) for every feasible π. Notice that (a) in (2.6.13)–(2.6.16), each decision variable appears

25 either only once, with coefficient −1 or 1, or exactly twice, once with coefficient −1 and once with coefficient 1, and (b) each row of the coefficient matrix corresponding to (2.6.9)–(2.6.12) contains only one nonzero entry which is ±1. Thus, the constraint matrix (2.6.9)-(2.6.16) is TU, and the result now follows; for details, see (Schrijver, 1998).

2.6.2 Proof of Proposition 2.4.1

Consider the example in Section 2.4.1. We ﬁrst ﬁnd a feasible policy, and then show that the ratio of the cost of the best integral policy and the cost of this feasible policy can be made as large as possible.

n Let q1 denote the ordering/shipping quantity at node n in period 1, for n ∈ {0, 1, ··· ,N}. These ordering/shipping quantities are non-negative decision variables:

n q1 ≥ 0 for n ∈ {0, 1, ··· ,N}. (2.6.18)

By material availability, the total quantity delivered in period 1 from node 0 is bounded from above by its available inventory. That is,

1 2 N q1 + q1 + ··· + q1 ≤ N − 2. (2.6.19)

0 Also, the quantity q1 is constrained by its ordering/shipping capacity 1. That is,

0 q1 ≤ 1. (2.6.20)

j Inequalities (2.6.18)-(2.6.20) constitute the constraints for q1, j ∈ {0, 1, ··· ,N}. In period 1, since all the unit holding and the unit ordering/shipping costs are 0, and there are no back-orders at any of the retail locations, the expected total cost incurred is 0. Thus, we only have to consider the cost incurred in period 2. Also, notice that the total holding cost is 0 in period 2; we therefore only have to consider the ordering/shipping and back-ordering costs at all the retail locations n, 1 ≤ n ≤ N, in period 2.

26 Let us now consider the following policy: In period 1, the ordering/shipping quantities are

∗ 0 1 N N−2 N−2 q1 = (q1, q1, ··· , q1 ) = (1, N , ··· , N ). In period 2, for any given n1, n2, ··· , nk, under

n1 n2 nk n the period-1 scenario that d1 = d1 = ··· = d1 = 1 and d1 = 0 for n∈ / {n1, n2, ··· , nk},

1 node nj, j ∈ {1, 2, ··· , k}, receives a delivery of k unit, and node 0 does not place any order. ∗ The feasibility of this policy can be easily verified; in particular, q1 satisfies (2.6.18)-(2.6.20), and therefore, is a feasible period-1 solution. It remains to find the cost of this policy.

n1 n2 nk n Consider the scenario, d1 = d1 = ··· = d1 = 1 and d1 = 0 for n∈ / {n1, n2, ··· , nk}. In

N−2 this scenario, the amount of on-hand inventory at the end of period 1 is 1 + N at retail N−2 location n for n∈ / {n1, n2, ··· , nk}, and N at retail location nj, for j ∈ {1, 2, ··· , k}. Then 1 in period 2, by receiving the delivery of k unit, the demand at retail location nj, 1 ≤ j ≤ k, N−2 1 will be satisﬁed (note that N + k = 1), and the demand at each of the remaining retail locations will also be satisﬁed. Thus the only cost incurred in period 2 is the sum of the

1 ordering/shipping costs across the retail locations. It is easy to see that this cost is kf( k ), and therefore, the expected total cost incurred over the horizon is

N 1 1 · δ · kf( ) = kf( ). k k k

We now ﬁnd the best (minimum-cost) integral policy. For every feasible integral policy

0 with q1 = 0 and satisfying (2.6.18)-(2.6.20), it is easy to verify that the cost incurred by this policy is greater than or equal to the cost incurred by the same policy with the modiﬁcation

0 0 0 that q1 = 1 (instead of q1 = 0), since there is no cost associated with this order (i.e., q1 = 1), and it may help reduce the back-ordering cost at the retail locations in period 2. Thus, we

0 restrict our attention to the feasible integral policies in which q1 = 1. Since the maximum demand at each retail location over the horizon is 2, the cumulative shipment to this retail

location is either 0 or 1 (note that each retail location already has 1 unit of inventory on

hand). Let x denote the on-hand inventory at node 0 immediately after delivering to the

retail locations in period 1. Then N − 2 − x is the total inventory received by the retail

27 locations in period 1 and the total inventory on hand at the end of period 1 is x + 1, where 0 ≤ x ≤ N − 2. Without loss of generality, let 1 unit be shipped to each of the ﬁrst N − 2 − x retail locations. We now consider two cases: Case 1: k − 1 ≤ x ≤ N − 2. Then, the inventory on hand at 0 at the beginning of period 2 is x + 1 ≥ k. Thus for each retail location n with N − 2 − x + 1 ≤ n ≤ N, if demand arises in period 1, then 1 unit of inventory will be delivered from node 0 to retail location n in period 2 to satisfy the demand (note that this shipment is always feasible since there is enough on-hand inventory at node 0 at the beginning of period 2), and, therefore, an ordering/shipping cost of 1 will be incurred at this retail location in period 2; otherwise, no ordering/shipping is needed and there will be no cost incurred. For each retail location n with 1 ≤ n ≤ N − 2 − x, no back-ordering or ordering/shipping costs will be incurred in period 2, since this retail location already has enough on-hand inventory at the beginning of period 2 and no ordering/shipping is needed.

For each i ∈ Z+ with 0 ≤ i ≤ k, the number of scenarios in period 1 in which demand arises at k − i of the ﬁrst N − 2 − x retail locations and at i of the remaining x + 2 retail

N−2−xx+2 n + locations is k−i i (Since m = 0 for n, m ∈ Z satisfying n < m, the analysis goes N−2−xx+2 through even when N − 2 − x < k − i). The corresponding cost incurred is k−i i · i. Then, the expected total cost incurred over the horizon is k k X N − 2 − xx + 2 X N − 2 − xx + 2 δ · · i = δ · · i k − i i k − i i i=0 i=1 k X N − 2 − x x + 1 = δ · · (x + 2) k − i i − 1 i=1 N − 1 = δ(x + 2) (2.6.21) k − 1 N − 1 ≥ δ(k + 1) , (2.6.22) k − 1 where equality (2.6.21) follows from the identity rn − r r n − r r n − r n + + ··· + = , 0 n1 1 n1 − 1 n1 0 n1

28 + for n, r, n1 ∈ Z ; for details, see (Feller, 1968). Furthermore, inequality (2.6.22) becomes an equality if and only if x = k − 1. Case 2: 0 ≤ x ≤ k − 2. Then the inventory on hand at 0 at the beginning of period 2

N−k+1 N−k+2 N n is x+1 ≤ k−1. Consider the period-1 scenario, d1 = d1 = ··· = d1 = 1, and d1 = 0 for n ∈ {1, 2, ··· ,N − k}, whose probability of realization is δ. To satisfy all the demand in period 2, min{x + 2, k} units of inventory has to be delivered from node 0; however, there are only x + 1 units at the beginning of period 2, which is strictly less than min{x + 2, k}. Thus, there exists a scenario in which the total back-ordering quantity across all the retail

1 N2 locations in period 2 is at least 1. Then, the expected total cost is at least δ · δ2 ≥ δ · k . N2 N−1 We claim that δ · k > δ(k + 1) · k−1 . This is because N2 N − 1 δ · − δ(k + 1) · k k − 1 ( ) N − 1 N N − 1 N − 2 N − k + 1 = δ · · N · · ····· − (k + 1) k − 1 k k k − 1 2 N − 1 ≥ δ · 2 · N − (k + 1) (2.6.23) k − 1 > 0, (2.6.24)

where (2.6.23) and (2.6.24) follow from k ∈ N and N = 2k. N−1 k+1 Thus, the minimum cost of an integral solution is δ(k + 1) · k−1 = 2 . The ratio of the minimum cost of an integral solution and the optimal cost is at least

k+1 2 k + 1 1 1 = · 1 . kf( k ) 2k f( k )

Notice that limx→0+ f(x) = 0. Thus, the ratio can be made as large as possible.

2.6.3 Proof of Proposition 2.4.2

Consider the example in Section 2.4.2. We ﬁrst ﬁnd a feasible policy, and then show that the ratio of the cost of any integral policy and the cost of this feasible policy can be made as large as possible.

29 n Let qt denote the ordering/shipping quantity at node n in period 1 for 0 ≤ n ≤ N.

n These quantities, q1 , 0 ≤ n ≤ N, form the decision variables in period 1. The feasibility constraints for these quantities are the same as those in Section 2.4.1. That is,

0 n 1 2 N 0 ≤ q1 ≤ 1; q1 ≥ 0 for n ∈ {1, 2, ··· ,N}; and q1 + q1 + ··· + q1 ≤ N − 1. (2.6.25)

Let us now consider the following policy: In period 1, the ordering/shipping quantities

∗ 0 1 N 1 N−1 N−1 are q1 = (q1, q1, ··· , q1 ) = ( N , N , ··· , N ). In period 2, for any givenn ˆ ∈ {1, 2, ··· ,N}, nˆ n under the period-1 scenario that d1 = 1 and d1 = 0 for n 6=n ˆ, noden ˆ receives a delivery

1 of N unit, and node 0 does not place any order. We can easily verify that (a) this policy is feasible, (b) the only cost incurred in period 1 under this policy is the holding cost at node 0,

1 which is N , and (c) the cost in period 2 under this policy is 0 for each demand realization 1 in period 1. Thus, the total cost incurred over the horizon is N . We now show that the cost of any integral solution is at least6 1. Note that for each demand realization in period 1, the on-hand inventory at node 0 at the end of period 1 (also at the beginning of period 2) is

0 0 1 2 N I2 ≡ N − 1 + q1 − q1 − q1 − · · · − q1 .

Consider the following two cases:

0 • Suppose I2 = 0. Then we have

0 1 2 N q1 = 0, and q1 + q1 + ··· + q1 = N − 1,

n which follows from (2.6.25). Under the requirement that q1 , 1 ≤ n ≤ N, is integral, j there must exist j with 1 ≤ j ≤ N such that q1 = 0. Let j = 1 without loss

1 2 N of generality. Then consider the scenario, d1 = 1 and d1 = ··· = d1 = 0, whose

6 0 1 N In fact, the reader can verify that (q1, q1, ··· , q1 ) = (1, 1, ··· , 1, 0) is a least-cost integral solution, whose cost is 1.

30 1 probability of realization is δ = N . The on-hand inventory is 0 at node 1 at the 0 end of period 1 in this scenario. Moreover, since I2 = 0, node 1 is unable to receive any replenishment in period 2. Recall that the demand at node 1 in period 2 is 1.

Therefore, the back-ordering quantity at node 1 in period 2 is 1. Thus, the total cost

1 1 incurred over the horizon is at least δ · δ2 = δ , which is greater than 1.

0 n 0 • Suppose I2 > 0. Under the requirement that q1 , 0 ≤ n ≤ N, is integral, we have I2 ≥ 1. Then, the total cost incurred over the horizon is at least 1 (since the holding cost at

node 0 in period 1 is at least 1).

Thus, for any integral solution, the total cost incurred over the horizon is at least 1. Then

the ratio of the minimum cost of an integral solution and the optimal cost is at least N.

Notice that this assertion holds for every N ∈ N with N ≥ 2. Thus, the ratio can be made arbitrarily large.

2.6.4 Proof of Theorem 2.4.1

Let ωt denote the realization of demands at the retail nodes in period t. Let ω = (ωt; t ∈ T ) denote a generic sample path of demand realizations from period 1 through period T . Let Ω

j denote the collection of all ω. Let dt,ω denote the demand realization at leaf node j in

period t under ω for j ∈ N∗, t ∈ T , and ω ∈ Ω. Let the following parameters be deﬁned as

j j j j j j j j in Section 2.3: I1 , S1, B1, Ut , ct , ht , and bt . The deﬁnitions of all the variables, i.e., zt,ω, j j j j Bt+1,ω, St+1,ω, It+1,ω, and qt,ω, are similar to those in Section 2.3, except that each of them has an additional index ω to indicate the dependence on the sample path ω, ω ∈ Ω. Let π

j j j j j denote a generic policy, i.e., the collection of variables zt,ω, Bt+1,ω, St+1,ω, It+1,ω, and qt,ω, for all j, t, and ω. Note that, for simplicity, we suppress the dependence of these variables on π

0 0 in the notation. For a pair of sample paths ω = (ωt; t ∈ T ) and ω = (ωt; t ∈ T ), deﬁne Tω,ω0

n 0 0 0o 0 as Tω,ω0 = max t : t ∈ T ∪ {0}; ω1 = ω1, ω2 = ω2, ··· , ωt = ωt . Then, ω and ω follow the

31 same demand realization until period Tω,ω0 . Note that π should satisfy non-anticipativity constraints. That is, for every pair of sample paths ω and ω0, we have

j j j j j j zt,ω = zt,ω0 ,Bt+1,ω = Bt+1,ω0 ,St+1,ω = St+1,ω0 , for j ∈ NΛ, t ∈ {1, ··· ,Tω,ω0 }; (2.6.26)

j j qt,ω = qt,ω0 , for j ∈ N , t ∈ {1, ··· ,Tω,ω0 }; (2.6.27)

j j It+1,ω = It+1,ω0 , for j ∈ N \NΛ, t ∈ {1, ··· ,Tω,ω0 }. (2.6.28)

Let pω denote the probability of the realization of ω, ω ∈ Ω. Let α, 0 ≤ α ≤ 1, denote the discount factor. Then, the total cost incurred over the horizon under π is

X X t−1h i TC(π) = pω α TBt(π; ω) + SCt(π; ω) + Ht(π; ω) . ω∈Ω t∈T

where TBt(π; ω), SCt(π; ω), and Ht(π; ω) are, respectively, the back-ordering, the ordering/shipping, and the holding costs, in period t under ω; they are deﬁned similarly as in

Section 2.6.1.

In addition to (2.6.26)–(2.6.28), the feasibility constraints for Problem P-SD are as

follows:

j It+1,ω ≥ 0, j ∈ N N∗, t ∈ T , ω ∈ Ω, (2.6.29)

j j St+1,ω,Bt+1,ω ≥ 0, j ∈ N∗, t ∈ T , ω ∈ Ω, (2.6.30)

j qt,ω ≥ 0, j ∈ N , t ∈ T , ω ∈ Ω, (2.6.31)

j j qt,ω ≤ Ut , j ∈ N , t ∈ T , ω ∈ Ω, (2.6.32)

j P i It,ω − i∈Succ(j) qt,ω ≥ 0, j ∈ N N∗, t ∈ T , ω ∈ Ω, (2.6.33)

j Pt j j P Pt i It+1,ω = τ=1 qτ,ω + I1 − i∈Succ(j) τ=1 qτ,ω, j ∈ N N∗, t ∈ T , ω ∈ Ω, (2.6.34)

j j Pt j Pt j St+1,ω = S1 + τ=1 qτ,ω − τ=1 zτ,ω, j ∈ N∗, t ∈ T , ω ∈ Ω, (2.6.35)

j j Pt j Pt j Bt+1,ω = B1 + τ=1 dτ,ω − τ=1 zτ,ω, j ∈ N∗, t ∈ T , ω ∈ Ω, (2.6.36)

j + zt,ω ∈ Z , j ∈ N∗, t ∈ T , ω ∈ Ω. (2.6.37)

32 The objective of Problem P-SD is to minimize TC(π) subject to the constraints (2.6.26)-

(2.6.37). We will prove Theorem 2.4.1 by (a) constructing a feasible integral policyπ ˜ by a

1 β rounding scheme applied on π, and (b) showing thatπ ˜ attains a cost at most max{ γ , 1+βΨ } times the cost under π.

We are now ready to constructπ ˜ from π. For every sample path ω, we do the following:

j j (a) Letz ˜t,ω = zt,ω for j ∈ N∗ and t ∈ T .

j Pt j Pt−1 j Pt j (b) Letq ˜t,ω = b τ=1 qt,ωc − b τ=1 qt,ωc for j ∈ N and t ∈ T . This implies τ=1 q˜τ,ω = Pt j b τ=1 qτ,ωc. That is, under each ω, the cumulative quantity ordered by node j in the ﬁrst t periods underπ ˜ is the same as that under π rounded down.

˜j ˜j ˜j The other decision variables, It+1,ω, St+1,ω, and Bt+1,ω are completely deﬁned due to (2.6.34)- (2.6.36). We now show thatπ ˜ is integral and satisﬁes (2.6.26)-(2.6.37).

• The fact thatπ ˜ is integral and satisﬁes (2.6.34)-(2.6.37) follows from its deﬁnition.

j j • The fact thatz ˜t,ω andq ˜t,ω satisfy (2.6.26)-(2.6.28) immediately follows from the fea- j j ˜j ˜j sibility of zt,ω and qt,ω with respect to (2.6.26)-(2.6.28). The fact that It+1,ω, St+1,ω, ˜j and Bt+1,ω satisfy (2.6.26)-(2.6.28) follows from (2.6.34)-(2.6.37).

• We then show thatπ ˜ satisﬁes (2.6.29) and (2.6.33). Plugging (2.6.34) into (2.6.33)

j P Pt i Pt−1 j for π gives I1 − i∈Succ(j) τ=1 qτ,ω + τ=1 qτ,ω ≥ 0 for j ∈ N N∗ and t ∈ T . Then

t−1 t t−1 t j X j X X i j X j X X i I1 + qτ,ω ≥ qτ,ω =⇒ I1 + qτ,ω ≥ qτ,ω τ=1 i∈Succ(j) τ=1 τ=1 i∈Succ(j) τ=1 t−1 t j X j X X i =⇒ I1 + qτ,ω ≥ qτ,ω (2.6.38) τ=1 i∈Succ(j) τ=1 t t−1 j X X i X j =⇒ I1 − q˜τ,ω + q˜τ,ω ≥ 0, i∈Succ(j) τ=1 τ=1

33 where (2.6.38) follows from the fact that

bx1 + x2 + ··· + xnc ≥ bx1c + bx2c + ··· + bxnc ∀n ∈ N, ∀x1, x2, ··· , xn ∈ R.

Thus, the feasibility ofπ ˜ with respect to (2.6.33) is established, and the feasibility of π˜ with respect to (2.6.29) naturally follows.

• The feasibility ofπ ˜ with respect to (2.6.31) and (2.6.32) follows from that of π and the following facts:

(a) bx + yc − bxc ≥ 0 and bx + yc − bxc ≤ dye for all x, y ≥ 0.

(b) dxe ≤ y for x ∈ R, y ∈ Z with x ≤ y.

• The feasibility ofπ ˜ with respect to (2.6.30) follows from that of π, (2.6.35), (2.6.36),

j j z˜t,ω = zt,ω, and the following fact:

if x − y ≥ 0 for x ∈ R, y ∈ Z, then bxc − y ≥ 0.

Thus, we have demonstrated the feasibility ofπ ˜. We now proceed to show TC(˜π) ≤

1 β max{ γ , 1+βΨ }· TC(π). It suﬃces to show the following claims for every ω: (i) the back- ordering cost underπ ˜ is equal to that under π, (ii) the holding cost underπ ˜ is at most

1 γ times the holding cost under π, and (iii) the ordering/shipping cost underπ ˜ is at most β max{1, 1+βΨ } times the ordering/shipping cost under π. To simplify notation, we replace j j j j j j j j j j zt,ω, qt,ω, It+1,ω, St+1,ω, and Bt+1,ω by zt , qt , It+1, St+1, and Bt+1, respectively. j j Claim (i) follows fromz ˜t = zt for j ∈ N∗ and t ∈ T . We now establish claim (ii). For every node j ∈ N , let AP(j) denote the collection of n o ˜j i all its (not necessarily immediate) predecessors. Let ht = max ht : i ∈ AP(j) ∪ {j} for j ∈ N and t ∈ T . It follows that

˜j j ˜j ˜i ht ≥ ht and ht ≥ ht for j ∈ N , i ∈ AP(j) ∪ {j}, and t ∈ T . (2.6.39)

34 It follows directly from Assumption LH that

1 h˜i ≤ hj for j ∈ N , i ∈ AP(j) ∪ {j}, and t ∈ T . (2.6.40) t γ t

We now compare the holding cost for t ∈ T . Notice that

Λ−1 1 X X 1 X 1 H (π; ω) = hjIj + hjSj γ t γ t t+1 γ t t+1 λ=0 j∈Nλ j∈N∗ Λ−1 X X ˜j j X ˜j j ≥ ht It+1 + ht St+1 (2.6.41)

λ=0 j∈Nλ j∈N∗ Λ−1 t t ! t t ! X X ˜j X j j X X i X ˜j j X j X j = ht qτ + I1 − qτ + ht S1 + qτ − zτ

λ=0 j∈Nλ τ=1 i∈Succ(j) τ=1 j∈N∗ τ=1 τ=1 (2.6.42)

Λ t Λ−1 t ! X X ˜j ˜Pred(j) X j X X ˜j j X ˜j j X j = ht − ht qτ + ht I1 + ht S1 − zτ ,

λ=0 j∈Nλ τ=1 λ=0 j∈Nλ j∈N∗ τ=1 (2.6.43)

Λ t Λ−1 t ! X X ˜j ˜Pred(j) X j X X ˜j j X ˜j j X j ≥ ht − ht q˜τ + ht I1 + ht S1 − z˜τ

λ=0 j∈Nλ τ=1 λ=0 j∈Nλ j∈N∗ τ=1 (2.6.44)

Λ−1 X X ˜j ˜j X ˜j ˜j = ht It+1 + ht St+1

λ=0 j∈Nλ j∈N∗ Λ−1 X X j ˜j X j ˜j ≥ ht It+1 + ht St+1 (2.6.45)

λ=0 j∈Nλ j∈N∗

= Ht(˜π; ω), where (2.6.41) follows from (2.6.40), (2.6.42) follows from (2.6.34) and (2.6.35), and (2.6.44)

Pred(0) and (2.6.45) follow from (2.6.39). In the above deduction, we also take ht = 0 for convenience, and make use of the following identity:

Λ−1 t Λ t X X ˜j X X i X X ˜Pred(j) X j ht qτ = ht qτ ,

λ=0 j∈Nλ i∈Succ(j) τ=1 λ=1 j∈Nλ τ=1

35 Claim (ii) then follows.

j j j j We now establish claim (iii). For j ∈ N and t ∈ T , let xt = qt − bqt c; then, xt ∈ [0, 1). It suﬃces to show that

T T X n β o X h i αt−1cjq˜j ≤ max 1, · αt−1 cjbqjc + f j(xj) (2.6.46) t t 1 + βΨ t t t t t=1 t=1

for each j ∈ N . Noticing that

$ t % $ t−1 % $ t % $ t−1 % j X h j j i X h j j i j X j X j q˜t = bqτ c + xτ − bqτ c + xτ = bqt c + xτ − xτ , τ=1 τ=1 τ=1 τ=1

we claim that (2.6.46) is equivalent to

T T $ t % $ t−1 %! T X X X X n β o X αt−1cjbqjc + αt−1cj xj − xj ≤ max 1, · αt−1cjbqjc t t t τ τ 1 + βΨ t t t=1 t=1 τ=1 τ=1 t=1 T n β o X + max 1, · αt−1f j(xj). 1 + βΨ t t t=1

Therefore, to establish (2.6.46), it suﬃces to show that

T $ t % $ t−1 %! T X X X n β o X αt−1cj xj − xj ≤ max 1, · αt−1f j(xj) (2.6.47) t τ τ 1 + βΨ t t t=1 τ=1 τ=1 t=1

j for each j ∈ N . From now on, to simplify notation, we use xt to represent xt for t ∈ T . Our objective is now to show that

T $ t % $ t−1 %! T X X X n β o X αt−1cj x − x ≤ max 1, · αt−1f j(x ), (2.6.48) t τ τ 1 + βΨ t t t=1 τ=1 τ=1 t=1

where xt ∈ [0, 1) for t ∈ T . We assume xt ∈ (0, 1) for each t, since the proof is similar for

the case in which xt ∈ (0, 1) for some values of t and xt = 0 for the other values of t. Since

j ft (·) is concave non-decreasing in [0, 1], we have

1 − x x − f j(x ) ≥ t f j() + t f j(1) for x ∈ (0, 1) and ∈ (0, x ). t t 1 − t 1 − t t t

36 j + ft (x) j j By letting → 0 , and recalling that limx→0+ j ≥ Ψ (Ψ ∈ [0, 1]) and ft (1) = ct , we have ft (1)

j j j ft (xt) ≥ (1 − xt) lim ft () + xtft (1) →0+

j j ≥ (1 − xt)Ψft (1) + xtft (1)

j = ct xt + Ψ(1 − xt)

for xt ∈ (0, 1). Then to show (2.6.48), it suﬃces to show that

T $ t % $ t−1 %! T X X X n β o X αt−1cj x − x ≤ max 1, · αt−1cjx + Ψ(1 − x ). t τ τ 1 + βΨ t t t t=1 τ=1 τ=1 t=1 (2.6.49)

jPt k jPt−1 k ˜ Since τ=1 xτ − τ=1 xτ ∈ {0, 1} for t ∈ T , we let T be the collection of t ∈ T such jPt k jPt−1 k jPt k jPt−1 k ˜ that τ=1 xτ − τ=1 xτ = 1. That is, (a) τ=1 xτ − τ=1 xτ = 1 for t ∈ T jPt k jPt−1 k ˜ and (b) τ=1 xτ − τ=1 xτ = 0 for t ∈ T \T . Without loss of generality, we assume ˜ T = {t1, t2, ··· , tk}, where k ∈ {0, 1, ··· ,T } and 1 < t1 < t2 < ··· < tk ≤ T . Then, to show (2.6.49), we only have to show

X n β o X αt−1cj ≤ max 1, · αt−1cjx + Ψ(1 − x ). (2.6.50) t 1 + βΨ t t t t∈T˜ t∈T

Before establishing (2.6.50), we ﬁrst claim that (xt)t∈T satisﬁes

t −1 t Xl Xl l − 1 < xτ < l ≤ xτ < l + 1 for l ∈ {1, 2, ··· , k}. (2.6.51) τ=1 τ=1 To see this, let us examine the sequence

t Xk x1, x1 + x2, x1 + x2 + x3, ··· , xτ . τ=1

Noticing that x1 < 1 and xt1 < 1, and by the deﬁnition of t1, we have

t −1 t X1 X1 0 < x1 < x1 + x2 < ··· < xτ < 1 ≤ xτ < 2. τ=1 τ=1

37 Thus, we have established (2.6.51) for l = 1. Suppose (2.6.51) holds for l = l0, i.e., we have

tl −1 tl X0 X0 l0 − 1 < xτ < l0 ≤ xτ < l0 + 1. τ=1 τ=1

Noticing that x < 1 and using the deﬁnition of t , we then have tl0+1 l0+1

tl tl +1 tl +2 tl +1−1 tl +1 X0 X0 X0 0X X0 l0 ≤ xτ < xτ < xτ < ··· < xτ < l0 + 1 ≤ xτ < l0 + 2. τ=1 τ=1 τ=1 τ=1 τ=1

Thus, we have established (2.6.51) for l = l0 + 1. By induction, we have established (2.6.51) for l ∈ {1, 2, ··· , k}. We now prove (2.6.50) by cases:

β Case 1: 1+βΨ ≥ 1. Then β(1 − Ψ) ≥ 1. It suﬃces to show

X t−1 j X t−1 j β α ct [xt + Ψ(1 − xt)] − (1 + βΨ) α ct ≥ 0. (2.6.52) t∈T t∈T˜

j j j j By Assumption LN, there exists c ≥ 0 such that ct ∈ [c , βc ] for t ∈ T . Then

X t−1 j X t−1 j β α ct [xt + Ψ(1 − xt)] − (1 + βΨ) α ct t∈T t∈T˜ ! X t−1 j X t−1 j X t−1 j X t−1 j =β(1 − Ψ) α ct xt − α ct + βΨ α ct − α ct t∈T t∈T˜ t∈T t∈T˜

X t−1 j X t−1 j X t−1 j X t−1 j =β(1 − Ψ) α ct xt + β(1 − Ψ) α ct xt − α ct + βΨ α ct t∈T \T˜ t∈T˜ t∈T˜ t∈T \T˜

X t−1 j X t−1 j X t−1 j =β α ct xt − α ct [1 − β(1 − Ψ)xt] + βΨ α ct (1 − xt) t∈T \T˜ t∈T˜ t∈T \T˜

X t−1 j X t−1 j ≥ α βct xt − α ct [1 − xt] (2.6.53) t∈T \T˜ t∈T˜

X t−1 j X t−1 j ≥ α βc xt − α βc [1 − xt] (2.6.54) t∈T \T˜ t∈T˜ j X t−1 X t−1 =βc α xt − α , t∈T t∈T˜

38 where (2.6.53) follows from β(1 − Ψ) ≥ 1 and xt ∈ (0, 1), and (2.6.54) follows from Assump- tion LN. It only remains to show that

X t−1 X t−1 ∆ = α xt − α ≥ 0. (2.6.55) t∈T t∈T˜

This follows by repeatedly applying (2.6.51).

β 1 Case 2: 1+βΨ ≤ 1. Then Ψ ≥ 1 − β . To prove (2.6.50), we only have to show that

X 1 X αt−1cj[x + (1 − )(1 − x )] − αt−1cj ≥ 0. t t β t t t∈T t∈T˜ In fact, we have

X 1 X αt−1cj[x + (1 − )(1 − x )] − αt−1cj t t β t t t∈T t∈T˜ j X h 1 1 i X c = αt−1cj 1 − + x − αt−1 t (1 − x ) t β β t β t t∈T \T˜ t∈T˜

X t−1 j X t−1 j ≥ α c xt − α c (1 − xt) (2.6.56) t∈T \T˜ t∈T˜ j X t−1 X t−1 = c α xt − α t∈T t∈T˜ ≥ 0, (2.6.57)

1 1 where (2.6.56) follows from Assumption LN and 1 − β + β xt ≥ xt. The proof of (2.6.57) is the same as that of (2.6.55). This completes the proof of Theorem 2.4.1.

39 CHAPTER 3

ON FINDING PROCESS CAPACITY

3.1 Introduction

The well-known “process view” of a firm considers the firm as a process that transforms inputs into outputs using a network of activities that consume resources; see, e.g., (Anupindi et al., 2014). The basic unit transformed by the process is commonly referred to as a flow unit. Three key measures that are typically used to assess the operational effectiveness of a process are flow time, flow rate (or throughput), and inventory. The capacity of a process is its maximum sustainable flow rate, i.e., the maximum number of completed flow units that can be produced per unit of time in the long run. Thus, the question of determining the capacity of a process is fundamental in Operations Management (OM).

Most OM textbooks use the following simple approximation to study capacity of a process that produces a single product: The capacity of each resource is ﬁrst calculated by examining that resource in isolation. Process capacity is then deﬁned as the capacity of a bottleneck resource, i.e., the resource with the least capacity; see, e.g., (Anupindi et al., 2014), (Cachon and Terwiesch, 2013), (Heizer and Render, 2014), (Stevenson, 2014). In a recent paper,

(Gurvich and Van Mieghem, 2015) show that, in the presence of simultaneous collaboration and multitasking of resources, and when resources are indivisible, the bottleneck formula can be incorrect and indeed signiﬁcantly inaccurate. In fact, there are examples for which the ratio of the capacity of a bottleneck resource to the capacity of the process is arbitrarily large. The authors provide a necessary and suﬃcient condition under which the bottleneck formula is valid. As a special case, if the collaboration network of a process has what they refer to as a “nested collaboration architecture”, the bottleneck formula indeed provides the true process capacity. The analysis in (Gurvich and Van Mieghem, 2015) naturally raises the following two questions:

40 1. How diﬃcult is it to compute process capacity?

2. Given that the bottleneck formula is unsatisfactory as an approximation of process

capacity, can it be replaced by another simple but close approximation of process

capacity?

We answer Question 1 by showing that the problem of ﬁnding the capacity of a single-product

process is strongly NP-hard. A key step in showing this result is to establish a connection

between process capacity and graph coloring. The collaboration graph Hc of a process P, as deﬁned in (Gurvich and Van Mieghem, 2015), is a graph where (i) each node corresponds

to an activity of the process, and (ii) there exists an edge between two nodes if and only

if the corresponding two activities share at least one resource. By restricting attention to

a subclass of processes where there is one unit of each resource and the processing time

of each activity is 1 time unit, we show (Lemma 3.3.3; Section 3.3.3) that the capacity of

the process equals the reciprocal of the fractional chromatic number of Hc – this number is deﬁned in Section 3.3.3. On the other hand, an arbitrary graph can be shown to be

the collaboration graph of an appropriately-deﬁned process with one unit of each resource

(Lemma 3.4.1; Section 3.4). These two results enable us to reduce the problem of ﬁnding

the fractional chromatic number of a graph to that of ﬁnding the capacity for processes

in the above-mentioned subclass. The hardness result (Theorem 3.4.1; Section 3.4) then

follows from the fact that the problem of ﬁnding the fractional chromatic number is strongly

NP-hard (Lund and Yannakakis, 1994).

It is also known that for any δ > 0, it is NP-hard to approximate the fractional chromatic

1 −δ number within a multiplicative factor of n 7 , where n is the number of nodes in the input graph (Lund and Yannakakis, 1994; Bellare et al., 1998; Kennedy, 2011). In particular, this rules out the possibility of an eﬃcient constant-factor approximation algorithm, for any

41 constant. Using Lemma 3.3.3 and Lemma 3.4.1, it is easy to see that the same inapproximability result holds for process capacity (Theorem 3.4.2; Section 3.4.1). Thus, the answer to Question (2) above is in the negative. Since it is unlikely that we can, in general, efficiently compute (or closely-approximate) process capacity, a natural quest is to identify subclasses of processes for which one can efficiently determine process capacity. Of course, one such subclass has already been identified in (Gurvich and Van Mieghem, 2015) – namely, processes for which the bottleneck formula correctly computes capacity. In Section 3.5, we show that capacity can be efficiently computed for processes for which the collaboration graph is a perfect graph (Theorems 3.5.1 and 3.5.2). This broadens the class of processes identified in (Gurvich and Van Mieghem, 2015) in the following precise sense: The collaboration graph of any process for which the bottleneck formula correctly computes capacity is a perfect graph (Theorem 3.5.3). From a practical viewpoint, our analysis results in a natural hierarchy of subclasses of policies that require an increasing amount of sophistication in implementation and management. We now elaborate. We show that, to determine process capacity, it is sufficient to restrict attention to the class of cyclic policies, i.e., policies that repeatedly execute a finite sequence of activities (Lemma 3.2.2; Section 3.2.2). The class of cyclic policies can be further organized into a hierarchy of subclasses of policies whose structure becomes in- creasingly complex. The simplest subclass in this hierarchy is that of 1-unit cyclic policies, namely those that repeatedly executes a finite sequence of activities such that during each execution exactly 1 unit of the product completes its processing. Cyclic policies of higher order, namely k-unit cyclic policies for k ≥ 2, are analogously defined. A general subclass Ck, k ≥ 1, of our hierarchy consists of all j-unit cyclic policies satisfying j ≤ k. While process capacity is the maximum long-term process rate achievable over all cyclic policies, we provide a precise expression for the maximum process rate over policies in each subclass Ck, k ≥ 1, (Lemma 3.3.1 and Theorem 3.6.1; Section 3.3), thus highlighting the tradeoff between operational difficulty and the achievable process rate.

42 3.1.1 A Brief Overview of the Related Literature

Apart from (Gurvich and Van Mieghem, 2015), the literature that is most closely related to our work is the one on the cyclic jobshop problem and its variants. A jobshop is a production system in which there are resources (also referred to as “machines”) and multiple types of products to be processed on these resources; the set of resources required as well as the sequence of processing through the resources may differ across product types. The production of each copy of a product is referred to as a job. In a cyclic schedule for a jobshop, the same set of jobs is repetitively processed to meet a certain production requirement. The smallest of such sets is referred to as the minimal part set (MPS; see (Hitz, 1980)). For example, if the production requirement is 1000 copies of product A, 3000 copies of product B, and 5000 copies of product C, then the MPS is (1A, 3B, 5C) and the processing of the MPS is repeated 1000 times. The objective of the cyclic jobshop scheduling problem is to find a schedule in which the MPS is repetitively produced over the infinite horizon, to maximize the flow rate. For complexity results on the cyclic jobshop scheduling problem and its variants, we refer the reader to (Levner et al., 2010) for details. In the special case of the repetitive production of a single job (i.e., a single product), the MPS has exactly 1 job, and the cyclic jobshop scheduling problem becomes one of obtaining the best 1-unit cyclic schedule. The problem we study is fundamentally different in that the capacity of a process is defined as the maximum sustainable flow rate over all the possible schedules (cyclic as well as acyclic). In particular, the cyclic schedules considered in determining process capacity include all k-unit cyclic schedules, k ≥ 1. Thus, the space of policies for determining process capacity is much larger. The problem of maximizing the flow rate over the subclass of 1-unit cyclic schedules is known to be NP-hard (Roundy, 1992). It is important, however, to note that this does not imply the hardness of the problem of optimizing over all k-unit cyclic schedules, k ≥ 1. Indeed, in optimizing over all k-unit cyclic schedules, it is not even clear upfront if the maximum flow rate (i.e., process capacity) is

43 achieved by a k-unit cyclic schedule, for some ﬁnite k. Our results in this paper show that it is indeed NP-hard to optimize over all k-unit cyclic schedules and that there always exists a ﬁnite k for which a k-unit cycle achieves the capacity of the process.

There are several other types of cyclic scheduling problems in the literature; for example, the cyclic flowshop problem (McCormick et al., 1987; McCormick an Rao, 1994), the cyclic robotic-cell scheduling problem (Dawande et al., 2007; Brauner, 2008), and the cyclic project scheduling problem (Levner and Kats, 1998; Kampmeyer, 2006). All these are significantly different from the problem of determining process capacity that we consider in this paper.

For brevity, we avoid a detailed comparison and refer the reader to the survey paper on complexity of cyclic scheduling problems by (Levner et al., 2010). We note that in the robotic-cell scheduling literature, the complexity of obtaining the best cyclic schedule (i.e., the best k-unit cyclic schedule over all k ≥ 1) is a well-known open problem (Dawande et al., 2007).

Our paper is also closely related to the recent literature on collaboration and multitasking. As mentioned earlier in this section, (Gurvich and Van Mieghem, 2015) study when and why the bottleneck formula correctly calculates process capacity. (Gurvich and Van

Mieghem, 2016) study the trade-oﬀ between task priorities and capacity. For ongoing empirical investigations, see, e.g., (Wang et al., 2016) and (Gurvich et al., 2016).

3.2 Preliminary Results

We ﬁrst formally deﬁne the capacity of a single-product process in Section 3.2.1. Then,

Section 3.2.2 establishes some preliminary results that will be useful for our subsequent analysis. In Section 3.2.3, we note the consistency of our deﬁnition of process capacity with the formulation in (Gurvich and Van Mieghem, 2015).

44 3.2.1 Formal Deﬁnition of Process Capacity

We consider a deterministic process that produces a single product. Let V represent the set of activities required to produce one ﬂow unit of the product. Without loss of generality, we assume that each ﬂow unit needs each activity v ∈ V exactly once (for otherwise, we can

re-name the activities to satisfy this property). Let τ : V → N denote the processing time, where N is the set of strictly positive integers. Let W represent the set of resources. For

w ∈ W , let Nw denote the number of units of resource w. Let Wv denote the set of resources that are required to perform activity v for v ∈ V ; activity v requires all the resources in

Wv simultaneously. To capture precedence relationships (if any) among the activities, let G = (V,A) denote the directed acyclic graph deﬁned as follows: Each node v ∈ V corresponds

to an activity, and the directed edge eij = (vi, vj) ∈ A represents the precedence relationship

vi → vj (i.e., activity vi must precede activity vj); thus, A = ∅ corresponds to the case without any precedence constraints. Deﬁne   1, if preemptive processing of the activities is not allowed, I :=  0, otherwise. A process is deﬁned by P = V, A, τ, W, {Nw; w ∈ W }, {Wv; v ∈ V }, I , consolidating this notation.

For j ∈ N, flow unit j refers to the jth flow unit that enters the process since time unit 1. We define a generic policy of implementing a process, π : N2 → {0, 1}|V |, as follows: For j, t ∈ N, let

v π(j, t) := (Ij,t; v ∈ V ), (3.2.1)

where for v ∈ V ,  1, if activity v is executed on ﬂow unit j during time unit t, v  Ij,t :=  0, otherwise. We consider the following constraints on π:

45 (a) Resource availability constraints. For t ∈ N and w ∈ W , we have

X X v Ij,t ≤ Nw. (3.2.2) j∈N v:w∈Wv That is, during any time unit t, the total number of units in use of a resource cannot exceed the total number of the available units of that resource.

(b) Activity constraints. For j ∈ N and v ∈ V , we have

X v Ij,t = τ(v). (3.2.3) t∈N That is, activity v is scheduled on ﬂow unit j exactly once.

vk vl max t : t ∈ N,Ij,t = 1 < min t : t ∈ N,Ij,t = 1 (3.2.4)

That is, if we have a precedence relationship vk → vl, then on any ﬂow unit j, activity vk

must be scheduled before activity vl.

(d) Non-preemption constraints. For j, T ∈ N and v ∈ V , we have " t T # v Y v Y v I· Ij,T · max t : t ≥ T, Ij,i = 1 − min t : t ≤ T, Ij,i = 1 − τ(v) = 0, i=T i=t (3.2.5)

v where we take Ij,t = 0 for t < 0 for convenience. That is, if preemption is not allowed v (I = 1) and activity v is on ﬂow unit j during time unit T (Ij,T = 1), then there must exist τ(v) consecutive time units in {T − τ(v) + 1,T − τ(v) + 2, ··· ,T + τ(v) − 1} during which activity v is on ﬂow unit j.

It is easy to see that (3.2.2)–(3.2.5) constitute the feasibility constraints1 on policy π.

Let Π denote the collection of all feasible policies π for process P. Let P0 represen- t the same process as P except without precedence and non-preemption constraints; i.e.,

1Note that, if two activities do not share any resource, our model allows the two activities to be scheduled on the same ﬂow unit in the same time unit. For the case in which more than one activity cannot be performed on the same ﬂow unit simultaneously (i.e., in the same time unit), all our results carry over.

46 P0 = V, ∅, τ, W, {Nw; w ∈ W }, {Wv; v ∈ V }, 0 . Let Π0 denote the collection of all feasible

policies π for process P0. Then, Π ⊆ Π0.

π For π ∈ Π0 and T ∈ N, let N [1,T ] represent the number of ﬂow units that enter the process after the beginning of time unit 1 and are completed by the end of time unit T under π. That is,

T π X v N [1,T ] := j : j ∈ N, Ij,t = τ(v) for v ∈ V . (3.2.6) t=1

π For π ∈ Π0, let r represent the process rate (also referred to as the throughput rate) under π. That is,

N π[1,T ] rπ := lim inf . (3.2.7) T →∞ T

Let

(Process Capacity) µ(P) := sup rπ (3.2.8) π∈Π

represent the capacity of the process. The objective is to determine µ(P).

3.2.2 Suﬃciency of Cyclic Policies and Redundance of Precedence and Non- Preemption Constraints

In this section, we ﬁrst deﬁne some notions and introduce the corresponding notation for our subsequent analysis. The main result of this section is Lemma 3.2.2, which establishes that precedence constraints and non-preemptive processing constraints can be ignored for the purpose of computing process capacity.

For k ∈ N, a policy π is said to be a k-unit cyclic policy if the following is satisﬁed: There exists Lπ ∈ N such that   π π π(j, t − nL ), if t ≥ nL + 1, π(nk + j, t) = for n ∈ N and j ∈ {1, 2, ··· , k}. (3.2.9)  0, otherwise.

47 That is, the schedule for ﬂow units (j−1)k+1, (j−1)k+2, ··· , jk, j ∈ N, is exactly the same as the schedule for ﬂow units 1, 2, ··· , k, except with a time shift of (j − 1)Lπ time units.

For i ∈ N, the ﬁnite sequence of activities in time units (i − 1)Lπ + 1, (i − 1)Lπ + 2, ··· , iLπ, is said to be a cycle of the process. The integer Lπ is said to be the cycle length of π. Notice that a k-unit cyclic policy can also be an nk-unit cyclic policy for any n ∈ N by deﬁnition. For k ∈ N, let Πk denote the collection of all feasible k-unit cyclic policies (i.e., satisfying constraints (3.2.2)-(3.2.5) and (3.2.9)) for process P.

A policy π is said to be cyclic if π is a k-unit cyclic policy for some k ∈ N. The result below is a direct implication of the activity constraints and the deﬁnition of a cyclic policy.

k π Lemma 3.2.1. Consider any k ∈ N and any π ∈ Π . There exists ns ∈ N that satisﬁes the following properties:

(a) Steady-State Property.

π (i) For n ≥ ns , we have

π π π nL ns L X X v X X v Ij,t = Ij,t for v ∈ V, (3.2.10) π π π j∈N t=(n−1)L +1 j∈N t=(ns −1)L +1 π π where L is the cycle length of π. That is, from cycle ns onwards, the amount of time spent on activity v, v ∈ V , is the same in every cycle.

π (ii) For n ≥ ns , exactly k ﬂow units exit the process (i.e., complete their processing) in the nth cycle. Thus, we have N π[1,T ] k rπ = lim = . (3.2.11) T →∞ T Lπ

(b) Balanced Property. We have

nLπ X X v π Ij,t = kτ(v) for v ∈ V, n ≥ ns . (3.2.12) j∈N t=(n−1)Lπ+1 π That is, from cycle ns onwards, the amount of time spent on activity v, v ∈ V , equals k times the duration of activity v in every cycle.

48 The proof of Lemma 3.2.1 is in Section 3.7.1. This result will be useful for the proof of

Lemma 3.2.2, and for the subsequent construction of graphs for diﬀerent classes of policies of a process in Sections 3.3 and 3.6.

Let Πcyclic denote the collection of all feasible cyclic policies for process P, i.e., Πcyclic =

∞ k cyclic cyclic π ∪k=1Π ; then, Π ⊆ Π. For any π ∈ Π , let ns be the smallest positive integer that satisﬁes statements (a) and (b) of Lemma 3.2.1. The process is said to be in steady state

π π π after the ﬁrst (ns − 1) cycles (i.e., in any time unit t for t ≥ (ns − 1)L + 1).

For any process P, recall that P0 is the same process as P except that it does not have any

cyclic precedence and non-preemption constraints. Let Π0 denote the collection of all feasible,

cyclic cyclic policies (i.e., satisﬁng (3.2.2)–(3.2.3) and (3.2.9)) for process P0. Then, Π ⊆

cyclic Π0 ⊆ Π0. The result below establishes that, to ﬁnd µ(P), we can restrict our attention to feasible, cyclic policies, and drop the precedence and non-preemption constraints, without loss of generality.

Lemma 3.2.2. We have

π µ(P) = µ(P0) = sup r . (3.2.13) cyclic π∈Π0

The proof of Lemma 3.2.2 is in the appendix. Our proof is constructive and illustrates

cyclic cyclic how to create an optimal policy π ∈ Π for process P, given an optimal policy π ∈ Π0 for process P0.

Due to Lemma 3.2.2, we will henceforth restrict our attention to process P0

cyclic and policies in Π0 . We also assume that there is one unit of each resource w, w ∈ W , i.e.,

Nw = 1 for w ∈ W. (3.2.14)

This assumption will be relaxed in Section 3.6.

49 k For k ∈ N and process P0, let Π0 : be the collection of all feasible, k-unit cyclic policies.

k j cyclic ∞ k k k j Let C be the union of Π0, 1 ≤ j ≤ k. Then, Π0 = ∪k=1Π0 and C = ∪j=1Π0. For

Process P0, we construct the corresponding expanded process Pexp through the following transformation: Every activity v, v ∈ V , with τ(v) > 1, is divided into τ(v) activities with each having a duration of 1 time unit. It is easy to see that the capacities of Processes P0 and Pexp are equal, i.e.,

µ(P0) = µ(Pexp). (3.2.15)

Let Vexp be the set of activities in Process Pexp.

3.2.3 Static Planning Problem for Collaboration

We now present the static planning problem for collaboration (SPPC) in (Gurvich and Van

Mieghem, 2015) and note its connection with our deﬁnition of process capacity.

An activity subset S ⊆ V is feasible if no two activities in S share a resource; i.e., all the activities in S can be scheduled simultaneously. Let C denote the collection of all feasible activity subsets: C = S ⊆ V : Wv ∩ Wv˜ = ∅ for v, v˜ ∈ S . The linear programming formulation of SPPC for one product under assumption (3.2.14) is as follows:

ρnet(λ) = min ρ X s.t. xS = λτ(v) for v ∈ V, (3.2.16) S:v∈S X xS ≤ ρ, xS ≥ 0 for S ∈ C, S∈C for λ > 0, where xS, S ∈ C, is interpreted as the long-run proportion of time allocated for S, and ρnet is interpreted as the minimal network utilization. The network capacity, λ∗, is such that ρnet(λ∗) = 1, which indicates that the network is fully utilized; it is easy to see that λ∗ is unique.

50 For the LP formulation of SPPC in general, i.e., without assumption (3.2.14), we refer

the reader to (Gurvich and Van Mieghem, 2015). Deﬁne

n ∗ X o FEAS = (λ; xS,S ∈ C) : 0 ≤ λ ≤ λ ; xS = λτ(v) for v ∈ V ; xS ≥ 0 for S ∈ C . S:v∈S

cyclic We note a “quasi” one-to-one correspondence between Π0 and FEAS:

cyclic (a) For any π ∈ Π0 , it is easy to see from the deﬁnition of a cyclic policy that the

π long-run proportion of time xS for S, S ∈ C, under π exists. By Lemma 3.2.1, we have

π π π P π (λ ; xS,S ∈ C) ∈ FEAS, where λ = S:v∈S xS for any v ∈ V .

(n) cyclic (b) For any (λ; xS,S ∈ C) ∈ FEAS, we can ﬁnd a sequence {π }n∈N ⊆ Π0 such that,

(n) (n) for S ∈ C, the long-run proportion of time xS under π approaches xS as n → ∞. This is because every real number is the limit of a sequence of rational numbers; for

details, see, e.g., (Rudin, 2013).

This quasi one-to-one correspondence implies the following result:

Theorem 3.2.1. (Gurvich and Van Mieghem, 2015) We have

sup rπ = λ∗. cyclic π∈Π0

This theorem, combined with Lemma 3.2.2, implies

Lemma 3.2.3. We have

∗ µ(P0) = λ .

net ∗ net ∗ Let Cexp, ρexp(λ), and λexp for Process Pexp be similarly deﬁned as C, ρ (λ), and λ for

Process P0.

51 3.3 Characterizing Process Rates for Subclasses of Cyclic Policies

The notion of cyclic policies results in a natural hierarchy of subclasses of these policies:

1 2 3 π C ⊆ C ⊆ C ⊆ · · · . From Lemma 3.2.2, it is immediate that the sequence {supπ∈Ck r } increases in k and converges to µ(P0), the process capacity. This hierarchy highlights the trade-oﬀ between the sophistication needed in managing/implementing a cyclic policy and the corresponding process rate achieved as compared to the true process capacity: The higher the value of k is, the closer the optimal process rate over Ck is as compared to process capacity, and yet, the more complicated it is to implement the corresponding policy.

In this section, we characterize the maximum process rate for each subclass Ck, k ∈ N, using the notion of graph coloring. The main result of this section is Theorem 3.3.1, which specifies an exact expression for the maximum process rate over Ck, k ∈ N. Combining this result with Lemma 3.3.3, which specifies an exact expression for process capacity, we can measure how far the maximum process rate over the policies in Ck is, relative to process capacity. To facilitate the understanding of Theorem 3.3.1, we first consider the special case of optimizing over policies in the simplest subclass C1, namely 1-unit cyclic policies, in Section 3.3.1.

3.3.1 The Maximum Process Rate Over 1-Unit Cyclic Policies

Let H = (V,E) be an arbitrary undirected graph, where V is the set of nodes and E ⊂ V ×V is the set of edges. A node coloring c : V → N of H = (V,E) is an assignment of a color (we use diﬀerent positive integers to indicate distinct colors, e.g., color 1 is diﬀerent from color 2) to each node in V such that no two endpoints of an edge e ∈ E have the same color, i.e.2, c(v) 6= c(˜v) if (v, v˜) ∈ E. The chromatic number of H = (V,E), denoted by χ(H), is the minimum number of distinct colors used in a coloring, over all node colorings of H = (V,E); see, e.g., (Bollobas, 2002).

2Note that (v, v˜) = (˜v, v) since the edges are undirected.

52 The collaboration graph (see, e.g., (Gurvich and Van Mieghem, 2015)), Hc = (V,Ec), for process P0 is deﬁned as follows: Each node v, v ∈ V , corresponds to an activity, and edge e = (v, v˜) ∈ Ec if activities v andv ˜ share at least one common resource, i.e., Ec = (v, v˜) ∈ V × V : Wv ∩ Wv˜ 6= ∅, v 6=v ˜ . (3.3.1)

It is easy to see that, if activities v andv ˜ share a common resource (in which case we say there is a conﬂict), then they cannot be performed simultaneously. The expanded graph,

Hexp = (Vexp,Eexp), for process P0 is deﬁned as the collaboration graph of the corresponding expanded process Pexp. Note that Hc = Hexp if τ(v) = 1 for v ∈ V .

1 By restricting our attention to Π0 (all feasible, 1-unit cyclic policies), we have

1 Lemma 3.3.1. Under Assumption (3.2.14), the maximum process rate over Π0 is

π 1 1 sup r = π = , 1 min 1 L χ(Hexp) π∈Π0 π∈Π0 where rπ and Lπ denote, respectively, the process rate and the cycle length under policy π.

Lemma 3.3.1 is a special case of Lemma 3.3.2. The proof of Lemma 3.3.2 is in Sec-

1 tion 3.7.2. This proof employs a correspondence between Π0 and the class of feasible node 1 colorings of Hexp: Given an arbitrary π ∈ Π0, we can construct a feasible node coloring c(·) out of the schedule of activities under π in a cycle in steady state (see Lemma 3.2.1 for the existence and deﬁnition of steady state), such that the number of colors of c(·) equals the cycle length of π; given a feasible coloring c(·), we can construct a feasible policy π such that the cycle length of π equals the number of colors of c(·).

Remark 3.3.1. Lemma 3.3.1 implies that the problem of determining the maximum process

1 rate over Π0 is NP-complete, since (i) the problem of ﬁnding the chromatic number of a graph is known to be NP-complete; see, e.g., (Garey and Johnson, 1979) and (ii) an arbitrary instance of the latter problem can be polynomially transformed in an instance of the former problem (see Lemma 3.4.1 below). Note, however, that this result does not imply that ﬁnding the process capacity (i.e., optimizing over all cyclic policies) is hard.

53 3.3.2 The Maximum Process Rate Over k-Unit Cyclic Policies

k k k For k ∈ N, we deﬁne graph Hexp(Vexp,Eexp) as follows:

1 2 k (a) For α ∈ {1, 2, ··· , |Vexp|}, we make k copies of activity vα ∈ Vexp; let vα, vα, ··· , vα

j denote these copies. That is, for each copy vα, the processing time and the set of

k resources needed to perform vα are the same as those for activity vα. Let

k n j o Vexp := vα : j ∈ {1, 2, ··· , k}, α ∈ {1, 2, ··· , |Vexp|} .

k That is, Vexp represents the collection of k copies of all the activities in Vexp.

(b) Let

k n i j o Eexp : = (vα, vα): i 6= j, i, j ∈ {1, 2, ··· , k}; α ∈ {1, 2, ··· , |Vexp|}

n i j o ∪ (vα, vβ): i, j ∈ {1, 2, ··· , k};(vα, vβ) ∈ Eexp .

That is, two nodes are connected by an edge if and only if the corresponding activities cannot be scheduled simultaneously.

This construction captures all the essential features that the schedule of activities should

k k satisfy in a cycle in steady state under an arbitrary policy π, π ∈ Π0: (a) The set Vexp captures the balanced property of π (see Lemma 3.2.1 for the steady-state and balanced

k properties), and (b) the set Eexp captures the resource availability constraints. It is easy to

1 1 1 see that Hexp(Vexp,Eexp) = Hexp(Vexp,Eexp), where Hexp(Vexp,Eexp) is the expanded graph of

k process P0. By restricting our attention to Π0, we have

k Lemma 3.3.2. Under assumption (3.2.14), the maximum process rate over Π0, k ∈ N, is

π k k sup r = π = k , (3.3.2) k min k L χ(H ) π∈Π0 π∈Π0 exp where rπ and Lπ, denote, respectively, the process rate and the cycle length under π.

54 By the deﬁnition of Ck, we immediately have

Theorem 3.3.1. Under assumption (3.2.14), the maximum process rate over Ck, k ∈ N, is

π j sup r = max j . π∈Ck 1≤j≤k χ(Hexp)

3.3.3 The Maximum Process Rate Over Cyclic Policies (Process Capacity)

For an arbitrary undirected graph H = (V,E), an independent set I of H is a subset of V such that (vi, vj) ∈/ E for each pair of nodes vi, vj ∈ I. Let I denote the collection of all the independent sets of H. Consider the following linear program:

0 X χ (H) = min xI , I∈I X s.t. xI ≥ 1 for v ∈ V, (3.3.3) I:v∈I

xI ≥ 0 for I ∈ I, where χ0(H) is referred to as the fractional chromatic number of H; see (Scheinerman and

Ullman, 2011) for details. Notice that if we restrict xI to be either 0 or 1, then the resulting linear integer program formulates the chromatic number of H. Thus, the fractional chromatic number is the value of the LP relaxation of this integer program.

Noticing the one-to-one correspondence between the set Cexp of feasible activity subsets of Vexp and the collection I of independent sets of Hexp, and comparing LP’s (3.2.16) and

(3.3.3), we have

net 0 ρexp(1) = χ (Hexp). (3.3.4)

55 For λ > 0, we also notice that

ρnet (λ) ρ exp = min λ λ X xS X xS ρ xS s.t. = 1 for v ∈ V , ≤ , ≥ 0 for S ∈ C , λ exp λ λ λ exp S:v∈S S∈Cexp

= min ρ0

X X 0 s.t. yS = 1 for v ∈ Vexp, yS ≤ ρ , yS ≥ 0 for S ∈ Cexp,

S:v∈S S∈Cexp

net = ρexp(1).

In particular,

net ∗ 1 ρexp(λexp) net ∗ = ∗ = ρexp(1). (3.3.5) λexp λexp Combining Lemma 3.2.3, (3.2.15), (3.3.4), and (3.3.5), we arrive at the following connection:

Lemma 3.3.3. Let P0 be any process satisfying Assumption (3.2.14). Let Hexp = (Vexp,Eexp) be its expanded graph. Then,

1 µ(P0) = 0 , (3.3.6) χ (Hexp)

0 where χ (Hexp) is the fractional chromatic number of Hexp.

This characterization will be used in Section 3.4 for establishing the hardness and inapproximability of process capacity in general, and in Section 3.5 for establishing the pseudo- polynomial-time solvability of process capacity for a special class of processes.

Remark 3.3.2. Let P0 be any process that satisﬁes Assumption (3.2.14) and where the

processing time of each activity is 1 time unit. Let Hc = (V,Ec) be its collaboration graph.

Notice that the expanded graph of P0 is also Hc. It follows from Lemmas 3.3.1 and 3.3.3 that

µ(P0) χ(Hc) π = 0 . supπ∈Π1 r χ (Hc)

56 χ(Gn) (Larsen et al., 1995) demonstrate a sequence {G } of graphs such that the ratio 0 → ∞ n χ (Gn)

as n → ∞. Since each of the graphs Gn is a collaboration graph of an appropriately-deﬁned process (Lemma 3.4.1 in Section 3.4), it follows immediately that there exist processes in which process capacity is arbitrarily higher than the maximum process rate over 1-unit cyclic policies. Consequently, optimizing over 1-unit cyclic policies (which are the simplest among all cyclic policies) will, in general, not give us a good approximation of the true capacity of

the process.

3.4 Hardness and Inapproximability Results

Using the characterization in Lemma 3.3.3 and an auxiliary result (Lemma 3.4.1, below), Theorem 3.4.1 shows the NP-hardness of the problem of ﬁnding process capacity. Theo- rem 3.4.2 establishes an inapproximability result for this problem.

Lemma 3.4.1. Given an arbitrary graph H = (V,E), we can construct (in polynomial time)

a process P0 satisfying assumption (3.2.14) such that H is precisely the collaboration graph of this process.

Theorem 3.4.1. The problem of ﬁnding process capacity is strongly NP-hard.

3.4.1 Inapproximability of Determining Process Capacity

For any δ > 0, it is strongly NP-hard to approximate the fractional chromatic number

1 −δ to within a multiplicative factor of n 7 , where n is the number of nodes in the input graph (Lund and Yannakakis, 1994; Bellare et al., 1998; Kennedy, 2011). Lemma 3.3.3 and Lemma 3.4.1 imply that the same inapproximability result holds for the problem of ﬁnding process capacity as well. Therefore, we have

Theorem 3.4.2. For any δ > 0, it is NP-hard to approximate process capacity to within a

1 −δ multiplicative factor of n 7 , where n is the number of activities in the input process.

57 As a special case, Theorem 3.4.2 implies that it is NP-hard to approximate process

capacity to within a constant factor, for any constant.

3.5 Good News

The successful establishment of the complexity results about process capacity (Theorem-

s 3.4.1 and 3.4.2, Section 3.4) consolidates the usefulness of the graph-coloring characteri-

zation in Lemma 3.4.1. In this section, we further employ this characterization by identi-

fying a special class of processes that is related to the notion of “perfect graphs” (deﬁni-

tion 3.5.1). The importance of this class and perfect graphs is two-folds: (a) There exists a

pseudo-polynomial-time3 algorithm to ﬁnd the capacity of every process in this class (The-

orem 3.5.2). (b) Perfect graphs include the collaboration graph of every process satisfying

the necessary and suﬃcient condition in (Gurvich and Van Mieghem, 2015), under which

the bottleneck formula correctly ﬁnds its capacity (Theorem 3.5.3). We begin our discussion

with the deﬁnition of perfect graphs.

For an arbitrary undirected graph H = (V,E), the complement graph H = (V, E) is a

graph such that (i) V = V , and (ii) (v, v˜) ∈ E iﬀ (v, v˜) ∈/ E, for v, v˜ ∈ V . An induced cycle,

HIC = (VIC ,EIC ), of H (if any) is a graph such that

(i) VIC = {v0, v1, ··· , vl} ⊆ V for some l ∈ N.

(ii) EIC = (v0, v1), (v1, v2), ··· , (vl−1, vl), (vl, v0) .

(iii) For v, v˜ ∈ VIC ,(v, v˜) ∈ E iﬀ (v, v˜) ∈ EIC .

The number of edges of HIC is called the length of HIC . There are several equivalent deﬁnitions of a perfect graph, and the following version is due to (Chudnovsky et al., 2006):

3Explain pseudo-polynomial-time algorithms

58 Deﬁnition 3.5.1. A perfect graph H = (V,E) is a graph such that neither H nor its complement H has an induced cycle whose length is odd and is at least 5.

For further information about perfect graphs, see (Alfonsin, 2001). The following lemma states some important properties of a perfect graph:

Lemma 3.5.1. We have

(a) ((Gr¨otschelet al., 1988)) There is an algorithm to ﬁnd the fractional chromatic number of a perfect graph in polynomial time.

(b) ((Lov´aszet al., 1972)) A graph H is perfect if and only if its corresponding expanded

graph Hexp is perfect.

Lemmas 3.4.1 and 3.5.1(a) combined immediately imply

Theorem 3.5.1. There is a polynomial-time algorithm to ﬁnd the capacity of every process satisfying the following requirements: (i) Assumption (3.2.14) is satisﬁed, (ii) the collaboration graph is perfect, and (iii) the processing time of each activity is 1 time unit.

By noticing that the construction of Hexp out of H to is pseudo-polynomial, and by making use of Lemmas 3.4.1, 3.5.1(a), and 3.5.1(b), we have

Theorem 3.5.2. There is a pseudo-polynomial-time algorithm to ﬁnd the capacity of every process satisfying the following requirements: (i) Assumption (3.2.14) is satisﬁed, and (ii) the collaboration graph is perfect.

We now explain the connection of perfect graphs to the bottleneck formula for ﬁnding process capacity. For Process P0, recall that W = w1, w2, ·, w|W | is the set of resources,

59 and V = v1, v2, ··· , v|V | is the set of activities. Deﬁne the resource-activity matrix A =

(aij)|W |×|V | as follows:   1, if resource wi is needed by activity vj, aij =  0, otherwise, for i ∈ 1, 2, ··· , |W | and j ∈ 1, 2, ··· , |V | . (Gurvich and Van Mieghem, 2015) provides the following parameter-independent (i.e., regardless of the number of units of each resource and the processing time of each activity) necessary and suﬃcient condition under which the bottleneck formula correctly ﬁnds process capacity:

|V | The polyhedron x ∈ R : Ax ≤ 1, x ≥ 0 is integral. (3.5.1)

We have

Theorem 3.5.3. For every process P0 satisfying (3.5.1), i.e., when the bottleneck formula is correct to ﬁnd the capacity of P0, the corresponding collaboration graph Hc is perfect.

The proof is in Section 3.7.4. Note that the inverse of Theorem 3.5.3 is not true. That is, there exists a process whose collaboration graph is perfect whereas whose capacity cannot be correctly found by the bottleneck formula. To see this, consider the following example in (Gurvich and Van Mieghem, 2015):

• V = {v1, v2, v3}; τ(v1) = τ(v2) = τ(v3) = 1.

• W = {w1, w2, w3}; Nw1 = Nw2 = Nw3 = 1.

• Wv1 = {w1, w2},Wv2 = {w1, w3},Wv3 = {w2, w3}.

Note that the collaboration graph of this process is a triangle, and is therefore, a perfect graph per Deﬁnition 3.5.1. However, it has been established in (Gurvich and Van Mieghem,

1 2015) that the capacity given by the bottleneck formula (i.e., 2 ) does not equal the true 1 process capacity (i.e., 3 ).

60 3.6 Extension: Relaxation of Assumption (3.2.14)

In this section, we relax assumption (3.2.14). In the observation that Theorems 3.2.1, 3.4.1,

3.4.2, and 3.5.3, hold regardless of assumption (3.2.14), we will only discuss/comment on the extension of Theorems 3.3.1, 3.5.1, and 3.5.2. In particular, we will derive an expression for the maximum process rate over Ck for k ∈ N (Theorem 3.6.1). We begin with the deﬁnitions of several notions around hypergraphs.

A hypergraph H is a pair H = (X , E), where X is the set of elements (nodes), and E is a set of non-empty subsets of X; the sets in E are referred to as hyperedges. That is,

E ⊆ 2X \{∅}, where 2X is the power set (i.e., the collection of all the subsets) of X.A node coloring c : V → N of a hypergraph H = (X , E) is an assignment of a color to each node of H for which there does not exist any hyperedge S ∈ E, |S| ≥ 2, such that all the nodes in S have the same color, i.e.,

c(v): v ∈ S ≥ 2 if S ∈ E and |S| ≥ 2.

The chromatic number of a hypergraph H = (X , E), denoted by χ(H), is the minimum number of distinct colors in a node coloring, over all node colorings of H.

The corresponding fractional chromatic number of H, denoted by χ0(H), as for χ(H), is similarly deﬁned as χ0(H) as for χ(H) (Section 3.3.3), where H is a graph. That is, χ0(H) is the value of the LP relaxation of the independent-set-related IP formulation for χ(H). For more details about hypergraphs, see (Berge, 1989). Note that an arbitrary graph H(V,E) can also be regarded as a hypergraph.

We now derive an expression for the maximum process rate over Ck for k ∈ N. Deﬁne

k k k hypergraph Hexp(Vexp, Eexp) as follows:

k (a) We ﬁrst deﬁne Vexp: For α ∈ {1, 2, ··· , |Vexp|}, we make k copies of activity vα ∈ Vexp;

1 2 k j let vα, vα, ··· , vα denote these copies. That is, for each copy vα, the processing time

61 k and the set of resources needed to perform vα are the same as those for activity vα. Let

k n j o Vexp := vα : j ∈ {1, 2, ··· , k}, α ∈ {1, 2, ··· , |Vexp|} .

k That is, Vexp represents the collection of k copies of all the activities in Vexp.

k k 0 (b) We then deﬁne Eexp: For v ∈ Vexp, let Wv denote the set of resources that are required

k k to perform v. For S ⊆ Vexp, we have S ∈ Eexp if and only if the following are satisﬁed:

(i) All the activities in S cannot be scheduled simultaneously (not necessarily on the same ﬂow unit). That is,

0 {v : v ∈ S, w ∈ Wv} > Nw for some w ∈ W. (3.6.1)

0 k 0 0 (ii) There does not exist S ⊆ Vexp such that S ( S and S satisﬁes (3.6.1). That is, S is a “minimal set” that satisﬁes (3.6.1).

This construction captures all the essential features that the schedule of activities should

k satisfy in a cycle in steady state under an arbitrary policy π, π ∈ Π0. It is easy to see that

k k k k k k Hexp(Vexp, Eexp) simpliﬁes to Hexp(Vexp,Eexp) for k ∈ N under assumption (3.2.14). We have

k Lemma 3.6.1. The maximum process rate over Π0, k ∈ N, is

π k k sup r = π = k , (3.6.2) k min k L χ(H ) π∈Π0 π∈Π0 exp where rπ and Lπ, denote, respectively, the process rate and the cycle length under π.

The proof of Lemma 3.6.1 is similar to that of Lemma 3.3.2, and is therefore, omitted for brevity. By the deﬁnition of Ck, we immediately have

Theorem 3.6.1. The maximum process rate over Ck, k ∈ N, is

π j sup r = max j . π∈Ck 1≤j≤k χ(Hexp)

62 Corresponding to Lemma 3.3.3, which relates the capacity of a process satisfying assump-

tion (3.2.14) to the fractional chromatic number of a graph, we have

Lemma 3.6.2. Let P0 be any process. Let V be the corresponding set of activities. For every k, k ∈ N, satisfying

k · τ(v) ≥ Nw for w ∈ Wv, v ∈ V, (3.6.3) we have

k µ(P0) = 0 k . (3.6.4) χ (Hexp)

The proof of Lemma 3.6.2 is in Section 3.7.5. Notice that condition (3.6.3) is necessary

for (3.6.4) to hold. To see this, consider the following simple process P0: V = {v}, τ(v) = 1,

0 1 1 W = {w}, Nw = 2, and Wv = {w}. Clearly, µ(P0) = 2. However, χ (Hexp) = 1 since Hexp

1 contains only one node, and therefore, µ(P0) 6= 0 1 . The failure of (3.6.4) for k = 1 χ (Hexp) 1 0 2 is because Hexp violates (3.6.3). On the other hand, it is easy to see that χ (Hexp) = 1,

2 and therefore, µ(P0) = 0 2 . The success of (3.6.4) for k = 2, instead, is because of the χ (Hexp) 2 satisfaction of (3.6.3) by Hexp.

3.7 Proofs of Some Results

3.7.1 Proof of Lemma 3.2.1

Consider an arbitrary k ∈ N and an arbitrary π ∈ Πk. By the activity constraints (3.2.3), there exists nj ∈ N such that ﬂow unit j exits the process (i.e., completes processing) in the

th π nj cycle, for j ∈ {1, 2, ··· , k}. Let ns = max{nj : j = 1, 2, ··· , k}. To simplify notation,

π π we use L and ns to represent L and ns , respectively. We now prove Lemma 3.2.1 using the following steps:

63 Proof of statement (a)(i).

By the definition of ns, flow units 1 through k will have exited the process by the end of time unit nsL. In general, by (3.2.9), flow units 1 through (n − ns + 1)k will have exited the process by the end of time unit nL, n ≥ ns. Thus, for n ≥ ns and v ∈ V , we have

nL nk nL nsk nL X X v X X v X X v Ij,t = Ij,t = Ij+(n−ns)k,t j∈N t=(n−1)L+1 j=(n−ns)k+1 t=(n−1)L+1 j=1 t=(n−1)L+1

nsk nL nsk nsL X X v X X v = Ij,t−(n−ns)L = Ij,t (3.7.1) j=1 t=(n−1)L+1 j=1 t=(ns−1)L+1

nsL X X v = Ij,t, j∈N t=(ns−1)L+1 where the ﬁrst equality of (3.7.1) follows from (3.2.9). Thus, we have established (3.2.10).

Proof of statement (a)(ii).

By the deﬁnition of nj, j ∈ {1, 2, ··· , k}, we know that ﬂow unit nk + j exits the process in

th the (nj + n) cycle for n ∈ N ∪ {0}. Then, by (3.2.9) and the deﬁnition of ns, exactly k ﬂow th units exit the process in the n cycle, n ≥ ns.

We now establish (3.2.11). For any T ∈ N, there uniquely exist bT , rT ∈ N such that

T = bT L + rT and 0 ≤ rT < L. Then,

π π N [1,T ] ≤ N [1, (bT + 1)L] ≤ (bT + 1)k. (3.7.2)

By the deﬁnition of ns, we have

π π N [1,T ] ≥ N [1, bT L] ≥ (bT + 1 − ns)k. (3.7.3)

Combining (3.7.2) and (3.7.3), we have

(b + 1 − n )k (b + 1 − n )k N π[1,T ] (b + 1)k (b + 1)k T s ≤ T s ≤ ≤ T ≤ T . (bT + 1)L T T T bT L By letting T → ∞, we establish (3.2.11).

64 Proof of statement (b).

Take an arbitrary v ∈ V . Notice that at most nk ﬂow units enter the process in the ﬁrst nL

time units for n ∈ N by (3.2.9). Then, we have

nL X X v Ij,t ≤ nkτ(v) for n ∈ N (3.7.4) j∈N t=1 by the activity constraints (3.2.3). Also notice that the ﬁrst k ﬂow units exit the process by

the end of time unit nsL. Then, we have

nL X X v Ij,t ≥ (n − ns + 1)kτ(v) for n ∈ N (3.7.5) j∈N t=1 by (3.2.9). Combining (3.7.4) and (3.7.5), we have

nL (n − n + 1)kτ(v) P P Iv nkτ(v) s ≤ j∈N t=1 j,t ≤ for n ∈ . n n n N

By letting n → ∞, we have

P PnL Iv lim j∈N t=1 j,t = kτ(v). (3.7.6) n→∞ n

Using statement (a)(ii) of Lemma 3.2.1, we have

nsL P PnL Iv X X v j∈N t=1 j,t Ij,t = lim = kτ(v), n→∞ n j∈N t=(ns−1)L+1

which concludes the proof.

3.7.2 Proof of Lemma 3.3.2

π k Take an arbitrary k ∈ . By statement (a) of Lemma 3.2.1, we have sup k r = π . N π∈Π0 inf k L π∈Π0 π π π Since L ∈ , we have inf k L = min k L . It remains to show that N π∈Π0 π∈Π0

min Lπ = χ(Hk ). k exp π∈Π0

65 π k k k We ﬁrst claim that L ≥ χ(Hexp) for π ∈ Π0. To see this, we take an arbitrary π ∈ Π0

k π and construct the following node coloring of Hexp: For l ∈ {1, 2, ··· ,L }, we color the nodes corresponding to the activities that are scheduled in the lth time unit in every cycle

k in steady state using color l. This is a feasible node coloring of Hexp. In fact, for any edge

k e = (v, v˜) ∈ Eexp, activities v andv ˜ must be scheduled in diﬀerent time units in every cycle

k per the construction of Hexp, and therefore, have diﬀerent colors by our node coloring. Since

π π k this coloring uses L colors, the claim that L ≥ χ(Hexp) follows.

π k Next, we claim that min k L ≤ χ(H ). To see this, consider the following schedule of π∈Π0 exp k k activities in every cycle in steady state: There exists a node coloring C(·) of Hexp with χ(Hexp)

k distinct colors. For l ∈ {1, 2, ··· , χ(Hexp)}, we perform the activities corresponding to the nodes that have been colored by color l per C(·) in the lth time unit in every cycle. This is a

feasible schedule because every pair of activities, v andv ˜, can be scheduled simultaneously

if and only if the corresponding nodes are not connected by an edge per the construction

k k π of H . Since this schedule needs χ(H ) time units in every cycle, we have min k L ≤ exp exp π∈Π0 k χ(Hexp).

3.7.3 Proof of Theorem 3.4.1

The problem of ﬁnding the fractional chromatic number of a graph is strongly NP-hard (Lund

and Yannakakis, 1994). Consider an arbitrary instance of this problem, speciﬁed by an input

graph H. Using Lemma 3.4.1, we deﬁne a process P such that H is the collaboration graph

of process P. The result now follows from Lemma 3.3.3.

3.7.4 Proof of Theorem 3.5.3

Let P0 be a process. Assume that the corresponding collaboration graph Hc is not perfect.

Then, either Hc or its complement Hc has an induced cycle whose length is odd and ad

|V | least 5. We only need to show that the polyhedron x ∈ R : Ax ≤ 1, x ≥ 0 is not

66 integral, i.e., there exists an extreme point that is fractional. For details, see (Schrijver,

1998). We prove by cases.

Case 1: Hc has an induced cycle whose length is odd and at least 5.

Let HIC = (VIC ,EIC ) be the induced cycle. Let 2n + 1 be the length of this cycle for some

n ∈ N. For i ∈ {1, 2, ··· , 2n + 1}, since (vi, vi+1) ∈ EIC (v2n+2 = v1 for convenience), there

exists wi ∈ W such that wi ∈ Wvi ∩ Wvi+1 . We also have wi ∈/ Wv for v ∈ VIC \{vi, vi+1} per the deﬁnition of an induced graph. Let A be the resource-activity matrix for P0. Let B be the submatrix of A whose rows correspond to resource wi, i ∈ {1, 2, ··· , 2n + 1}, and whose columns correspond to activity vj, j ∈ {1, 2, ··· , 2n + 1}. Then,   1 1 0 0 ··· 0 0      0 1 1 0 ··· 0 0      0 0 1 1 ··· 0 0       B =  0 0 0 1 ··· 0 0    ......   ......       0 0 0 0 ··· 1 1     1 0 0 0 ··· 0 1 per the deﬁnition of A. That is, there are two 1’s in each row of B. We now solve Bx = 1, where x = (xj : j ∈ {1, ··· , 2n + 1}). Summing all the equations and dividing both sides of the resulting equation by 2 gives

2n + 1 x + x + ··· + x = . 1 2 2n+1 2

1 By subtracting equation 2i, i ∈ {1, 2, ··· , n}, from this equation, we obtain x1 = 2 . By 1 symmetry, we have x1 = x2 = ··· = x2n+1 = 2 . Thus, the extreme point corresponding to B is fractional.

67 Case 2: Hc has an induced cycle whose length is odd and at least 5.

Let HIC = (VIC ,EIC ) be this induced cycle. Let 2n + 1 be its length for some n ≥ 2. Without loss of generality, let

VIC = {v1, ··· , v2n+1} and EIC = (v1, v2), ··· , (v2n, v2n+1), (v2n+1, v1) .

Let HIC = (V IC , EIC ) be the complement graph of HIC , i.e.,

V IC = VIC , and EIC = (vi, vj) : 1 ≤ i < j ≤ 2n + 1 (vi, vi+1) : 1 ≤ i ≤ 2n + 1 ,

where v2n+2 = v1 for convenience. It is easy to see that, for v, v˜ ∈ V IC ,(v, v˜) ∈ EIC iﬀ

(v, v˜) ∈ Ec. We now introduce some notions that will be useful for subsequent analysis. ˜ ˜ For a subset V , V ⊆ V IC , a bond is a resource w, w ∈ W , such that w ∈ Wv for some ˜ ˜ v ∈ V . The number of such v is called the length of bond w for V . A set K, K ⊆ V IC , is

complete if |K| ≥ 2 and every two nodes of K are connected by an edge in HIC . It is easy to see that

|K| ≤ n. (3.7.7)

This is because out of every two nodes, vi and vi+1, of HIC , i ∈ {1, 2, ··· , 2n+1}, at most one

2n+1 can be in K, and therefore, |K| ≤ 2 . It is also easy to see that there exists a complete set K with |K| = n. In fact, {v1, v3, ··· , v2n−1} is such a set. The following is a direct

|V | implication of the integrality of x ∈ R : Ax ≤ 1, x ≥ 0 :

For a complete set K,K ⊆ V IC , there exists a bond of length |K|. (3.7.8)

We prove (3.7.8) by induction on |K|. The proof is trivial for |K| = 2. Assume that (3.7.8) is true for |K| ≤ m (induction hypothesis), where m < n. Let us examine an arbitrary

0 0 0 complete set K with |K | = m + 1. Without loss of generality, let K = {vi1 , vi2 , ··· , vim+1 },

0 where 1 ≤ i1 < i2 < ··· < im+1 ≤ 2n + 1. Let Kj = K \{vij } for j ∈ {1, 2, ··· , m + 1}. By

68 the induction hypothesis, there exists a bond wj of length m for Kj. That is, wj ∈ Wv for

0 v ∈ Kj. Now, suppose (3.7.8) fails for K , i.e., any bond has a length of at most m. Then,

wj ∈/ Wvj . Let B be the submatrix of A whose rows correspond to resource wj, and whose columns correspond to activity vij , for j ∈ {1, 2, ··· , m + 1}. Per the deﬁnition of A, there is one 0 in each row of B, and all the 0’s are on the diagonal. We now solve Bx = 1, where x = (xj; j ∈ {1, ··· , m + 1}). Summing all the equations and dividing both sides of the resulting equation by m gives m + 1 x + x + ··· + x = . 1 2 m+1 m

1 By subtracting equation 1 from this equation, we obtain x1 = m . By symmetry, we have 1 x1 = x2 = ··· = xm+1 = m . Thus, the extreme point corresponding to B is fractional, which |V | contradicts the integrality of x ∈ R : Ax ≤ 1, x ≥ 0 . This contradiction indicates that claim (3.7.8) must be true for |K0| = m + 1. The validity of (3.7.8) then follows from the induction principle. Let K be the collection of complete sets K such that |K| = n. By (3.7.8), there is a

resource w(K), K ∈ K, such that w(K) ∈ Wv for v ∈ K. By (3.7.7), w(K) ∈/ Wv for v∈ / K. We now examine the following system of linear equations: X xv = 1,K ∈ K. (3.7.9) v:v∈K

Let BK be the coeﬃcient matrix of (3.7.9). It is easy to see that BK is a submatrix of A, with

rows corresponding to resource w(K), K ∈ K, and columns corresponding to activity vj,

1 j ∈ {1, 2, ··· , 2n + 1}. In what follows, we will show that x1 = x2 = ··· = x2n+1 = n is the |V | only solution to (3.7.9). This fact will contradict the integrality of x ∈ R : Ax ≤ 1, x ≥ 0 and conclude the proof.

Let f(n) be the number of K, K ∈ K, such that v1 ∈ K, where n indicates the cardinality of K. Note that n ≥ 2. We claim

f(n) = n. (3.7.10)

69 The proof is as follows. For such a K, notice that v2, v2n+1 ∈/ K since v1 ∈ K. Then, f(n) equals the number of ways of choosing n − 1 elements from the (2n − 2)-elements set

{v3, v4, ··· , v2n}, such that at most one of vi and vi+1 can be chosen for 3 ≤ i ≤ 2n − 1. Given this equivalence, f(2) = 2 easily follows. We now ﬁnd a recursive relation for f(n).

(i) If v2n is chosen, then v2n−1 can no longer be chosen. Out of {v3, v4, ··· , v2n−2}, n − 2 elements still need to be chosen. The number of ways of doing this is f(n − 1).

(ii) If v2n is not chosen, then all the n − 1 elements need to be chosen from the (2n − 3)-

elements set {v3, ··· , v2n−1}. There is one way of doing this: picking v3, v5, ··· , v2n−1.

Then, f(n) = f(n − 1) + 1. Combining with f(2) = 2, we establish (3.7.10). We now

ﬁnd |K|. Let us enumerate the complete sets K such that |K| = n and vj ∈ K, for j ∈ {1, 2, ··· , 2n + 1}. We have in total (2n + 1)f(n) such K. Since the same K, K ∈ K, appears n times in this enumeration, it follows that

(2n + 1)f(n) |K| = = 2n + 1. (3.7.11) n

We are now ready to solve (3.7.9). Summing the equations over K, we have

2n+1 X nxvj = 2n + 1, j=1 where the coeﬃcients on the left-hand-side follow from (3.7.10), and the right-hand-side follows from (3.7.11). Then,

2n+1 X 2n + 1 x = . vj n j=1 By subtracting the following equations from this equation,

xv1 + xv3 + ··· + xv2n−1 = 1 and xv2 + xv4 + ··· + xv2n = 1,

1 1 we have xv2n+1 = n . By symmetry, xvj = n for j ∈ {1, 2, ··· , 2n + 1}. This fact contradicts |V | the integrality of x ∈ R : Ax ≤ 1, x ≥ 0 . We conclude the proof.

70 3.7.5 Proof of Lemma 3.6.2

Let P0 = V, ∅, τ, W, {Nw; w ∈ W }, {Wv; v ∈ V }, 0 be a process. Let k be an arbitrary integer satisfying condition (3.6.3), i.e.,

k · τ(v) ≥ Nw for w ∈ Wv, v ∈ V.

˜ We now ﬁx k. Let P0 = V, ∅, τ,˜ W, {Nw; w ∈ W }, {Wv; v ∈ V }, 0 be exactly the same

˜1 k ˜ process as P0 exceptτ ˜(v) = kτ(v) for v ∈ V . Let Hexp (resp., Hexp) be a hypergraph for P0 ˜1 k (resp., P0) as deﬁned in Section 3.6. It is easy to see that Hexp and Hexp are essentially the same, and therefore,

0 ˜1 0 k χ (Hexp) = χ (Hexp). (3.7.12)

We now present the LP formulation of SPPC for the capacity of P0 in (Gurvich and Van Mieghem, 2015). This formulation will facilitate the subsequent analysis. Let V = {v1, v2, ··· , v|V |} without loss of generality. For i ∈ 1, 2, ··· , |V | , let ai ∈ N ∪ {0}. Then, a = (a1, ··· , a|V |) is a conﬁguration vector. Let A = (Aij)|W |×|V | be the resource-activity matrix for P0. Then, a is feasible if

|V | X Aijaj ≤ Nwi for i ∈ 1, 2, ··· , |W | . j=1

Let A be the collection of all feasible a. The LP formulation of SPPC for P0 is as follows:

ρnet(λ) = min ρ X s.t. ajx(a) = λτ(vj) for j ∈ 1, 2, ··· , |V | ,

a=(a1,··· ,a|V |)∈A X x(a) ≤ ρ, a∈A x(a) ≥ 0 for a ∈ A,

71 for λ > 0. The network capacity, λ∗, is such that ρnet(λ∗) = 1. It is easy to see that λ∗ is unique. Note that this formulation simpliﬁes to the one in Section 3.2.3 under assump-

net ˜∗ ˜ tion (3.2.14). Letρ ˜ (λ) and λ be similarly deﬁned for P0. Using the same logic as in Section 3.3.3, we have

1 µ(P ) = . (3.7.13) 0 ρnet(1)

We further notice that

ρ˜net(1) = kρnet(1). (3.7.14)

We claim that it only remains to show

0 ˜1 net χ (Hexp) =ρ ˜ (1). (3.7.15)

In fact, combining (3.7.12)–(3.7.15), we have

1 k k k µ(P0) = = = = , net net 0 ˜1 0 k ρ (1) ρ˜ (1) χ (Hexp) χ (Hexp) which is exactly the conclusion of Lemma 3.6.2. We now establish (3.7.15).

˜1 ˜ 1 An independent set I of Hexp is a subset of Vexp such that all the activities with indices in I can be scheduled simultaneously (not necessarily on the same ﬂow unit). That is,

0 {v : v ∈ I, w ∈ Wv} ≤ Nw for some w ∈ W. (3.7.16)

˜ 0 ˜1 Let I be the collection of all I. The LP formulation for χ (Hexp), similarly as deﬁned in Section 3.3.3, is as follows:

0 ˜1 X χ (Hexp) = min xI , I∈I˜ X ˜ 1 s.t. xI = 1 for v ∈ Vexp, I:v∈I,I∈I˜ ˜ xI ≥ 0 for I ∈ I,

72 0 ˜1 We now ﬁnd an equivalent formulation for χ (Hexp) that will be useful for our subsequent ˜ ˜ analysis. For j ∈ 1, ··· , |V | , let Vj be the set of activities in the expanded process Pexp ˜ corresponding to activity vj in P0. That is,

V˜ = v : W 0 = W , τ(v ) = 1 for l = 1, ··· , kτ(v ) for j ∈ 1, ··· , |V | . j j,l vj,l vj j,l j

˜ 1 |V | ˜ ˜ ˜ Then, Vexp = ∪j=1Vj, and Vj1 ∩ Vj2 = ∅ for j1 6= j2. For a = (a1, ··· , a|V |) ∈ A, deﬁne

˜ ˜ Ia = I : I ∈ I; aj elements of I are in Vj, for j = 1, ··· , |V | .

˜ ˜ ˜ ˜ 0 ˜ Then, I = ∪a∈AIa, and Ia ∩ Ia0 = ∅ for a 6= a . Such Ia must exist by the deﬁnition of k and (3.7.16). We now have our equivalent formulation:

0 ˜1 X X χ (Hexp) = min xI ,

a∈A I∈I˜a X X ˜ s.t. xI = 1 for vl ∈ Vj, l ∈ 1, ··· , kτ(vj) , ˜ a:aj ≥1,a∈A I:vj,l∈I,I∈Ia

and j ∈ 1, ··· , |V | ,

˜ xI ≥ 0 for I ∈ I.

For convenience, we will use the following equivalent formulation forρ ˜net(1):

X ρ˜net(1) = min x(a) a∈A X s.t. ajx(a) = kτ(vj) for j ∈ 1, 2, ··· , |V | ,

a=(a1,··· ,a|V |)∈A

x(a) ≥ 0 for a ∈ A.

Our argument below essentially relies on the correspondence between x(a) and P x∗. I∈Iã I 0 ˜1 net ∗ ˜ 0 ˜1 We first show that χ (Hexp) ≥ ρ˜ (1). Let (xI : I ∈ I) be an optimal solution for χ (Hexp). Define x(a) = P x∗ for a ∈ A. Since χ0(H˜1 ) = P P x∗ = P x(a), we I∈Iã I exp a∈A I∈Iã I a∈A

73 only need to show that (x(a): a ∈ A) is a feasible solution forρ ˜net(1). Take an arbitrary

P P ∗ j ∈ 1, ··· , |V | . Summing both sides of ˜ x = 1 over l gives a:aj ≥1,a∈A I:vj,l∈I,I∈Ia I

kτ(vj ) ! X X X ∗ xI = kτ(vj). ˜ a:aj ≥1,a∈A l=1 I:vj,l∈I,I∈Ia

˜ ˜ ∗ Notice that, every I, I ∈ Ia, contains aj elements in Vj; therefore, xI is added for aj times in the brackets on the left-hand side of the above equation. We then have

X X ∗ ajxI = kτ(vj),

a:aj ≥1,a∈A I:I∈I˜a but this is exactly

X ajx(a) = kτ(vj).

a=(a1,··· ,a|V |)∈A

That is, (x(a): a ∈ A) is feasible.

0 ˜1 net ∗ We then show that χ (Hexp) ≤ ρ˜ (1). Let (x (a): a ∈ A) be an optimal solution for

net 1 ∗ ˜ ˜ ˜ ρ˜ (1). Define xI = x (a) for I ∈ Ia and a ∈ A, where |Ia| is the cardinality of Ia. Since |Iã| ρñet(1) = P x∗(a) = P P x , we only need to show that (x : I ∈ I˜ , a ∈ A) is a a∈A a∈A I∈Iã I I a 0 ˜1 feasible solution for χ (Hexp). We arbitrarily take j ∈ 1, ··· , |V | and l ∈ 1, ··· , kτ(vj) . Then,

˜ X X X I : vj,l ∈ I,I ∈ Ia x = x∗(a). I ˜ ˜ |Ia| a:aj ≥1,a∈A I:vj,l∈I,I∈Ia a:aj ≥1,a∈A

˜ Q|V | kτ(vi) ˜ It is easy to see that |Ia| = , since |Ia| equals the number of ways of picking ai ele- i=1 ai

˜ ˜ kτ(vj )−1 Q kτ(vi) ments out of Vi, i ∈ 1, ··· , |V | . Similarly, I : vj,l ∈ I,I ∈ Ia = , aj −1 i6=j ai ˜ since the left-hand-side equals the number of ways of picking ai elements out of Vi, i ∈

74 ˜ 1, ··· , |V | with the restriction that vj,l must be picked out of Vj. Thus,

˜ X X X I : vj,l ∈ I,I ∈ Ia x = x∗(a) I ˜ ˜ |Ia| a:aj ≥1,a∈A I:vj,l∈I,I∈Ia a:aj ≥1,a∈A

kτ(vj )−1 Q kτ(vi) X aj −1 i6=j ai = x∗(a) Q|V | kτ(vi) a:aj ≥1,a∈A i=1 ai

X aj = x∗(a) kτ(vj) a:aj ≥1,a∈A

= 1.

˜ 0 ˜1 That is, (xI : I ∈ Ia, a ∈ A) is a feasible solution for χ (Hexp).

75 CHAPTER 4

ANALYSIS OF SCRIP SYSTEMS

4.1 Introduction

This essay is motivated by the analysis of scrip systems in (Johnson et al., 2014). A scrip (or token) system refers to a market where scrips play the role of real currency – the beneﬁciary of a service “pays” scrips to the provider, who in turn can “purchase” services using scrips.

One classical example of a scrip system is that of a baby-sitting co-operative: A group of couples, each having at least one child, forms a co-op in which scrips are used to pay for babysitting services. Each couple is assigned a certain number of scrips at the time the co-op is started. Every time a couple needs someone to baby-sit their child, they pay for the service using scrips. To earn scrips, a couple has to provide babysitting service. For details of the Capitol Hill Baby-Sitting Co-op, see (Sweeney and Sweeney, 1977; Krugman, 1999), and (Johnson et al., 2014). Several other examples of scrip systems in resource exchange and online resource allocation environments are discussed in (Johnson et al., 2014).

(Johnson et al., 2014) use an infinite-horizon game with discounting to model a scrip system. Players are assumed to be homogeneous. The unit cost (sacrifice of utility) incurred in providing service and the unit benefit (gain of utility) from obtaining service are both assumed to be constants; clearly, the latter is greater than the former. In each period, a randomly-chosen player requests service; all the other players have a choice of whether or not to volunteer to provide service. Among the players who volunteer, the service provider is chosen according to the minimum-scrip service-provider selection rule: A player with the minimum number of scrips is chosen as the service provider. Throughout our paper, we will refer to this rule simply as the “minimum-scrip rule”.

(Johnson et al., 2014) study what they refer to as the “always-trade” strategy for a player, i.e., the strategy of a player always willing to provide service, regardless of the distribution of

76 scrips among the players. A key result of their work is that, under the minimum-scrip rule,

for any number of players and for every discount factor close enough to one, there exists a

Nash equilibrium in which each player plays the always-trade strategy. Thus, the centralized

optimal social welfare is achieved in equilibrium. This result, however, is established under

the assumption that any player who refuses to provide service is punished: If a player selected

to provide service refuses to do so even once, then that player will be forever banned from

participating in the system, thus losing all potential future beneﬁt. We now take a closer

look at this assumption.

The threat of punishment seems reasonable when the number of players in the system is

small; in this case, it is perhaps feasible to detect and punish non-contributors. For large

scrip systems, however, (Johnson et al., 2014) admit that the assumption of punishment is

unsatisfactory. In a large system, it becomes diﬃcult to detect players who refuse service and

to verify the reason for their refusal. Detection, as the authors state, could be diﬃcult “if the

system relies on the service requester to report whether trade occurred, which could open

the possibility of malicious players getting others kicked out of the system”. Verifying the

reason(s) for refusal can also be challenging in large systems – for instance, a player may have

refused service because she is genuinely unwell. Such diﬃculties in detection and veriﬁcation

may result in players being unfairly banned from participation, thus hurting social welfare.

Motivated by this concern, (Johnson et al., 2014) suggest an important direction of future

research – to investigate whether or not the always-trade strategy is a Nash equilibrium –

under the same conditions as above – without the punishment assumption.

This question was earlier studied in (Friedman et al., 2006). A major departure in (John- son et al., 2014) from (Friedman et al., 2006) is that the former considers the minimum-scrip rule, while the latter works with the random provider-selection rule – the service provider

is randomly selected from among the players other than the service requester. Without the

assumption of punishment and under the random-provider selection rule, (Friedman et al.,

77 2006) show that, when the number of players is large enough and the discount factor is close enough to one, there exists an -Nash equilibrium in which each player plays a threshold policy, i.e., a player volunteers to provide service only if the number of scrips she has is lower than a threshold. As (Johnson et al., 2014) suggest, there is a fundamental departure between the minimum-scrip rule and the random provider-selection rule. Under the random provider selection rule, the evolution of the scrip system satisﬁes certain properties that enable (Friedman et al., 2006) to employ powerful techniques from statistical mechan- ics (for example, see (Jaynes, 1978)). These properties do not necessarily hold under the minimum-scrip rule. In this essay, we address the open question in (Johnson et al., 2014) for their model – without using the assumption of punishment, we establish that when the number of players is large enough and the discount factor is close enough to one, there exists an -Nash equilibrium in which each player plays the always-trade strategy. The model we study is, in fact, a generalization of that in (Johnson et al., 2014), in the sense that we allow for the possibility that players might genuinely not be available to provide service (say, due to sickness in the example of the baby-sitting co-operative). Precisely, a player is available to provide service in a period only with a certain probability. This generalization is important due to the above- mentioned issues about genuine non-availability and the diﬃculty associated with verifying genuineness of claimed non-availability.

4.2 Model

We deﬁne our generalization of the game in (Johnson et al., 2014) below. Let N, N ≥ 2, be the number of players in the game. Assume that the players are homogeneous. Let m be the number of scrips initially assigned to each player. The horizon is inﬁnite and the discount factor is δ ∈ (0, 1). Let β ∈ (0, 1] be a constant. In each period, for each player, the probability that she is available is β, and the probability that she is not

78 available is 1−β. We assume that whether a particular player is available or not in a period is independent of the availability of all the other players in the same period and is independent of what happens in the other periods. The sequence of events in each period t, t ≥ 1, is as follows:

1 2 N 1. Let xt , xt , . . . , xt denote the numbers of scrips for players 1 through N, respectively.

1 Each player is randomly selected as the service requester with probability N . Let i i denote the service requester in this period, without loss of generality. If xt = 0, then player i cannot request service and no trade will occur in this period; otherwise, the system goes to step 2.

2. For j ∈ {1, 2, ··· ,N}\{i}, player j is available to provide service in this period with probability β, independent of the availabilities of all the other players. Let V denote the set of players who are available and choose to volunteer. If V = ∅, then no trade will occur in this period; otherwise, the system goes to step 3.

3. Within the group V, the player with the minimum number of scrips is chosen as the service provider; ties are broken evenly. Let j (j 6= i) denote the service provider, without loss of generality.

4. Player i (the service requester) earns a utility of u, and pays one scrip to player j (the service provider). Player j incurs a cost (a negative utility) of c and earns one scrip from player i.

Note that β = 1 corresponds to the model analysed in (Johnson et al., 2014).

4.3 Analysis and Results

Our main result for the above model is that, when the number of players is large enough and the discount factor is close enough to one, there exists an -Nash equilibrium in which each

79 player plays the always-trade strategy. In Section 4.3.1, we deﬁne an -Nash equilibrium,

provide a formal statement of the main result, and present an outline of its proof. We need

several intermediate results for establishing the main result; these are stated in Section 4.3.2.

The proof of the main result is in Section 4.3.3. For better readability, the proofs of all the

intermediate results are placed in Section 4.4.

4.3.1 The Main Result

We begin by introducing some useful notation below.

Let s = (x1, ··· , xN ) denote a generic distribution of the scrips, where the jth entry represents the number of scrips that player j has on hand for j ∈ {1, 2, ··· ,N}. Then, s satisﬁes x1 + x2 + ··· + xN = Nm and xj ≥ 0 for j ∈ {1, 2 ··· ,N}. Let S denote the

1 N collection of all states s; then, it is easy to see that S is ﬁnite. Let st = (xt , ··· , xt ) ∈ S denote the distribution of the scrips at the beginning of period t, t ≥ 1. Let πt denote a generic single-period strategy for a player (who is not requesting service and who is available) in period t, i.e., the strategy of whether or not to volunteer in period t, as a function of st.

Let π = {πt}t≥1 denote the policy of playing πt in period t for t ≥ 1. Let Π denote the collection of all policies π. A policy π is called stationary if πt remains the same for all t.

Let Πs denote the collection of all stationary policies. We note that Πs is ﬁnite because S is ﬁnite.

Let π~ = (π1, ··· , πN ) denote the strategy proﬁle of all the players when player i plays

πi ∈ Π for i ∈ {1, 2, ··· ,N}. Let π−i denote the strategy proﬁle of all the players other

δ than player i. Let Vi (π~ ) denote the expected discount payoﬀ function of player i. An -

i Nash equilibrium in our context is deﬁned as follows: For any > 0, a policy π∗ for player i

−i δ i −i δ i −i is called an “-best response” to a joint strategy π∗ if Vi (π∗, π∗ ) > (1 − )Vi (π , π∗ )

i i for π ∈ Π. If π∗ is an -best response for player i, i ∈ {1, 2, ··· ,N}, then the strategy proﬁle π~ ∗ is called an -Nash equilibrium. This notion is similar to the one in (Friedman

80 et al., 2006). We note that other notions of an -Nash equilibrium are also available in the literature; see, e.g., (Maschler et al., 2013).

Let πv ∈ Πs denote the always-trade strategy, i.e., a player is always willing to provide

service if she is not requesting service and is available. Our goal is to show that π~ v =

(πv, ··· , πv) is an -Nash equilibrium. By deﬁnition and by symmetry of all the players, it

suﬃces to show that πv is an -best response for player 1 when all the other players play πv.

δ δ δ For each π ∈ Π, we deﬁne V (π) := V1 (π, πv, ··· , πv) for short; that is, V (π) represents the expected discounted payoﬀ of player 1 when player 1 plays π and all the other

players play πv. The main result of the paper is as follows:

Theorem 4.3.1. Assume that players 2 through N play the always-trade strategy πv. For

∗ ∗ ∗ any constant > 0, there exist N > 0 and δ,N ∈ (0, 1) such that for N > N and ∗ δ ∈ (δ,N , 1), we have

δ δ V (πv) > (1 − )V (π) for π ∈ Π.

That is, π~ v is an -Nash equilibrium.

The proof of Theorem 4.3.1 is in Section 4.3.3. To establish this result, we need several intermediate results – Lemmas 4.3.1 through 4.3.4. Before stating these lemmas below, we provide a brief explanation of how they are used in the proof of Theorem 4.3.1.

Outline of the proof of Theorem 4.3.1: Statement (a) of Lemma 4.3.1 shows the optimal-

ity of a stationary policy π∗ ∈ Πs for all suﬃciently-large discount factors δ. Statement (b)

avg δ of Lemma 4.3.1 establishes, for π ∈ Πs, a connection between V (π) and V (π), where V avg(π) denotes the long-run average payoﬀ of player 1 when she plays π and all the other

players play πv. We will establish Theorem 4.3.1 using the following steps: (i) We ﬁrst ﬁnd

avg an explicit expression of V (π) for π ∈ Πs via Lemmas 4.3.2 and 4.3.3. This, along with

δ Lemma 4.3.1, will be useful in deriving an upper bound on V (π) for π ∈ Π\{πv}. (ii) We then establish Lemma 4.3.4, which, together with Lemma 4.3.3, will be useful for obtaining

δ a lower bound on V (πv). These two bounds will be combined to establish Theorem 4.3.1.

81 4.3.2 Intermediate Results

Lemma 4.3.1 below is from (Bertsekas, 2012).

Lemma 4.3.1. Since S is a ﬁnite state space, we have

(a) There exists a stationary policy π∗ ∈ Πs and a constant δ ∈ (0, 1) such that π∗ is optimal for δ ∈ (δ, 1). That is,

δ δ V (π∗) ≥ V (π) for π ∈ Π, for δ ∈ (δ, 1).

(b) For any stationary policy π, π ∈ Πs, of player 1, we have

V avg(π) = lim(1 − δ)V δ(π). δ→1

We ﬁrst introduce Lemma 4.3.2. It is easy to see from the sequence of events that,

for π ∈ Πs, {st; t ≥ 1} is a Markov chain on S if player 1 plays π and all the other players

play πv. For t ≥ 1, deﬁne  1, if x1 = 0, 0  t It = (4.3.1)  1 0, if xt 6= 0.

0 That is, It indicates whether or not player 1 has no scrip on hand in period t. Then, we have

Lemma 4.3.2. For π ∈ Πs, let

PT 0 t=1 EIt µN (0; π) = lim . (4.3.2) T →∞ T denote the long-run proportion of time when player 1 has no scrip on hand, if player 1 plays π and all the other players play πv; the subscript N indicates the number of players in the system. Then, µN (0; π) exists and satisﬁes µN (0; π) > 0.

82 We now use Lemma 4.3.2 to establish an explicit expression for V avg(π).

Lemma 4.3.3. For π ∈ Πs, we have

1 − µ (0; π) V avg(π) = N (u − c)1 − (1 − β)N−1. N

Next, we derive a limiting result on µN (0; πv).

Lemma 4.3.4. For β ∈ (0, 1], we have limN→∞ µN (0; πv) = 0.

The proofs of Lemmas 4.3.2–4.3.4 are in Section 4.4. We are now ready to establish the main result.

4.3.3 Proof of Theorem 4.3.1

∗ ∗ Consider an arbitrary constant > 0. By Lemma 4.3.4, there exists N such that for N > N , we have

µ (0; π ) < . (4.3.3) N v 2

∗ ∗ Take an arbitrary constant N > N . We only have to show that there exists δ,N ∈ (0, 1)

∗ such that for δ ∈ (δ,N , 1), we have

δ δ V (πv) > (1 − )V (π) for π ∈ Π.

avg By applying Lemma 4.3.1(b) to πv and using the expression of V (πv) in Lemma 4.3.3, ˆ we know that there exists δ,N ∈ (0, 1) such that

1 − µ (0; π ) − (1 − δ)V δ(π ) > N v 2 (u − c)1 − (1 − β)N−1 for δ ∈ (δˆ , 1). v N ,N

Combining with (4.3.3), we have

1 − (1 − δ)V δ(π ) > (u − c)1 − (1 − β)N−1 for δ ∈ (δˆ , 1). (4.3.4) v N ,N

83 δ Thus, we have obtained a lower bound on V (πv). We now proceed to ﬁnd an upper bound on V δ(π).

avg For π ∈ Πs, by applying Lemma 4.3.1(b) to π, using the expression of V (π) in

π Lemma 4.3.3, and noticing that µN (0; π) > 0, we know that there exists δ,N ∈ (0, 1) such that

1 (1 − δ)V δ(π) < (u − c)1 − (1 − β)N−1 for δ ∈ (δπ , 1). N ,N

˜ π ˜ Noticing that Πs is ﬁnite, we let δ,N = maxπ∈Πs δ,N . Then, δ,N ∈ (0, 1), and we have

1 (1 − δ)V δ(π) < (u − c)1 − (1 − β)N−1 for π ∈ Π , for δ ∈ (δ˜ , 1). (4.3.5) N s ,N

By Lemma 4.3.1(a), there exists π∗ ∈ Πs and δN ∈ (0, 1) such that

δ δ V (π) ≤ V (π∗) for π ∈ Π, for δ ∈ (δN , 1). (4.3.6)

∗ ˆ ˜ ∗ Let δ,N = max{δ,N , δ,N , δN }. Then δ,N ∈ (0, 1). Combining (4.3.4)–(4.3.6), we have, for π ∈ Π,

(1 − ) 1 V δ(π ) > (u − c)1 − (1 − β)N−1 v (1 − δ) N

δ > (1 − )V (π∗)

≥ (1 − )V δ(π),

which concludes the proof.

4.4 Proofs of the Intermediate Results – Lemmas 4.3.2 – 4.3.4

4.4.1 Proof of Lemma 4.3.2

Consider an arbitrary π ∈ Πs. Following (Ross, 1995), the following properties hold for S because of its ﬁniteness:

84 • There is a unique way to partition S into a ﬁnite number (which we denote by n + 1)

of sets, S0,S1, ··· ,Sn, that satisfy the following properties:

n (a) S = ∪j=0Sj,

(b) Si ∩ Sj = ∅ for i 6= j and i, j ∈ {0, 1, ··· , n},

(d) Sj, j ∈ {1, 2, ··· , n}, is a single, irreducible, positive recurrent communication

class, i.e., for every pair of states, i, l ∈ Sj, state l can be reached within a ﬁnite number of periods (in expectation) from state i with probability 1, and vice versa.

• Starting from any s ∈ S, the Markov chain will reach some Sj, j ∈ {1, 2, ··· , n}, with probability 1 in a ﬁnite number of periods (in expectation), and will evolve

within Sj from then on. The Markov chain will become a single, irreducible, and

positively recurrent Markov chain since then. For j ∈ {1, 2, ··· , n}, let Pj(s) de-

note the probability that the Markov chain will eventually reach Sj starting from s.

Then, P1(s) + ··· + Pn(s) = 1.

• In each Sj, j ∈ {1, 2, ··· , n}, there must exist a state where player 1 has no scrip on

1 N 1 hand. In fact, for any s = (x , ··· , x ) ∈ Sj, such a state can be reached from s in x periods (i.e., it is possible for player 1 to be the service requester and for all the other players to be available in x1 consecutive periods).

1 N 1 • Notice that, for any s = (x , ··· , x ) ∈ Sj, j ∈ {1, 2, ··· , n}, with x = 0, the one- step transition from s to s can happen with a strictly positive probability, because player 1 could be the service requester in the current period. Then, the evolution of

the Markov chain on Sj is aperiodic. Let µN (0; πj) denote the long-run proportion of

time when player 1 has no scrip on hand, if the starting state is some s ∈ Sj (and

therefore, the Markov chain will evolve within Sj). Then, µN (0; πj) exists and satisﬁes

µN (0; πj) > 0; see, e.g., Theorem 6 in Chapter 3 of (Wolﬀ, 1989) for details.

85 Pn Since µN (0; π) = j=1 Pj(s)µN (0; πj), µN (0; π) exists and satisﬁes µN (0; π) > 0. By deﬁnition, we have

PT 0 t=1 EIt µN (0; π) = lim , T →∞ T

which concludes the proof.

4.4.2 Proof of Lemma 4.3.3

Consider an arbitrary π ∈ Πs. For t ≥ 1, deﬁne   1, if player 1 is the service requester and some other player is available in period t, It =  0, otherwise. (4.4.1)

We also deﬁne  1, if player 1 provides service in period t, c  It = (4.4.2)  0, otherwise,

0 c 0 for t ≥ 1. The payoff of player 1 in period t is u · It[1 − It ] − c · It , where It is defined by (4.3.1). The average payoff of player 1 in the first T periods, T ≥ 1, is

u · PT I (1 − I0) − c · PT Ic t=1 t t t=1 t . (4.4.3) T

Since player 1 can request service only if she has scrip(s) on hand, we have

T T X 0 X c It(1 − It ) ≤ It + m. (4.4.4) t=1 t=1 Since the total number of scrips in the system is Nm, we have

T T X 0 X c It(1 − It ) ≥ It − Nm. (4.4.5) t=1 t=1

86 Combining (4.4.3)-(4.4.5), we have

PT I (1 − I0) Nm u · PT I (1 − I0) − c · PT Ic t=1 t t (u − c) − ≤ t=1 t t t=1 t T T T PT I (1 − I0) m ≤ t=1 t t (u − c) + . T T

1 N−1 0 By taking expectation, and noticing that EIt = N [1 − (1 − β) ] and It and It are independent, we have

T − PT EI0 1 Nm t=1 t (u − c) [1 − (1 − β)N−1] − T N T " # u · PT I (1 − I0) − c · PT Ic ≤ E t=1 t t t=1 t T T − PT EI0 1 m ≤ t=1 t (u − c) [1 − (1 − β)N−1] + . T N T

By letting T → ∞ and using (4.3.2), we have

1 − µ (0; π) h i V avg(π) = N (u − c) 1 − (1 − β)N−1 , N

which concludes the proof.

4.4.3 Proof of Lemma 4.3.4

Assume that all the players play πv. We ﬁrst prove Lemma 4.3.4 when β = 1. j For j ∈ {1, 2 ··· ,N}, let P xt = 0 denote the probability that player j has no scrip in

period t, t ≥ 1. Recall that the initial state is s1 = (m, ··· , m). By the minimum-scrip rule, it is easy to see that the Markov chain will never reach a state with more than one players having no scrip. That is,

1 2 N n N j o P xt = 0 + P xt = 0 + ··· + P xt = 0 = P ∪j=1 xt = 0 ≤ 1 for t ≥ 1.

Then, for T ≥ 1, we have

PT P x1 = 0 PT P xN = 0 t=1 t + ··· + t=1 t ≤ 1 for t ≥ 1. (4.4.6) T T

87 Notice that

PT 1 t=1 P xt = 0 µN (0; πv) = lim . T →∞ T

Then, by letting T → ∞ in (4.4.6) and using the fact that the players are symmetric, we

1 obtain µN (0; πv) ≤ N . The assertion follows. We then prove Lemma 4.3.4 when β ∈ (0, 1) as below.

(a) We show that {st; t ≥ 1} is a single, irreducible, aperiodic, positive recurrent Markov chain on S.

For ease of analysis, we deﬁne the distance D(s, s0) for every pair of states, s =

(x1, ··· , xN ) ∈ S and s0 = (y1, ··· , yN ) ∈ S by D(s, s0) = |x1 − y1| + |x2 − y2| +

··· + |xN − yN |; it is easy to see that D(s, s0) is even. To establish the above s-

tatement, we first claim that {st; t ≥ 1} is a single, irreducible, positive recurrent Markov chain on S. Since S is finite, it suffices to show that for every pair of states,

s = (x1, ··· , xN ) ∈ S and s0 = (y1, ··· , yN ) ∈ S, one can reach s0 starting from s

with a strictly positive probability; for details, see (Ross, 1995). To see this, consider

the following procedure: If D(s, s0) = 0, we are done. If D(s, s0) > 0, then there

exist i, j ∈ {1, 2, ··· ,N} with i 6= j such that xi − yi > 0 and xj − yj < 0. Since the

scenario that player i is the service requester, player j is available, and all the other

players are not available, can happen with a strictly positive probability, one can reach

a new state s00 with D(s00, s0) = D(s, s0) − 2.

Noticing that the above procedure ends in a ﬁnite number of steps (because D(s, s0) is

ﬁnite), the above claim follows.

Next, we observe that the evolution of {st; t ≥ 1} on S is aperiodic. Consider the state s = (0, x2, ··· , xN ) ∈ S. Since the one-step transition from s to s can happen

with a strictly positive probability, this claim follows from (Ross, 1995).

88 (b) We introduce some notation that will be useful for our subsequent analysis.

0 Let P be the transition matrix of the Markov chain, {st; t ≥ 1}. Let p(s, s ) be a generic element of P for all s, s0 ∈ S; that is, p(s, s0) is the probability that the Markov chain transits from s to s0 in one step. Then, it is easy to see from the sequence of events that p(s, s0) = 0 for s, s0 ∈ S with D(s, s0) > 2. For S0 ⊆ S and for s ∈ S, let p(s, S0) denote the probability that the Markov chain transits from s to some state in

0 0 P 0 S in one step; then, p(s, S ) = s0∈S0 p(s, s ).

1 N For k ∈ {0, 1, ··· ,N − 1}, let Sk denote the collection of states s = (x , ··· , x ) ∈ S

1 N such that exactly k elements of (x , ··· , x ) are zeros. Then, Si ∩ Sj = ∅ for i, j ∈ N−1 ˜ {0, 1 ··· ,N − 1} with i 6= j, and S = ∪k=0 Sk. For k ∈ {1, ··· ,N − 1}, let Sk denote the collection of states s = (x1, ··· , xN ) ∈ S such that x1 = 0 and exactly k − 1

2 N ˜ elements of (x , ··· , x ) are zeros; then, Sk ⊆ Sk for k ∈ {1, 2, ··· ,N − 1}.

For s ∈ S, let µN (s) denote the long-run proportion of time when the Markov chain is in s, where the subscript N indicates the number of players in the system. For a

0 0 subset S ⊆ S, let µN (S ) denote the long-run proportion of time when the Markov

0 chain is in S . Then, µN (0; πv) satisﬁes

˜ ˜ µN (0; πv) = µN (S1) + ··· + µN (SN−1). (4.4.7)

˜ k (i) For k ∈ {1, 2, ··· ,N − 1}, we have µN (Sk; πv) = N µN (Sk; πv).

1 k (ii) For k ∈ {2, 3, ··· ,N − 1}, we have p(s, Sk−1; πv) ≥ N 1 − (1 − β) for s ∈ Sk.

k (iii) For k ∈ {1, 2, ··· ,N − 2}, we have p(s, Sk+1; πv) ≤ (1 − β) for s ∈ Sk.

(iv) We have

β(1 − β)N−2 µ (S ; π ) ≤ µ (S ; π ). (4.4.8) N N−1 v 1 − (1 − β)N−1 N N−2 v

89 We prove the above statements as follows:

• Proof of statement (i): This result is intuitive; however, for completeness, we provide a formal argument. For s = (x1, ··· , xN ) ∈ S, deﬁne the “ordered state” of s by σ(s) = (x(1), ··· , x(N)), where x(1), ··· , x(N), is a permutation of x1, ··· , xN , such that x(1) ≤ · · · ≤ x(N). Then, σ(s) ∈ S. We write s ∼ s˜ ifs ˜

0 is the ordered state of s; then, s ∼ σ(s). By symmetry, we have µN (s) = µN (s ) for s, s0 ∈ S such that σ(s) = σ(s0).

1 N We ﬁx the value of k. Take an arbitrary s = (x , ··· , x ) ∈ Sk. Let l ∈ N 1 N denote the number of diﬀerent values in x , ··· , x . Let aj denote the number of

1 N th elements in {x , ··· , x } that are equal to the j smallest value. Then, a1 = k

and a1 + a2 + ··· + al = N. By a simple counting argument, we have

0 0 0 N! µN {s : s ∼ σ(s), s ∈ Sk} = µN (s). k!a2! ··· al!

Similarly, we have

0 0 0 ˜ (N − 1)! µN {s : s ∼ σ(s), s ∈ Sk} = µN (s). (k − 1)!a2! ··· al! ˜ Combining the above two equalities and noticing that Sk and Sk have the same collection of ordered states, we conclude statement (i).

• Proof of statement (ii): We ﬁx the value of k. Take an arbitrary s =

1 2 N 1 2 N 1 (x , x , ··· , x ) ∈ Sk. Then, there are k zeros in x , x , ··· , x . Since x +

N j Nm ··· + x = Nm, we have max{x ; 1 ≤ j ≤ N} ≥ N−k > m ≥ 1. Then, consider the scenario: the player with max{xj; 1 ≤ j ≤ N} scrips on hand is chosen as the service requester, and at least one of the k players with no scrip on hand is available. In this scenario, the Markov chain will reach a state with k − 1 zeros according to the minimum-scrip rule. Since the probability of this scenario is

1 k N [1 − (1 − β) ], we conclude statement (ii).

90 • Proof of statement (iii): We ﬁx the value of k. Take an arbitrary s =

1 2 N 1 2 N (x , x , ··· , x ) ∈ Sk. Then, there are k zeros in {x , x , ··· , x }. To reach a state with k + 1 zeros in one step, it must happen that all the players with no scrip on hand are not available, according to the minimum-scrip rule. Thus, we conclude statement (iii).

• Proof of statement (iv): Let ~µ denote the vector of the long-run proportions of time of all the states s ∈ S. Then, ~µ satisﬁes X ~µ = ~µP and µN (s) = 1; (4.4.9) s∈S for details, see (Ross, 1995). It follows that N − 1 1 µ (Nm, 0, ··· , 0) = µ (Nm, 0, ··· , 0) + (1 − β)N−1 N N N N 1 + (N − 1)µ (Nm − 1, 1, ··· , 0) β(1 − β)N−2, N N where we use the fact that

µN (Nm − 1, 1, 0, ··· , 0) = µN (Nm − 1, 0, 1, 0, ··· , 0) = ··· = µN (Nm − 1, 0, ··· , 0, 1).

By simpliﬁcation, we have β(1 − β)N−2 µ (Nm, 0, ··· , 0) = (N − 1)µ (Nm − 1, 1, 0, ··· , 0), (4.4.10) N 1 − (1 − β)N−1 N

Noticing that SN−1 = {(Nm, 0, ··· , 0), (0, Nm, 0, ··· , 0), ··· , (0, ··· , 0, Nm)}, we have

µN (SN−1) = NµN (Nm, 0, ··· , 0) (4.4.11)

by symmetry. Also notice that there are N(N −1) states in SN−2 where one player has 1 scrip on hand, one player has Nm − 1 scrips on hand, and the remaining players have no scrip on hand. By symmetry, we have

µN (SN−2) ≥ N(N − 1)µN (Nm − 1, 1, 0, ··· , 0). (4.4.12)

Combining (4.4.10)-(4.4.12), we conclude statement (iv).

91 (d) We are ready to establish Lemma 4.3.4 when β ∈ (0, 1). We begin with the following √ 2 b Nc ∗ easily veriﬁable fact: limN→∞ 8N (1 − β) = 0. Thus, there exists N > 0 such that for N > N ∗, we have √ √ √ 1 (a) 8N 2(1 − β)b Nc < 1, (b) N − 2 ≥ b Nc, and (c) 1 − 2(1 − β)b Nc > . 2 (4.4.13)

Throughout the proof, we fix N > N ∗. We proceed by finding an upper bound on ˜ ˜ ˜ µN (0; πv). Notice that µN (0; πv) = µN (S1) + µN (S2) + ··· + µN (SN−1). We first find

˜ ˜ √ an upper bound on µN (S1) + ··· + µN (Sb Nc). By statement (i) of Part (c), we have √ 1 b Nc ˜ ˜ √ √ µN (S1) + ··· + µN (Sb Nc) = µN (S1) + ··· + µN (Sb Nc) N√ N b Nc ≤ µ (S ) + ··· + µ (S √ ) N N 1 N b Nc 1 ≤ √ , (4.4.14) N

√ where (4.4.14) follows from the fact that µN (S1) + ··· + µN (Sb Nc) ≤ 1. It remains to ˜ √ ˜ ﬁnd an upper bound on µN (Sb Nc+1) + ··· µN (SN−1).

Consider an arbitrary k ∈ {2, 3, ··· ,N − 2}. For s ∈ Sk, it follows from (4.4.9) that

X 0 0 X 0 0 X 0 0 µN (s) = µN (s )p(s , s) + µN (s )p(s , s) + µN (s )p(s , s). 0 0 0 s ∈Sk−1 s ∈Sk s ∈Sk+1

By summing these linear equations across Sk, we obtain

X µN (Sk) = µN (s)

s∈Sk

X 0 0 X 0 0 = µN (s )p(s ,Sk) + µN (s )p(s ,Sk) 0 0 s ∈Sk−1 s ∈Sk

X 0 0 + µN (s )p(s ,Sk) (4.4.15) 0 s ∈Sk+1 n 1 o ≤ µ (S )(1 − β)k−1 + µ (S ) 1 − 1 − (1 − β)k N k−1 N k N

+ µN (Sk+1), (4.4.16)

92 where (4.4.15) follows from P p(s0, s) = p(s0,S ) for s0 ∈ S and for k ∈ {0, ··· ,N − s∈Sk k 1}, and (4.4.16) follows from statements (ii) and (iii) of Part (c), and (4.4.9). From

(4.4.16), we have the following recursive relation

1 1 − (1 − β)kµ (S ) − µ (S ) ≤ µ (S )(1 − β)k−1. (4.4.17) N N k N k+1 N k−1

In what follows, we will establish

1 √ µ (S ) < µ (S ) for k ∈ {b Nc, ··· ,N − 2} (4.4.18) N k+1 2N N k by induction using (4.4.8), (4.4.13), and (4.4.17). In fact, from (4.4.8), we have

β(1 − β)N−2 µ (S ) ≤ µ (S ) N N−1 1 − (1 − β)N−1 N N−2 √ β(1 − β)b Nc ≤ µ (S ) (4.4.19) 1 − 1/2 N N−2 √ (1 − β)b Nc ≤ µ (S ) 1 − 1/2 N N−2 1 < µ (S ), (4.4.20) 2N N N−2 where (4.4.19) follows from (4.4.13)(b), (4.4.13)(c), and the fact that β ∈ (0, 1), and

(4.4.20) follows from (4.4.13)(a). Thus, (4.4.18) is true for k = N − 2. Assume that

1 µ (S ) < µ (S ), (4.4.21) N k+2 2N N k+1 which is our induction hypothesis. Then, from (4.4.17), we have

1 µ (S )(1 − β)k ≥ 1 − (1 − β)k+1µ (S ) − µ (S ), N k N N k+1 N k+2 1 1 > 1 − (1 − β)k+1 − µ (S ), (4.4.22) N 2N N k+1 1 > µ (S ), (4.4.23) 4N N k+1

93 where (4.4.22) follows from (4.4.21), and (4.4.23) follows from (4.4.13)(a) and the fact that β ∈ (0, 1). Then,

1 µ (S ) < µ (S ) · 4N(1 − β)k < µ (S ), N k+1 N k 2N N k which follows from (4.4.13)(a). By the induction principle, we have thus established (4.4.18). It follows that

˜ √ ˜ √ µN (Sb Nc+1) + ··· µN (SN−1) ≤ µN (Sb Nc+1) + ··· µN (SN−1)

1 1 1 √ < + + ··· + √ µN (S ) 2N (2N)2 (2N)N−1−b Nc b Nc

1 1 √ < 1 µN (Sb Nc) 2N 1 − 2N 1 < . (4.4.24) N

Combining (4.4.14) and (4.4.24), we have

1 1 µN (0; πv) ≤ √ + b Nc N for every N satisfying (4.4.13), which concludes the proof.

94 APPENDIX

PROOFS OF LEMMA 3.2.2

a Let P1 denote the same process as P except without precedence constraints, i.e., P1 =

cyclic V, ∅, τ, W, {Nw : w ∈ W }, {Wv : v ∈ V }, I . Let Π1 denote the collection of all feasible,

cyclic cyclic cyclic cyclic policies π for process P1. Then, Π ⊂ Π1 ⊆ Π0 . We will establish the following claims in the rest of the proof; it is clear that these claims imply Lemma 3.2.2:

π (a) Claim 1: µ(P) = supπ∈Πcyclic r .

π π (b) Claim 2: supπ∈Πcyclic r = sup cyclic r . π∈Π1

π π (c) Claim 3: sup cyclic r = sup cyclic r . π∈Π1 π∈Π0

π (d) Claim 4: µ(P0) = sup cyclic r . π∈Π0

Proof of claim 1

Since Πcyclic ⊆ Π, we have

sup rπ ≤ µ(P). π∈Πcyclic

Then, we only have to show that

sup rπ ≥ µ(P) − for > 0. (A.0.1) π∈Πcyclic

Take an arbitrary constant > 0. By (3.2.8), there exists π∗ ∈ Π and T ∗ ∈ N such that

N π∗ [1,T ∗] > µ(P) − . (A.0.2) T ∗

95 ∗ π∗ ∗ Let k = N [1,T ]. Let {n1, n2, ··· , nk∗ } denote set of the indices of the ﬂow units that exit the process by the end of time unit T ∗ under π∗. We construct a policyπ ˜ as follows:   ∗ ∗ ∗ π (nj, t − nT ), if t ≥ nT + 1, ∗  ∗ π˜(nk + j, t) := for n ∈ N ∪ {0} and j ∈ {1, 2, ··· , k }.  0, otherwise, (A.0.3)

It is easy to see thatπ ˜ ∈ Πcyclic. By applying statement (b) of Lemma 3.2.1 and using (A.0.2), we have

k∗ rπ˜ = > µ(P) − . T ∗

which implies (A.0.1). Thus, we have established claim 1.

Proof of claim 2

cyclic cyclic Since Π ⊆ Π1 , we have

sup rπ ≤ sup rπ. π∈Πcyclic cyclic π∈Π1

Then, we only have to show that

sup rπ ≥ sup rπ. π∈Πcyclic cyclic π∈Π1

For the set of precedence constraints, A, we note that there exists another set of precedence

constraints, Atotal, that satisﬁes (i) Atotal = {(vi1 , vi2 ), (vi2 , vi3 ), ··· , (vi|V |−1 , vi|V | )} for some

permutation {i1, i2, ··· , i|V |} of {1, 2, ··· , |V |}, and (ii) Atotal respects all the precedence constraints in A. This is because A represents a partial order, and for every partial order,

there exists a total order that respects all the ordering relationships of this partial order; for

details, see, e.g., (Schroeder, 2016). We assume Atotal = {(v1, v2), (v2, v3), ··· , (v|V |−1, v|V |)}

96 cyclic cyclic without loss of generality. Let Πtotal be the collection of all π ∈ Π such that π satisﬁes

cyclic cyclic the precedence constraints Atotal. Then, Πtotal ⊆ Π , and we have

sup rπ ≤ sup rπ. cyclic π∈Πcyclic π∈Πtotal

Thus, to establish claim 2, we only have to show that

sup rπ ≥ sup rπ. (A.0.4) cyclic cyclic π∈Πtotal π∈Π1

cyclic Take an arbitrary π1 ∈ Π1 . Then, there exists k1 ∈ N such that π1 is a k1-unit cyclic

π1 policy. Let L1 denote the length of the corresponding cycle. Let ns indicate the ﬁrst cycle cyclic in steady state. To show (A.0.4), we only have to construct a policyπ ˆ ∈ Πtotal that satisﬁes

rπˆ ≥ rπ1 .

We ﬁrst illustrate the construction using an example.

Example 1: Consider the following process: V = {v1, v2}; τ(v1) = 1, τ(v2) = 2; W =

{w1, w2}; Nw1 = 1,Nw2 = 2; Wv1 = {w1},Wv2 = {w2}; On each ﬂow unit, activity v1

cyclic precedes activity v2. Suppose that a policy π1 ∈ Π1 is as shown in Figure A.1. It is

time unit 1 2 3 4 5 6 …

activity v 1 1 2 3 4 5 6

activity v 2 1 1 2 3 4 5 2 3 4 5 6

Figure A.1. Policy π1 in Example 1 easy to see that π1 is a 1-unit cyclic policy and under π1, (i) the cycle length is 1 time unit,

k1 i.e., L1 = 1, and therefore, (ii) the process rate is = 1 by statement (a) of Lemma 3.2.1. L1 cyclic Notice that π1 does not satisfy the precedence constraint v1 → v2, i.e., π1 ∈/ Πtotal .

97 cyclic We now construct a policyπ ˆ ∈ Πtotal that is a 1-unit cyclic policy (Figure A.2) as follows:

(a) We first define the schedule for flow unit 1. Let the schedule of activity v1 on flow unit 1 underπ ˆ copy the schedule of activity v1 on flow unit 1 under π1. Notice that, under π1, flow unit 2 is the first flow unit such that activity v2 on it begins after activity v1 on flow unit 1 ends. Then, we let the schedule of activity v2 on flow unit 1 underπ ˆ copy the schedule of activity v2 on flow unit 2 under π1. (b) We now completely defineπ ˆ: The schedule for flow units j, j ≥ 2, is exactly the same as the schedule for flow unit 1, except with a time shift

πˆ π of j − 1 time units. It is easy to see that r = 1 = r 1 .

time unit 1 2 3 4 5 6 …

activity v 1 1 2 3 4 5 6

activity v 2 1 1 2 3 4 2 3 4 5

Figure A.2. Policyπ ˆ in Example 1

cyclic For π1 ∈ Π1 , the formal construction of a k1-unit cyclic policyπ ˆ is as follows:

1. We first define the schedule for flow units 1, 2, ··· , k1. Let V = {v1, v2, ··· , v|V |}

vx without loss of generality. For j, t ∈ N, let π1(j, t) = (Ij,t ; x ∈ {1, 2, ··· , |V |}) be deﬁned similarly as in (3.2.1).

The schedule for ﬂow units 1, 2, ··· , k1 underπ ˆ is as follows:

(a) We first define the schedule for flow unit 1.

1 i. Let n1 = 0. The schedule of activity v1 on ﬂow unit 1 underπ ˆ copies the

1 schedule of activity v1 on ﬂow unit n1 + 1 under π1. That is,

ˆv1 v1 I1,t = I 1 for t ∈ N. n1+1,t

98 2 ii. Under π1, let n1 be any positive integer such that activity v2 on ﬂow unit

2 1 2 n1k1 + 1 begins after activity v1 on ﬂow unit n1k1 + 1 ends. That is, n1 satisﬁes

v2 v1 min{t : I 2 = 1} > max{t : I 1 = 1}. n1k1+1,t n1k1+1,t

Then, let the schedule of activity v2 on ﬂow unit 1 underπ ˆ copy the schedule

2 of activity v2 on ﬂow unit n1k1 + 1 under π1. That is,

ˆv2 v2 I1,t = I 2 for t ∈ N. n1+1,t

x ˆvx x+1 iii. In general, once we have deﬁned n1 and I1,t, we let n1 be any positive integer such that

vx+1 vx min{t : I = 1} > max{t : I x = 1}. x+1 n k1+1,t n1 k1+1,t 1

Then, let the schedule of activity vx+1 on ﬂow unit 1 underπ ˆ copy the schedule

x+1 of activity vx+1 on ﬂow unit n1 k1 + 1 under π1.

ˆvx+1 vx+1 I1,t = I x+1 for t ∈ N. n1 k1+1,t

x ˆvx iv. The above procedure ends when we have deﬁned n1 and I1,t for x ranging from 1–|V |.

(b) For j ∈ {2, 3, ··· , k1}, the definition of the schedule for flow unit j is similar to that for flow unit 1. The procedure is as follows:

1 i. Let nj = 0 and let

ˆv1 v1 Ij,t = I 1 for t ∈ N. nj k1+j,t

x ˆvx x+1 ii. Once we have deﬁned nj and Ij,t , we let nj be any positive integer such that

vx+1 vx min{t : I = 1} > max{t : I x = 1}. x+1 n k1+j,t nj k1+j,t j

99 Then, let

ˆvx+1 vx+1 Ij,t = I x+1 for t ∈ N. nj k1+j,t

x ˆvx iii. The above procedure ends when we have deﬁned nj and Ij,t for x ranging from 1–|V |.

To sum up, we have

ˆvx vx I = I x for x ∈ {1, 2, ··· , |V |} and j ∈ {1, 2, ··· , k1}. j,t nj k1+j,t

2. We now completely deﬁneπ ˆ:   πˆ(j, t − nL1), if t ≥ nL1 + 1, πˆ(nk1 + j, t) := for n ∈ N ∪ {0} and j ∈ {1, 2, ··· , k1}.  0, otherwise,

That is, the schedule for ﬂow units (η − 1)k1 + 1 through ηk1, η ≥ 2, is exactly the

same as the schedule for ﬂow unit 1 through k1, except with a time shift of (η − 1)L1 time units.

cyclic It is easy to see thatπ ˆ ∈ Πtotal . By statement (a) of Lemma 3.2.1, we have

k rπˆ = 1 = rπ, L1 which implies (A.0.4). Claim 2 follows.

Proof of claim 3

cyclic cyclic Since Π1 ⊆ Π0 , we have

sup rπ ≤ sup rπ. cyclic cyclic π∈Π1 π∈Π0

100 Then, we only have to show that

sup rπ ≥ sup rπ. (A.0.5) cyclic cyclic π∈Π1 π∈Π0

cyclic Take an arbitrary π0 ∈ Π0 . There exists k0 ∈ N such that π0 is a k0-unit cyclic policy.

π0 Let L0 denote the length of the corresponding cycle. Let ns indicate the ﬁrst cycle in steady cyclic state. To show (A.0.5), we only have to construct a policyπ ˇ ∈ Π1 that satisﬁes

rπˇ ≥ rπ0 .

We ﬁrst illustrate the construction using an example.

cyclic Example: Suppose a policy π0 ∈ Π0 is as shown in Figure A.3. It is easy to see that π0

time unit 1 2 3 4 5 6 …

activity v 1 1 2 3 4 5 6

activity v 2 1 2 1 2 3 4 3 4 5 6

Figure A.3. Policy π0 in Example 2

is a 1-unit cyclic policy and under π0, (i) the cycle length is 1 time unit, i.e., L0 = 1, and

k0 therefore, (ii) the process rate is = 1 by statement (a) of Lemma 3.2.1. Notice that π0 L0 cyclic does not satisfy the non-preemption constraint, i.e., π0 ∈/ Π1 . cyclic We will construct a policyπ ˇ ∈ Πtotal (Figure A.4) as follows: (a) Let M be the least common multiple of τ(v1) and τ(v2); then, M = 2. For i ∈ N, activities v1 and v2 (if any) in time unit i under π0 are scheduled in time units 2i − 1 and 2i underπ ˇ, i.e., the pattern in any time unit under π0 repeats itself twice underπ ˇ. (b) We ﬁrst deﬁne the schedule for

flow units 1 and 2. Corresponding to activity v1 that is scheduled on flow unit 1 under π0, we schedule activity v1 on flow unit 1 in time unit 1 and on flow unit 2 in time unit 2 under

101 st nd πˇ. Notice that, under π0, the 1 (resp., 2 ) time unit when activity v2 is scheduled on

flow unit 1 is time unit 1 (resp., 3). Then, underπ ˇ, we schedule activity v2 on flow unit 1 (resp., 2) in time units 1 and 2 (resp., 5 and 6). (d) The schedule for flow units 2j −1 and 2j is exactly the same as those for flow units 1 and 2, except with a time shift of 2(j − 1) time units.

time unit 1 2 3 4 5 6 7 8 9 10 11 12 …

activity v 1 1 2 3 4 5 6 7 8 9 10

activity v 2 1 1 3 3 2 2 4 4 6 6 8 8 5 5 7 7 9 9

Figure A.4. Policyπ ˇ in Example 2

πˇ π It is easy to see thatπ ˇ is a 2-unit cyclic policy, and r = r 0 . In general, the policyπ ˇ will be constructed in the following manner: In every time unit t, we repeat the activities under π0 M times underπ ˇ. This is to make sure that it satisﬁes the non-preemption constraints. The formal construction of a policyπ ˇ as follows:

(a) Let M denote the least common multiple of τ(v), v ∈ V . That is, M 0 M = min{M 0 : M 0 ∈ , ∈ for v ∈ V }. N τ(v) N

v For j, t ∈ N, let π0(j, t) = (Ij,t; v ∈ V ) be deﬁned similarly as in (3.2.1).

(b) We now deﬁne the schedule for ﬂow units 1, 2, ··· , Mk0. For v ∈ V , let M m = . v τ(v)

1 2 τ(v) For j ∈ {1, 2, ··· , k0} and v ∈ V , let iv,j, iv,j, ··· , iv,j be the indices of the time units

in which activity v is on ﬂow unit j under π0. That is,  1, if t ∈ {i1 , i2 , ··· , iτ(v)} v  v,j v,j v,j Ij,t =  0, otherwise.

102 1 2 τ(v) We assume that iv,j < iv,j < ··· < iv,j without loss of generality. For j ∈ {1, 2, ··· , k0} and v ∈ V , deﬁne   1, if ξ ∈ {1, 2, ··· , τ(v)}, ˇv I β = M(j−1)+(β−1)m +θ,M(i −1)+(θ−1)τ(v)+ξ v v,j  0, otherwise,

for β ∈ {1, 2, ··· , τ(v)} and θ ∈ {1, 2, ··· , mv}. That is, activity v is scheduled

β on ﬂow unit M(j − 1) + (β − 1)mv + θ in time units M(iv,j − 1) + (θ − 1)τ(v) + 1 β through M(iv,j − 1) + θτ(v).

for n ∈ N and j ∈ {1, 2, ··· , Mk0}.

cyclic It is easy to see thatπ ˆ ∈ Πtotal . By statement (a) of Lemma 3.2.1, the process rate is

Mk0 k0 rπˇ = = = rπ0 . ML0 L0

Thus, we have established claim 3.

Proof of Claim 4

The proof of Claim 4 is similar to that of Claim 1; we therefore avoid providing the details for brevity.

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107 BIOGRAPHICAL SKETCH

Yang Bo was born in the city of Baotou, Inner Mongolia, China. He received his bachelor’s degree in Mechanical Engineering from Tsinghua University, China. He then joined the PhD Program in Operations Management at The University of Texas at Dallas (UTD). Later this year, he will join the faculty at the Chinese University of Hong Kong as an Assistant Professor of Operations Management.

108 CURRICULUM VITAE

Yang Bo May 16, 2017

Contact Information: Department of Operations Management Email: [email protected] The University of Texas at Dallas 800 W. Campbell Rd. Richardson, TX 75080-3021, U.S.A. Educational History: BS, Mechanical Engineering, Tsinghua University, 2012 PhD, Operations Management, The University of Texas at Dallas, 2017 Essays on Inventory Management, Capacity Management, and Resource-Sharing Systems Ph.D. Dissertation Operations Management, The University of Texas at Dallas Advisors: Dr. Milind Dawande and Dr. Ganesh Janakiraman

Professional Memberships: The Institute for Operations Research and the Management Sciences, 2013–present The Manufacturing and Service Operations Management Society, 2014–present