modeling, analysis, and algorithmic development of some ... · the biomass logistics problem is a...

Modeling, Analysis, and Algorithmic Development of SomeScheduling and Logistics Problems Arising in Biomass Supply

Chain, Hybrid Flow Shops, and Assembly Job Shops

Sanchit Singh

Dissertation submitted to the Faculty of theVirginia Polytechnic Institute and State University

in partial fulfillment of the requirements for the degree of

Doctor of Philosophyin

Industrial & Systems Engineering

Subhash C. Sarin, ChairRobert H. SturgesMichael R. Taaffe

Manish Bansal

April 22, 2019Blacksburg, Virginia

Keywords: Lot Streaming, Hybrid Flow Shop, Biomass Supply Chain, NestedBenders Decomposition, Mass Customization, Assembly Job Shop, Lagrangian

RelaxationCopyright 2019, Sanchit Singh

Modeling, Analysis, and Algorithmic Development of SomeScheduling and Logistics Problems Arising in Biomass Supply

Chain, Hybrid Flow Shops, and Assembly Job Shops

Sanchit Singh

ABSTRACT

In this work, we address a variety of problems with applications to ‘ethanol produc-tion from biomass’, ‘agile manufacturing’ and ‘mass customization’ domains. Ourmotivation stems from the potential use of biomass as an alternative to non-renewablefuels, the prevalence of ‘flexible manufacturing systems’, and the popularity of ‘masscustomization’ in today’s highly competitive markets. Production scheduling anddesign and optimization of logistics network mark the underlying topics of our work.In particular, we address three problems, Biomass Logistics Problem, Hybrid FlowShop Scheduling Problem, and Stochastic Demand Assembly Job Scheduling Prob-lem.

The Biomass Logistics Problem is a strategic cost analysis for setup and operationof a biomass supply chain network that is aimed at the production of ethanol fromswitchgrass. We discuss the structural components and operations for such a net-work. We incorporate real-life GIS data of a geographical region in a model thatcaptures this problem. Consequently, we develop and demonstrate the effectivenessof a ‘Nested Benders’ based algorithm for an efficient solution to this problem.

The Hybrid Flow Shop Scheduling Problem concerns with production scheduling ofa lot over a two-stage hybrid flow shop configuration of machines, and is often en-countered in ‘flexible manufacturing systems’. We incorporate use of ‘lot-streaming’in order to minimize the makespan value. Although a general case of this problem isNP-hard, we develop a pseudo-polynomial time algorithm for a special case of thisproblem when sublot sizes are treated to be continuous. The case of discrete sublotsizes is also discussed for which we develop a branch-and-bound-based method andexperimentally show its effectiveness in obtaining a near-optimal solution.

The Stochastic Demand Assembly Job Scheduling Problem deals with the schedul-ing of a set of products in a production setting where manufacturers seek to fulfillmultiple objectives such as ‘economy of scale’ together with achieving the flexibilityto produce a variety of products for their customers while minimizing delivery leadtimes. We design a novel methodology that is geared towards these objectives andpropose a Lagrangian relaxation-based algorithm for efficient computation.

Modeling, Analysis, and Algorithmic Development of SomeScheduling and Logistics Problems Arising in Biomass Supply Chain,

Hybrid Flow Shops, and Assembly Job Shops

Sanchit Singh

GENERAL AUDIENCE ABSTRACT

In this work, we organize our research efforts in three broad areas - Biomass SupplyChain, Hybrid Flow Shop, and Assembly Job Shop, which are separate in termsof their application but connected by scheduling and logistics as the underlyingfunctions. For each of them, we formulate the problem statement and identify thechallenges and opportunities from the viewpoint of mathematical decision making.We use some of the well known results from the theory of optimization and linearalgebra to design effective algorithms in solving these specific problems within areasonable time limit. Even though the emphasis is on conducting an algorithmicanalysis of the proposed solution methods and in solving the problems analytically,we strive to capture all the relevant and practical features of the problems duringformulation of each of the problem statement, thereby maintaining their applicability.

The Biomass Supply Chain pertains to the production of fuel grade ethanol fromnaturally occurring biomass in the form of switchgrass. Such a system requiresestablishment of a supply chain and logistics network that connects the productionfields at its source, the intermediate points for temporary storage of the biomass,and bio-energy plant and refinery at its end for conversion of the cellulosic contentin the biomass to crude oil and ethanol, respectively. We define the components andoperations necessary for functioning of such a supply chain. The Biomass LogisticsProblem that we address is a strategic cost analysis for setup and operation of such abiomass supply chain network. We focus our attention to a region in South CentralVirginia and use the detailed geographic map data to obtain land use pattern in theregion. We conduct survey of existing literature to obtain various transportationrelated cost factors and costs associated with the use of equipment. Our ultimateaim here is to understand the feasibility of running a biomass supply chain in theregion of interest from an economic standpoint. As such, we represent the BiomassLogistics Problem with a cost-based optimization model and solve it in a series of

smaller problems.

A Hybrid Flow Shop (HFS) is a configuration of machines that is often encounteredin the flexible manufacturing systems, wherein a particular station of machines canexecute processing of jobs/tasks simultaneously. In our work, we approach a specifictype of HFS, with a single machine at the first stage and multiple identical machinesat the second stage. A batch or lot of jobs/items is considered for scheduling oversuch an HFS. Depending upon the area of application, such a batch is either allowedto be split into continuous sections or restricted to be split in discrete sizes only. Theobjective is to minimize the completion time of the last job on its assigned machineat the second stage. We call this problem, Hybrid Flow Shop Scheduling Problem,which is known to be a hard problem in literature. We aim to derive the resultswhich will reduce the complexity of this problem, and develop both exact as well asheuristic methods in order to obtain near-optimal solution to this problem.

An Assembly Job Shop is a variant of the classical Job Shop which considers schedul-ing a set of assembly operations over a set of assembly machines. Each operationcan only be started once all the other operations in its precedence relationship arecompleted. Assembly Job Shop are at the core of some of the highly competitivemanufacturing facilities that are principled on the philosophy of Mass Customization.Assuming an inherent nature of demand uncertainty, this philosophy aims to achieve‘economy of scale’ together with flexibility to produce a variety of products for thecustomers while minimizing the delivery lead times simultaneously. We incorporatesome of these challenges in a concise framework of production scheduling and callthis problem as Stochastic Demand Assembly Job Scheduling Problem. We design anovel methodology that is geared towards achieving the set objectives and proposean effective algorithm for efficient computation.

iv

Dedication

I truly dedicate my dissertation to my mother and father. For all that I am, andhope to be, I owe it to them.

v

Acknowledgments

I would first like to thank my advisor, Dr. Subhash C. Sarin, whose expertise wasinvaluable in the formulating of the research topic and methodology in particular.His constant encouragement allowed me to cultivate my passion for unconstrainedthinking towards problem-solving freely. I have benefited immensely from his reviewof my work and many insightful discussions about research. His calm demeanor andhigh moral values will continue to inspire me to become a better person myself. Sir,you have not only been my guru but a mentor as well! You have always encouragedme to be driven and remain focused in the face of uncertainty and hardship. Iconsider myself extremely lucky to come to know you and will continue to look up toyou for your advice and encouragement in my future endeavors too. I thank Dr. JohnCundiff for his kind support and co-operation, leading to significant development inmy research. I am also grateful to my other committee members, Dr. Robert H.Sturges, Dr. Michael R. Taaffe, and Dr. Manish Bansal, who were very supportive,and provided many useful suggestions. I want to thank the Grado Department ofIndustrial and Systems Engineering of Virginia Tech for their continued support inthe form of assistantships besides providing for a great learning opportunity througha valuable course curriculum. During my stay in Blacksburg, I have made manyfriends. In particular, I would like to thank Almas Khan, Anupam Gupta, BadriNarayanan, Harsh Chaturvedi, and Rikin Gupta, among many others, for beingthere and enriching my life experience. Lastly, I would like to acknowledge the moralsupport and unconditional love that I have received from my beloved parents, sisters,and brother-in-law throughout my time spent while writing this dissertation.

vi

Contents

1 Introduction 1

1.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Contribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Design of a Biomass Logistics System for Cellulosic Ethanol Pro-duction using Nested Benders Decomposition 7

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Biomass Logistics System . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2.1 Land-use scenarios and operations at production fields . . . . 10

2.2.2 Operations at SSLs and BePs . . . . . . . . . . . . . . . . . . 14

2.2.3 Related Work . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2.4 Model Formulation . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Nested Benders decomposition . . . . . . . . . . . . . . . . . . . . . 21

2.4 Results and Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.4.1 Case of a single BeP . . . . . . . . . . . . . . . . . . . . . . . 32

2.4.2 Case of multiple BePs . . . . . . . . . . . . . . . . . . . . . . 35

2.5 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

vii

3 Single Lot, 1+m Hybrid Flow Shop Lot-streaming Problem 38

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.2 Model Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.3 Solution Methodology For Continuous Sublot Sizes . . . . . . . . . . 43

3.3.1 Determination of optimal schedule when the number of sublotsis fixed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.3.2 Determination of optimal schedule when the number of sublotsis not specified . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.4 Case of Discrete Sublot Sizes . . . . . . . . . . . . . . . . . . . . . . . 49

3.4.1 Branch-and-bound-based method . . . . . . . . . . . . . . . . 49

3.5 Computational Investigation . . . . . . . . . . . . . . . . . . . . . . . 52


4 A New Production Methodology Towards Achieving Mass Cus-tomization 58

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.2 Without Timing Aspect . . . . . . . . . . . . . . . . . . . . . . . . . 62


4.2.2 A Numerical Example . . . . . . . . . . . . . . . . . . . . . . 70

4.2.3 Numerical Example - Analysis of Results . . . . . . . . . . . 76

4.3 With Timing Aspect . . . . . . . . . . . . . . . . . . . . . . . . . . . 78


4.3.2 Solution Methodology . . . . . . . . . . . . . . . . . . . . . . 84

4.3.3 Algorithm To Solve ATO-CST . . . . . . . . . . . . . . . . . 97

4.4 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100


viii

5 Concluding Remarks and Ideas for Future Work 106

Bibliography 110

Appendices 118

A Proof of Theorem 3.1 119

B Determination of s1 127

B.1 Determination of s1, case: p 6= m . . . . . . . . . . . . . . . . . . . . 127

B.1.1 Linear homogenous recurrence sequence with constant coeffi-cients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

B.1.2 Determination of βn,k and ωn, ∀k = 1, . . . ,m− 1 . . . . . . . 132

B.2 Determination of s1, case: p = m . . . . . . . . . . . . . . . . . . . . 133

C Proof of Corollary 3.1 136

D Proof of Corollary 3.2 137

E Proof of Theorem 3.2 139

F Proof of Theorem 3.3 141

G Proof of Corollary 3.3 144

H Proof of Theorem 3.4 145

I Data Files Used For Table 4.16 and Table 4.17 146

ix

List of Figures

2.1 Existing rail network and distance to refinery. . . . . . . . . . . . . . 11

4.1 An illustration of the implementation of ATO-CS-D. . . . . . . . . . 67

4.2 Example 4.1-BOM of products. . . . . . . . . . . . . . . . . . . . . . 70

4.3 Example 4.1-Unique representation of each sub-assembly (node) in theBOM of products. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

4.4 Example 1: Comparison of expected cost (production and inventoryloss) for MTO, MTS, ATO and ATO-CS. . . . . . . . . . . . . . . . . 77

A.1 Proof of Theorem 3.1, Case A2.1: A schedule where ˆsr+1 = sr+1 +x, x < 0. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

A.2 Proof of Theorem 3.1, Case A2.2.1: A schedule where ˆsr+1 = sr+1 +x, x > 0, x− (vr + . . .+ vz) ≥ 0,∀r ≤ z ≤ r + 2−m. . . . . . . . . . 123

A.3 Proof of Theorem 3.1, Case A2.2.2: A schedule where ˆsr+1 = sr+1 +x, x > 0, x−(vr+. . .+vz) < 0, for atleast one value of z, r ≤ z ≤ r+2−m.125

x

List of Tables

1.1 Problems and their features, techniques used to solve them, and ourcontribution. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.1 Description of the land-use scenarios for switchgrass production andtotal available land area for 48-Km region around Gretna, VA. . . . . 12

2.2 Annual cost for a single equipment system, e0. . . . . . . . . . . . . . 15

2.3 At-plant size reduction for ‘densification’ equipment system. . . . . . 16

2.4 Experimental results for Gretna region: 48-Km (without rail trans-portation). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.5 Experimental results for overall demand at refinery with two potentialBeP regions - Gretna, VA and Bedford, VA. . . . . . . . . . . . . . . 36

3.3 Results for 1 + m TSHFS-LSP in the case of discrete sublot sizes(U = 100, t = 0.20 secs). . . . . . . . . . . . . . . . . . . . . . . . . . 54





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4.1 Products Data. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

4.2 Composition of sub-assemblies. . . . . . . . . . . . . . . . . . . . . . 71

4.3 Example 1: Cost of production by Method 1 (sub-assembly-specific)for deterministic demand (d1 = 10, d2 = 10). . . . . . . . . . . . . . . 72

4.4 Example 1-Cost of production by Method 2 (product-specific) for de-terministic demand (d1 = 10, d2 = 10). . . . . . . . . . . . . . . . . . 72

4.5 Example 1: Cost of production by ATO-CS for deterministic demand(d1 = 10, d2 = 10). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

4.6 Example 1: Cost of production and inventory losses in MTS systemfor stochastic demand (d1 ∼ Unif(9, 11), d2 ∼ Unif(9, 11)). . . . . . . . 74

4.7 Example 1: Cost of production and inventory losses in MTO systemfor stochastic demand (d1 ∼ Unif(9, 11), d2 ∼ Unif(9, 11)). . . . . . . . 74

4.8 Example 1: Cost of production and inventory losses in ATO systemfor stochastic demand (d1 ∼ Unif(9, 11), d2 ∼ Unif(9, 11)). . . . . . . . 75

4.9 Example 1: Cost of production and inventory losses in ATO-CS systemfor stochastic demand (d1 ∼ Unif(9, 11), d2 ∼ Unif(9, 11)). . . . . . . . 76

4.14 Analytic comparison between MTS, MTO, and ATO and the proce-dure to represent them as a special case of ATO-CST. . . . . . . . . . 101

4.15 Comparison of MTS, MTO, ATO, and ATO-CST. . . . . . . . . . . . 102

4.16 Results for direct solution by CPLEX R©. . . . . . . . . . . . . . . . . 104

4.17 Results for Algorithm 4.1 (subgradient method). . . . . . . . . . . . . 104

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Chapter 1

Introduction

Scheduling and logistics are important functions that are encountered in myriadsof application domains. With change in global trends pertaining to consumption,manufacturers and production planners need to continually adapt too. This has ledto the advent of ‘agile manufacturing’ and ‘mass customization’ in the recent past.There is also seen an urgency to develop renewable sources of energy to replace thefossil fuels. New and challenging problems pertaining to scheduling and logistics arisethat must be properly addressed for effective and efficient design and implementationof these systems. In this dissertation research, we address some pressing problemsthat arise in these systems and develop novel approaches for their solution. In par-ticular, we solve such problems pertaining to shipment of feed-stock to a bio-energyplant, delivery of parts and sub-assemblies from a facility at Stage 1 to parallel fa-cilities at Stage 2 (in ‘agile manufacturing’ or a ‘mass customization’ environment),and scheduling production of sub-assemblies that are common to various products,a situation often encountered in a ‘mass customization’ setting.

In the remainder of this chapter, we present a brief motivation for addressing theseproblems, definition of the problems discussed, and the contributions made.

1.1 Motivation

Biomass Logistics Problem. Amidst a rise in concern over the use of fossil fuelsbecause of their impact on global climate change, and in order to reduce dependenceon petroleum, the U.S. Congress adopted the Renewable Fuel Standard (RFS) in

1

Sanchit Singh Chapter 1. Introduction 2

2005 and expanded it under The Energy Independence and Security Act (EISA),2007, with the focus to increase production of advanced biofuels. Among its manyguidelines and directives, this Act specifies domestic production targets on advancedbiofuels to be 21.0 billion gallons by 2022, of which 16.0 billion gallons must beproduced in the form of cellulosic biofuel.

A biofuel is a liquid fuel that is used primarily for transportation. Besides biodiseal(a fuel that is typically made from soybean, canola, or other vegetable oils), ethanolis another important category of biofuel. Most of the ethanol produced in the U.S.is distilled from corn kernels. Contemporary research indicates that this process notonly results in higher food prices but also is insufficient in meeting desired productiongoals. There is a need to incorporate the second generation of biofuel that canbe produced from the cellulosic feed-stock that is available in different geographicregions.

Hybrid Flow Shop Scheduling Problem. A Hybrid Flow Shop (HFS) con-figuration is often encountered in flexible manufacturing systems that work on theprinciple of ‘agile manufacturing’. Primarily, these systems facilitate in effectively re-sponding to changing customers’ demand, and where distinct components producedat Stage 1 are configured into various products on the parallel machines at Stage2. We can generally split the application of HFS into two categories: continuousprocessing industries such as textile, food processing, chemicals and pharmaceutical,and processing of discrete products such as electronics, furniture, and steel indus-tries. Another typical application involves scheduling of a program on a parallel-taskcomputer where the external memory buffers data in a sequential manner at Stage 1while the CPU executes the program concurrently on the available parallel processorsat Stage 2.

Assembly Job Shop Scheduling Problem. Manufacturers have discovered thatthey can no longer capture market share and gain higher profits by producing largevolumes of a standard product for a mass market. They face the challenge of provid-ing as much variety as possible for the market without having to alter the manufac-turing system continually. Also, providing the finished goods to their customers withthe smallest possible lead time (i.e., the time between placement of an order and itsfabrication or delivery) becomes vital to ensure customer satisfaction and maintain acompetitive edge. Such a manufacturing mode is called ‘mass customization’, whichaims at meeting customers’ diverse requirements without a corresponding increasein cost and lead time via economy of scale.

Two common approaches that have been implemented by some of the leading manu-


facturers, such as Black and Decker, Hewlett-Packard, and Volkswagen are PlatformProduct Design and Platform Architecture. These approaches aim at creating a nec-essary product variety for competitive success in the market, and at the same timereducing both the production cost and time-to-market to a competitive level. Theunderlying idea is to produce common components of a variety of products on a massscale and add or remove parts to form final products as their demand occur.

We believe that design efforts leading to consideration of commonality within prod-ucts are crucial to a successful implementation of the ‘mass customization’ system.However, the current research in the literature on the design of platforms and theirproduction scheduling makes some unreasonable and simplistic assumptions as partof their problem statement. Most of the studies on this subject represent a productor product family as a set of components. There is little to no attention given tothe detail of the assembly structure of a product. In our work, we widen the scopeby considering a Bill-of-materials (BOM) for high-level assembly structure of all theproducts. We also assume the sub-assemblies to be common among different prod-ucts. We focus on production scheduling in the domain of ‘assembly job shop’, inthe presence of stochastic demand.

1.2 Problem Description

The Biomass Logistics Problem (BLP) that we address belongs to the productionof ethanol from a seasonal harvest of switchgrass (a cellulosic feed-stock) in UpperSoutheast of U.S.A. The operations include hauling of switchgrass from the pro-duction fields to Satellite Storage Locations (SSLs), loading/unloading of biomassat SSLs, transportation of switchgrass from SSLs to Bio-energy Plants (BePs), andtransportation of bio-crude oil from the BePs to a refinery via rail. We use a detailedland-use pattern from the Geographic Information System (GIS) data for a partic-ular location for establishing BeP operations and consider this as a base model forother locations in its vicinity. We study the availability of biomass in this region fora total of six harvest scenarios.

The Hybrid Flow Shop (HFS) that we consider consists of two stages of productionwith a single machine at Stage 1 and m parallel machines at Stage 2. We incorporatethe idea of ‘lot-streaming’ within 1 + m HFS. ‘Lot-streaming’ refers to a processwhere a production lot is split into sublots, which can be processed on machines ortransferred over from one machine to the next in an overlapping manner. It is widelyknown to offer benefits of lower completion times and higher machine utilization in


flow shops, job shops, and related configurations, when compared with processingof a single lot as a whole at a machine and then transferring it to a downstreammachine. The HFS that we consider processes a single lot in sublots (consisting ofmultiple identical items), where each sublot is processed first on a single machine atStage 1 and then on one of the m identical parallel machines at Stage 2. A sublotincurs fixed removal time before being transferred from the Stage 1 machine to one ofthe machines at Stage 2. The problem is to find an optimal number of sublots, sublotsizes, and to allocate sublots to the machines at Stage 2, to minimize makespan. Wecall this problem as Two-stage Hybrid Flow Shop Lot-streaming Problem (1 + mTSHFS-LSP). We obtain an optimal solution for both the cases of continuous anddiscrete sublot sizes.

The Stochastic Demand Assembly Job Shop Scheduling Problem (SD-AJSSP) thatwe address can be defined as follows: Given a set of products with stochastic demand,each of which consists of sub-assemblies that are common across products, and a setof machines for performing assembly operations, determine an optimal schedule forin-house production, in order to minimize the total cost incurred due to production,inventory, and delay in products’ order fulfillment. As mentioned earlier, there is acommonality of sub-assemblies across products. Each contiguous assembly operationincurs a setup cost and time and unit production/assembly cost and time. The sub-assemblies can either be produced/assembled from its constituent sub-assemblies andsubsequently be stocked up as inventory in anticipation of stochastic demand (Stage1), or assembled once the products’ demand is realized (Stage 2). The demand foreach product occurs at the onset of Stage 2. If any sub-assembly that is produced isnot utilized towards the fulfillment of products’ demand, it incurs a unit inventorycost. First, we relax the timing aspect, i.e., the assembly times of the sub-assembliesover the machines are not taken into account, and we only determine optimal quan-tities of sub-assemblies to be produced at both the stages of production. Then, weincorporate the timing aspect, thereby considering a full-fledged ‘assembly job shop’.For this case, we determine optimal quantities of sub-assemblies to be produced atboth the stages of production as well as their schedule on the machines for the ob-jective of minimizing the total cost of production, excess inventory, and due to delayin products’ order fulfillment (which is proportional to the lead time of a product’sorder delivery for all products).


1.3 Contribution

Table 1.1 presents an overview of the problems addressed in this dissertation. Wealso list the domain, decisions to be made, objective, techniques used to obtain thesolution, and our contributions and the novelty factor in our approach for each ofthese problems.

Table 1.1: Problems and their features, techniques used to solve them, and ourcontribution.

Problem Domain & Scope Decisions Objective Techniques Contribution & NoveltyFactor

BiomassLogisticsProblem

Logistics + Schedul-ing; Strategic + Op-erational

Selection of fields, SSLs,BePs, allocation of fields toSSLs, routing of equipment

Minimize equipment owner-ship cost, SSL setup cost,transportation cost via roadand rail

Nested Bendersdecomposition

Optimality cuts at integersolutions

1+m TSHFS-LSP

Scheduling; Opera-tional

Number of sublots, sublotsizes, allocation of sublots tomachines

Minimize makespan Recurrence se-quences, Branch& bound

Polynomial closed-formexpressions for continuoussublot sizes

SD-AJSSP Scheduling; Opera-tional

Quantities of sub-assembliesproduced at two stages;scheduling of sub-assemblieson machines

Minimize total productioncost, inventory loss and de-lay cost

Lagrangianrelaxation,subgradientmethod, Bendersdecomposition

Proposed ATO-CST thatcan leverage commonality ofsub-assemblies across prod-ucts

We develop a mathematical model for a comprehensive Biomass Logistics Problem(BLP). We decompose this model into multiple smaller problems based on the ‘NestedBenders decomposition’ technique. The novelty of this decomposition scheme isthat the sub-problems can be solved as integer programs rather than as their LPrelaxations. As such, we devise optimality cuts that can capture solution value atan integer solution for the sub-problem(s). We also show the validity of these cutswhich ensures the convergence of the decomposition approach.

While the two-stage ‘flow shop lot-streaming’ problems, as well as two-stage HFSproblems, have been studied extensively, there is little work done on the use of‘lot-streaming’ in an HFS setting, and the one that is presented is often restrictivein scope due to either: (1) being based on the assumptions such as equal sublotsizes, a fixed number of sublots or both, or (2) using a solution method which doesnot guarantee optimality such as the use of a sub-optimal heuristic method or asimulation model. We show that 1 +m TSHFS-LSP is NP-hard, and develop closed-form expressions and algorithms that can effectively solve a special version of thisproblem. In particular, we present closed-form expressions for optimal sublot sizesfor a given number of sublots, n, when the sublot sizes are relaxed to be continuous.We obtain these expressions by using recurrence sequences that lead to solving theproblem in polynomial time. We then present a pseudo-polynomial time algorithm


to determine optimal number of sublots and optimal schedule for processing sublotson the Stage 2 machines, for the case of continuous sublot sizes. Next, we addressthe case of discrete sublot sizes, wherein we develop a heuristic method based on a‘branch-and-bound’ scheme, following which, we can obtain optimal number and sizesof sublots and their schedule on the Stage 2 machines. We compare the performanceof this method with that obtained by directly solving the proposed model formulationfor the 1 +m TSHFS-LSP by CPLEX R©.

Assembly Job Shop Scheduling Problem (AJSSP) is a strongly NP-hard problemeven when the demand is deterministic. Most of the current research uses dispatch-ing rules to solve this problem to circumvent its complexity. Stochastic demandadds to the complexity of AJSSP. On the operational level, we propose to handlethis problem by having production in two stages, one prior to products’ demandrealization and the other after demand has been realized. As such, we propose a newproduction methodology, Assemble-To-Order with Commonality of Sub-assembliesand Timing Aspect (ATO-CST) to solve the SD-AJSSP. The ATO-CST is a cost-based production scheduling method which splits the production plan into two stages.We compare its performance against some of the other commonly used productionmethodologies that can be applied to solve the problem: (1) Make-To-Stock (MTS)- all the sub-assemblies are produced in anticipation of the worst-case scenario priorto demand realization and stocked up as inventory, (2) Make-To-Order (MTO) -all the sub-assemblies are produced after demand realization, and (3) Assemble-To-Order (ATO) - only the final assemblies representing the top level in products’Bill-of-materials are produced after demand realization, and the rest of the items areproduced prior to it. Our computational investigation as well an analytical compar-ison reveals that ATO-CST offers the least total cost among all four methodologiesconsidered for all the test cases used. We present a simple demonstration of theuse of our methodology (by relaxing some features) through an example. We alsodevelop an algorithm that can effectively solve the mathematical model correspond-ing to ATO-CST for large-sized data-set as well. This algorithm is based on thesolution of a ‘Lagrangian dual’ problem using the ‘subgradient method’, wherein theinner ‘Lagrangian relaxation’ problem is solved using the ‘Benders decomposition’technique. Towards the end, we present computational results demonstrating the ef-fectiveness of the proposed algorithm in obtaining lower cost solutions for test casesof large-sized data-sets over the direct solution of the mathematical model of theproblem using state-of-the-art commercial solver CPLEX R©. For all the test casesconsidered, the algorithm not only out-performs the direct solution method in termsof total cost, but also attains a lower optimality gap.

Chapter 2

Design of a Biomass LogisticsSystem for Cellulosic EthanolProduction using Nested BendersDecomposition

2.1 Introduction

The U.S. Congress adopted Renewable Fuel Standard (RFS) in 2005 which wasfurther extended under The Energy Independence and Security Act, 2007 (110thCongress [2007]). The legislation requires oil companies to blend renewable fuelsinto transportation fuels in increasing amounts each year, culminating with 36.0billion gallons (BG) in 2022. The applicable volume of advanced biofuel (defined asa renewable fuel that can reduce greenhouse gas emissions by at least 50%) for theyear 2022 is 21.0 BG, of which 16.0 BG must be produced in the form of cellulosicbiofuel (defined as a biofuel produced from grasses, wood, algae, or other plants).

On March 26th, 2010, the U.S. Environmental Protection Agency (EPA) expandedthe original RFS with the announcement of the final rule (Agency [2017]). It is note-worthy to point out that in 2016, the U.S. consumed 143.22 BG of finished motorgasoline (U.S. Energy Information Administration [2019b]), 10% of which came fromethanol (used as a blending agent in gasoline). In the same year, the U.S. produced15.40 BG of ethanol (U.S. Energy Information Administration [2019a]) with a com-

7

Sanchit Singh Chapter 2. Biomass Logistics Problem 8

bined bio-refinery capacity of 16.05 BG (almost entirely from corn-based production,i.e., using corn kernels). Out of this, the estimated combined total capacity for pro-duction of cellulosic ethanol (based on non-starchy feed-stock such as corn stover andswitchgrass) remained only at 121.00 million gallons (Renewable Fuels Association[2016]) against the statuary applicable volume of 4.25 BG in cellulosic biofuel in theyear 2016 (110th Congress [2007]).

Resop et al. [2011] assert that there is a concern with the use of grains for ethanolproduction (such as corn starch-based ethanol) as it tends to increase food prices,in order to justify the requirements in EISA w.r.t. cellulosic ethanol production tar-gets. They propose that the land that is unsuitable for food production and thusnot actively managed for growing crops be used for fuel production as an alterna-tive. As such, we focus our attention on South Central Virginia (SCV) in UpperSoutheast. Historically, this has been a region with tobacco as the main cash crop.Since tobacco production is declining (Capehart [2004]), the local farm economyneeds new opportunities. Some of the cropland fields can be used to grow perennialwarm-season grass (referred to as switchgrass) owing to the prevalent poor qualitysoil which is not suitable for grain production otherwise. Fike et al. [2006] haveperformed an extensive study on switchgrass in Upper Southeast, further justifyingconsideration of this particular species as a suitable biomass candidate for cellulosicethanol production.

2.2 Biomass Logistics System

The biomass is hauled from production fields to satellite storage locations (SSLs),which serve as intermediate storage places. It is loaded on the racks/trailers by aspecialized equipment system located at an SSL, and subsequently carried to a bio-energy plant (BeP) by delivery trucks. At a BeP, the cellulosic content in the biomassis converted into bio-crude oil. This bio-crude oil is then transported from BeP to arefinery for final conversion and use as an energy product such as fuel ethanol.

The cost incurred for transporting biomass from the production fields until its con-version to an energy product at a refinery is known to constitute a large portionof the total cost of biofuel production. Fales et al. [2007] report that the trans-portation cost amounts to 35− 60% of the total price of ethanol biofuel in its entiresupply chain. Raw biomass not only has a relatively low haul-density, but also, theequipment designed for hauling biomass in the field is inefficient on the roadways ascompared with the use of tractor-trailer trucks. It is for this purpose that we propose


the use of satellite storage locations (SSLs), distributed throughout the productionregion (Cundiff et al. [2007]), and the use of a specialized equipment system to un-load SSLs. An SSL is developed on a piece of an existing field itself. We also permitthe equipment system to move from its current SSL location to the next once it getsemptied of stored biomass.

Several previous studies have used intermediate points in between fields and BeP forstoring biomass, and they have attested to their effectiveness. Morey et al. [2010] andZhu et al. [2011] call these facilities ”local storage” and ”warehouses”, respectively.Cundiff et al. [2007] define SSL to be a dedicated uncovered gravel piece of landwithin a field in close proximity to the highways or secondary roads, which can serveas an intermediate storage location for round bales of switchgrass harvested from oneor more production fields.

Problem Statement: The Biomass Logistics Problem (BLP) that we address inthis chapter can be stated as follows: For a given set of production fields and po-tential SSL locations, design an optimal logistics network consisting of productionfields and SSLs, together with the assignment of production fields to their respectiveSSLs. Additionally, determine the optimal locations of the BePs from a set of prede-termined candidate locations in order to minimize total cost on an annual basis fortransportation of: (1) biomass from each chosen production field to its correspondingSSL, (2) biomass from each chosen SSL to a BeP, and (3) bio-crude oil from eachBeP to the refinery. Moreover, determine the minimal cost of routing of the mobileequipment system(s) among the SSLs.

In practice, an end-to-end biomass logistics system would involve a single refinerybeing fed by multiple BePs situated in locations where the biomass is in abundance.To reduce the cost of transportation of bio-crude from BePs to refinery, it is prudentto fix the locations of BePs close to existing rail network. As an example, consider fewpotential locations for the BePs in Virginia and North Carolina along a rail corridorsuch as Bedford, Keysville and Gretna in Virginia, and Reidsville and Liberty inNorth Carolina (see Figure 2.1 (a))).

For reasons that the BePs do not compete with each other for minimum fixed biomassintake, they are kept wide spread. As such, all BeP candidate locations operate withmutually exclusive clusters of farms, SSLs, and equipment systems.

We assume a BeP to be located in Gretna, Virginia. Production fields within an areaof 48-Km radii around BeP are considered for supplying the biomass (switchgrass).One more candidate BeP location is considered at Bedford, VA. However, withoutloss of generality, the biomass distribution, choice of potential SSLs and number of


equipment systems dedicated to this region are all assumed to be similar to thatconsidered for Gretna region. The closest refinery that can process bio-crude oil inethanol is Marathon Petroleum Corporation refinery located at Catlettsburg, KY,at a distance of 297 km via the rail line from Bedford, VA and 364 Km from Gretna,VA (see Figure 2.1 (b) and (c), respectively). We did not obtain any statement ofinterest from either Marathon or Norfolk Southern, and consider their role from ahypothetical viewpoint in our study.

2.2.1 Land-use scenarios and operations at production fields

The procedure of ”Land Conversion Scenarios and Data-set Development” is de-scribed in Resop et al. [2011], which we will refer to as Resop study hereafter. Wepresent it briefly here. Existing plantation in individual fields is classified in oneof the Land Use/ Land Cover (LULC) categories namely scrubland, grassland, pas-tureland, and cropland. Table 2.1 describes six scenarios that are considered forswitchgrass production ranging from most conservative to most optimistic (ha is thefield size in hectares). We also present the total potential land area thus calculatedfor Gretna region in Resop study (column 3) and in our work (column 4). Note thatthe former uses density grid and raster analysis for calculation whereas the latterdirectly evaluates all 29,889 fields in the original data-set for LULC and switchgrassproduction scenarios. Hence, there is a slight difference in the values in columns 3and 4.


(a) System overview. Norfolk Southern Corp. [2019] (fair use).

Map data ©2019 Google 50 mi

6 h 17 min294 miles

via US-52 S6 h 17 min without tra�c

Drive 294 miles, 6 h 17 minKenova, WV to Bedford, Virginia

(b) Route between Kenovo, WV to Bedford, VA. Mapdata: Google, Data provider: Google [2019b] (fairuse).

Map data ©2019 Google 20 mi

Drive 65.3 miles, 1 h 43 minBedford, Virginia to Gretna, VA

(c) Route between Bedford, VA toGretna, VA. Map data: Google,Data provider: Google [2019a] (fairuse).

Figure 2.1: Existing rail network and distance to refinery.


Table 2.1: Description of the land-use scenarios for switchgrass production and totalavailable land area for 48-Km region around Gretna, VA.

Scenario DescriptionSwitchgrass production area (ha)Resop study Our work

1 Scrubland (80%) + Grassland (80%) 46,993 47,0012 Scrubland (80%) + Grassland (80%) +

Cropland (20%)49,724 49,747

3 Scrubland (80%) + Grassland (80%) +Cropland (40%)

52,455 52,479

4 Scrubland (80%) + Grassland (80%) +Cropland (40%) + Pastureland (5%)

57,139 57,151


61,823 61,839


76,653 76,686

The Resop study used Geographic Information System (GIS) data for selection ofpotential SSLs based on the criteria such as minimum switchgrass production crite-rion obtained using raster analysis and density grids for each LULC category for allsix different scenarios within a 3.2-Km radial circle of each cell considered as a po-tential SSL, proximity to either state highways or secondary roads in the region, anda minimum distance of 6.5-Km between any two adjacent SSLs. Based upon thesecriteria, they manually selected 99 SSLs in the 32-Km radial circle and another 100SLSs in the annular disc within 32-Km and 48-Km circles with Gretna at its center.Resop study is fairly simplistic in the sense that it only assesses the availability ofbiomass from the region for ethanol production without considering any operationalcosts. As such, it devised a heuristic rule that, for each of the 199 chosen SSLs,switchgrass harvested from production fields that lie within a radius of 3.2-Km, istransported to and stored in that SSL.

Judd et al. [2012] refer to the same spatial data-set for production fields as developedin Resop study. Unlike Resop study, they solve a whole logistics problem from fieldsto SSLs to BeP by considering operational costs. They consider 29,782 productionfields initially which are merged to reduce the level of detail. They incorporate asingle land-use scenario only, wherein the resultant fields are randomly selected toachieve 6% land utilization rate, resulting into 3,655 production fields and 43,434 haof switchgrass production area for the 48-Km Gretna data-set. It is capable of feeding


an 83 Mg/h BeP (assuming an annual switchgrass yield of 15 Mg/h) located at thecenter of Gretna region. In their optimization model, they list four requirements forthe selection of potential SSLs with few differences compared to the Resop study.Overall, they chose 589 potential SSLs. The data aggregation procedure and SSLsselection procedure is described in detail in §2.3.1 in Judd et al. [2012].

We consider spatial data-set for production fields as developed in Resop study for48-Km Gretna region. Six different switchgrass production scenarios are considered(see Table 2.1), for which the original 29,889 number of fields are aggregated usingk-means weighted clustering algorithm resulting into 1,000 fields. We argue that aproduction field (upon clustering) does not retain the same level of granularity asthe original ones that constitute it. Therefore, we forgo the requirements listed inJudd et al. [2012] for the choice of a field as a potential SSL. As such, without loss ofgenerality, we consider all 1,000 fields as potential SSLs in our work. For benchmarkpurpose, we generate a separate data-set wherein the locations of the 199 SSLs chosenin Resop study are matched to those in the 1,000 potential SSLs described above.For this data-set, we only use 199 SSLs from among 1,000 in total to generate thesolution to their problem. The cost of transporting the biomass from a field to anSSL is considered in our work even though the field owner is responsible for it withhis own equipment.

Fike et al. [2006] studied switchgrass production in Upper Southeast, where theyobserved “great variability” in average annual yield for different regions under variousfactors considered. It ranged from 10.4 Mg/ha in Jackson, TN to 19.1 Mg/ha inBlacksburg, VA. We assume a more conservative figure of 11.2 Mg/ha as the annualyield, Y , of switchgrass in Gretna region, which is the same as that used in Resopstudy. Cundiff et al. [2009] assume the in-field haul density to be 4.5 Mg/haul (ten450 Kg, 1.5-m and 5-ft round bales). The cost of hauling a unit Mg of biomass fromfield i to SSL j, fij, is obtained as follows (Cundiff et al. [2009]).

fij = f fI + f vI dij

= 3.3538 + 0.5856dij,

where f fI and f vI are fixed and variable cost parameters for in-field hauling, and dijis the travel distance between field i and SSL j in Km.


2.2.2 Operations at SSLs and BePs

The loading and unloading of biomass at each SSL is done using one of the threespecialized equipment systems, two of which utilize rack concept first described inCundiff et al. [2004] - ‘rear-loading’ and ‘side-loading’ systems, and the third referredto as ‘densification’ system (Morey et al. [2010]). Each set/unit of any of thesesystems is permitted to move across multiple SSLs for biomass processing. Roundbales of switchgrass are either loaded directly or are first chopped and densified toincrease haul density before being loaded onto the trucks meant for highway haulingon way to BeP.

The cycle of operations at each SSL, specialized equipment needed, number of work-ers required at either SSL or BeP and the cost analysis for all the three systems isdescribed in detail in Judd et al. [2011]. Table 2.2 lists the required units of equip-ment and workers in column 1 and the annual cost of ownership and operation ofequipment or the associated labor cost in column 2 for a single unit of each of thethree systems in operation at an SSL. The cost associated with operation of trucks isnot included in current analysis, together with that of trailers or racks that remainon the trucks while they are in motion in between SSLs and BeP.

The two rack systems,‘rear-loading’ and ‘side-loading’ have identical theoretical op-eration capacity of 21.6 Mg/h, and that for “densification” system is 22.7 Mg/h. Theoperating hours for any of the equipment systems at an SSL is 10 h/d, 6 d/wk and47 wk/yr totaling 2,820 h/yr.

A ‘densification’ system also requires at-plant size reduction equipment and workers,the cost of which is given in Table 2.3. Here, x denotes the number of size reductionunits that would be required at each BeP. It is calculated as follows.

x =

⌈BeP net Mg/h

22.7 Mg/h× 1

0.90

⌉,

where BeP net is the total biomass collected in the region in Mg/h. The theoreticalcapacity of equipment at SSLs is scaled down to 70% for any of the three systemsand 90% for at-plant size reduction units to account for operational delays. All thesystems for handling biomass at BeP are assumed to be operational for 24 h/d, 7d/wk and 47 wk/yr totaling 7,896 h/yr.

We will add the net cost given in Table 2.3 separately for the ‘densification’ systemto that obtained from the solution of optimization model(s), in order to make a fair


Table 2.2: Annual cost for a single equipment system, e0.

Equipment systemEquipment/Worker (units) Annual Cost ($)

Rear-loading rack systemTelehandler/grapple (1) 29,054 + 9.65× 2,820Rack loader (1) 16,410 + 4.56× 2,820SSL forklift (1) 22,635 + 15.24× 1,410Racks (4) (9,664/8 + 0.26/8× 2,820)× 4Workers (2) (20× 2,820)× 2Net 247,658.20

Side-loading rack systemTelehandler (1) 27,790 + 9.65× 2,820Drop-deck trailers (2) (16,107/2 + 3.19/2× 2,820)× 2Racks (4) (11,114/8 + 0.29/8× 2,820)× 4Workers (1) 20× 2,820Net 142,471.70

Densification systemSkid steer (1) 11,685 + 8.61× 2,820Tub-grinder (1) 134,160 + 111.35× 2,820Roll press (1) 103,200 + 15.60× 2,820Drop-deck trailers (2) (16,107/2 + 3.19/2× 2,820)× 2Workers (2) (20× 2,820)× 2Net 769,227.00

comparison with the other two rack systems.

The cost expenditure of using the rack systems and the densification system forhauling a unit Mg of biomass with trucks from SSL j to BeP l (with their respectivehauling capacity of 14.4 Mg/haul and 22.7 Mg/haul, respectively) is obtained asfollows.

fjl = f fT + f vTdjl

=

{1.667 + 0.1381djl, rack systems,

(1.1747 + 0.0974djl, densification system,

where f fT and f vT are fixed and variable cost parameters for highway hauling withtrucks, and djl is the travel distance between SSL j and BeP l in Km. Thus, the totalcost for transporting biomass from field i to BeP l through SSL j, fijl is calculatedas follows.

fijl = (fij + fjl)hai × Y


Table 2.3: At-plant size reduction for ‘densification’ equipment system.

Equipment/Worker (units) Annual Cost ($)Tub-grinder (x) (202,249 + 1.43× 7,896)xElectric motor (x) (16,638 + 26.07× 7,896)xDust control tech. (x) (53,691 + 2.80× 7,896)xWorker (2x) (20× 7,896)(2x)Net 827,666.8x

An SSL land rental fee of cR = $0.36/Mg is used in our study (Morey et al. [2010]),which is kept identical for all the SSLs. A fixed setup/take-down cost, cS = $600 isincurred to prepare an SSL for the equipment system that visits it. We assume abi-annual harvesting season for switchgrass cultivation in the SCV region. Therefore,it is assumed that the equipment system departs from its current SSL once that isemptied of stored biomass, and moves on to the next SSL on its route. Each SSLon its route is revisited in the same manner during the second half of the year. Assuch, the optimization model(s) will incorporate twice the cS cost as given above.

The road transportation cost for moving any of the equipment systems from SSL jto SSL k is considered to be, cjk = $2.29/Km.

With the SSL loading operation done using either of the two rack systems at 70%efficiency, the minimum and maximum SSL size is calculated assuming 3 days and30 days of operation, respectively, amounting to smin = 40.18 ha and smax = 401.79ha (= 21.6Mg/h× 10h/day× 30days× 70%/11.2Mg/ha), respectively.

We fix the minimum amount of biomass required at each chosen BeP, BePmin toensure that the BeP operates at a certain economy of scale. BePmin is fixed at42,752, 45,176, 47,601, 51,700, 55,799 and 68,847 ha for six scenarios respectively(the same as used in Resop study). This amount is capable of supplying a BeP with61 Mg/h to 98 Mg/h of switchgrass, which amounts to an annual production of 134.4ML to 216.4 ML of bio-crude oil after conversion (based on an estimate of 330 L/dryMg and 15% moisture content in biomass; DiPardo [2000]).

The cost incurred for hauling a unit Mg of bio-crude oil via rail from BeP l to refineryis obtained as follows (Heinrich et al. [2009]).

fl = f fR + f vRdl

= 9.30 + 0.0472dl,

where f fR and f vR are fixed and variable cost parameters, and dl is the rail traveldistance from BeP l to the refinery in Km.


2.2.3 Related Work

In their study, Judd et al. [2012] considered only a single hypothetical BeP in themiddle of Gretna, VA with a varying size of biomass catchment area, full modelbeing that of 48-Km as originally developed by Resop study. They considered threesystems for loading/unloading of switchgrass at SSLs, namely, ‘side-loading‘, ‘rear-loading‘ and ‘densification‘. Their objective function includes biomass transportationcost from fields to BeP, cost incurred at SSLs for both the cases of equipment setsbeing stationed at each SSL or mobile, together with the cost to move equipment setsamong SSLs. They showed mobile equipment system(s) to incur less cost than thatfor stationary equipment. Among the different systems used, side-loading results inthe most cost savings. We extend the work done by Judd et al. [2012] to multipleBePs. Even for the case of single BeP, there is a significant difference betweentheir and our work that lies in the methods used. They proposed a Mixed IntegerProgramming (MIP) model, BLP-M, that struggled to find an optimal solution whena mid to full size data-set is used. As such, they proposed a decomposition-basedheuristic model by solving a two-stage MIP model iteratively. We propose an exactmethod that can solve to optimality current industrial-sized problem in a reasonableamount of computational time, besides emphasizing on the theoretical as well aspractical implementation issues.

Zhu et al. [2011] developed an integer model formulation for the solution of a sim-ilar biomass logistics system as used in our work, with the assumption of a singlepredetermined BeP location. However, their model required the data used as inputsto be overly aggregated in order to keep the models size tractable. Therefore, itis limited in its use with the industrial scale spatial data considered in our work.Shastri et al. [2011] also developed a model that is based on data aggregation at thecounty level. In the formulations of large-sized problems, it is prudent to providea balance between the amount of detail captured in the model and the amount ofdetail provided by the data. We keep this in mind while developing our model(s).

2.2.4 Model Formulation

Consider the following notation.

L Set of potential BePs.I Set of potential fields.J Set of potential SSLs.


T Set of tours (a single pass route among assigned SSLs pertaining to each setof equipment system is alternatively referred to as a tour, and is denoted by t).

We make the following assumptions: (1) Each potential BeP location, l ∈ L, oper-ates with its own mutually exclusive set of fields, Il, SSLs, Jl and equipment sys-tems/tours, Tl. (2) An SSL, j ∈ J , is only permitted to procure biomass from fieldsthat lie within a radius of 20-Km (12-mi) around it. This set of fields is denoted byNIj. Conversely, NI i represents a set of all the SSLs, j ∈ J , such that field, i ∈ NIj.

The rest of the notation is as follows.

fijl Cost for transporting biomass from field, i, to BeP, l, through SSL, j.cjk Total cost for moving an equipment system from SSL j to SSL k.e0 Annual cost of an equipment system.cS Fixed setup/take-down cost of an equipment system.cR Land rental fee (per Mg of biomass stored on it) for the SSLs.hai Size of production field i (in hectares).smin Minimum size of an SSL (3 days worth of unloading in hectares).smax Maximum size of an SSL (30 days worth of unloading in hectares).Emax Maximum processing capacity for each equipment system/tour (in hectares).BePmin Minimum biomass intake for each BeP (in hectares).R Annual biomass requirement at the refinery (in hectares).

Decision variables:

yij = 1, if field i transports biomass to SSL j; 0 otherwise.yijt = 1, if field i transports biomass to SSL j which is allocated to tour t; 0

otherwise.ql = 1, if BeP l is utilized; 0 otherwise.zj = 1, if SSL j is utilized; 0 otherwise.zjt = 1, if SSL j is utilized and is allocated to tour t; 0 otherwise. Note that,

z0t is an indicator variable if tour t is utilized or not.bepl Amount of biomass (in hectares) processed at BeP l.sl Auxiliary variable that will be used in Nested Benders decomposition

based model.xjkt = 1, if SSL k is preceded by SSL j on tour t associated with an equipment

system; 0 otherwise.ujt Order/ranking variable for SSL j on tour t.wjt Auxiliary variable to eliminate subtours (under the assumption that there

is no fixed depot when routing equipment systems).


The objective function value for any model formulation, m, will be denoted by g(m).Also, a model and its corresponding decision variables are denoted with an “ ′ ” assuperscript, whenever the model is solved by relaxing the integrality constraints.For example, g(BLP-SSP

′

l) denotes the objective function value of relaxed ModelBLP-SSPl.

We now present Model BLP, for the Biomass Logistics Problem as follows.

BLP:

Minimize 2cS∑l∈L

∑j∈Jl

zj +∑l∈L

∑j∈Jl

∑i∈NIj

fijlyij + cR∑l∈L

∑j∈Jl

∑i∈NIj

haiyij

+ Y∑l∈L

flbepl + e0

∑l∈L

∑t∈Tl

z0t + 2∑l∈L

∑j∈Jl

∑k∈Jl,k 6=j

∑t∈Tl

cjkxjkt (2.1)

subject to:

zj ≤ ql, ∀l ∈ L, j ∈ Jl (2.2)

yij ≤ zj, ∀j ∈ J, i ∈ NIj (2.3)∑j∈NIi

yij ≤ 1, ∀i ∈ I (2.4)

∑i∈NIj

haiyij ≥ sminzj, ∀j ∈ J (2.5)

∑i∈NIj

haiyij ≤ smaxzj, ∀j ∈ J (2.6)

bepl =∑j∈Jl

∑i∈NIj

haiyij ≥ BePminql, ∀l ∈ L (2.7)

∑l∈L

bepl ≥ R (2.8)

bepl ≤ |Tl|Emax, ∀l ∈ L (2.9)

yjj = zj, ∀j ∈ J (2.10)∑t∈Tl

zjt = zj, ∀l ∈ L, j ∈ Jl (2.11)

zjt ≤ z0t, ∀l ∈ L, j ∈ Jl, t ∈ Tl (2.12)∑t∈Tl

yijt = yij, ∀l ∈ L, j ∈ Jl, i ∈ NIj (2.13)

yijt ≤ zjt, ∀l ∈ L, j ∈ Jl, i ∈ NIj, t ∈ Tl (2.14)


∑j∈Jl

∑i∈NIj

haiyijt ≤ Emax, ∀l ∈ L, t ∈ Tl (2.15)

∑j∈Jl,j 6=k

xjkt = zkt, ∀l ∈ L, k ∈ Jl, t ∈ Tl (2.16)∑j∈Jl,j 6=k

xkjt = zkt, ∀l ∈ L, k ∈ Jl, t ∈ Tl (2.17)∑j∈Jl

wjt = z0t, ∀l ∈ L, t ∈ Tl (2.18)

wjt ≤ zjt, ∀l ∈ L, j ∈ Jl, t ∈ Tl (2.19)

j−wj−,t ≥ jwjt, ∀l ∈ L, j ∈ Jl\J1l , t ∈ Tl (2.20)

ujt − ukt − |Jl|(wjt + wkt) + |Jl|xjkt + (|Jl| − 2)xkjt

≤ |Jl| − 1, ∀l ∈ L, j ∈ Jl, k ∈ Jl, k 6= j, t ∈ Tl (2.21)

ujt ≥ zjt − wjt, ∀l ∈ L, j ∈ Jl, t ∈ Tl (2.22)

ujt ≤∑k∈Jl

zkt − z0t, ∀l ∈ L, j ∈ Jl, t ∈ Tl (2.23)

ql, bepl ∈ {0, 1}, ∀l ∈ L (2.24)

zj ∈ {0, 1}, ∀j ∈ J (2.25)

yij ∈ {0, 1}, ∀j ∈ J, i ∈ NIj (2.26)

zjt ∈ {0, 1}, ∀l ∈ L, j ∈ Jl ∪ {0}, t ∈ Tl (2.27)

yijt ∈ {0, 1}, ∀l ∈ L, j ∈ Jl, i ∈ NIj, t ∈ Tl (2.28)

xjkt ∈ {0, 1}, ∀l ∈ L, j ∈ Jl, k ∈ Jl, k 6= j, t ∈ Tl (2.29)

ujt ≥ 0, ∀l ∈ L, j ∈ Jl, t ∈ Tl (2.30)

wjt ∈ {0, 1}, ∀l ∈ L, j ∈ Jl, t ∈ Tl (2.31)

The objective function in (2.1) consists of equipment systems’ setup cost, biomasstransportation cost from fields to BeP (in-field hauling from a field to SSl andtrailer/truck from SSL to BeP subsequently), SSLs’ rental cost, biomass transporta-tion cost from BePs to refinery by rail, equipment ownership cost, and travel costfor routing equipment systems through SSLs, respectively. Constraints (2.2) assert aBeP to be utilized only if it is used. Similarly, constraints (2.3) assert an SSL to beutilized only if it is used. Constraints (2.4) capture the fact that every productionfield is allocated to at most one SSL. Constraints (2.5) and (2.6) provide a lower andan upper bound, respectively, on the size of each SSL. Constraints (2.7) capture theamount of biomass processed at each BeP, together with providing a lower bound


on this amount. Constraints (2.8) ensure that enough biomass is collected throughall the BePs to fulfill a fixed annual requirement at the refinery. Constraints (2.9)provide an upper bound on the amount of biomass processed at each BeP, based onthe total capacity of all the equipment systems combined. Constraints (2.10) ensurethat a field that is used for the development of an SSL is allocated to that SSL itself,if used. Constraints (2.11) allow at most one tour to be allocated to an SSL if used.Constraints (2.12) assert a tour/equipment system to be utilized only if it is usedfor loading/unloading at one of the SSLs. Constraints (2.15) ensure tour balancingby restricting the maximum amount of biomass that a single equipment system canprocess. Constraints (2.16) and (2.17) are degree constraints for the underlying TSPat each tour. Constraints (2.18) and (2.19) ensure that exactly one SSL is consid-ered to act as a makeshift depot for a given tour, so that the other SSLs that belongto the same tour do not form a subtour. This is ensured by subtour eliminationconstraints (2.21) that were first proposed by Miller et al. [1960] (as such they arepopularly known as MTZ SECs), and subsequently tightened by Desrochers and La-porte [1991]. We modify the MTZ SECs to incorporate requirement that there isno fixed depot in any tour. Constraints (2.20) are symmetry breaking constraints.Constraints (2.22) and (2.23) not only provide lower and upper bound on the rank-ing variables respectively, but also tighten the LP convex hull of the formulation.Constraints (2.24) - (2.31) impose logical restrictions on the decision variables.

2.3 Nested Benders decomposition

In a conventional application of Benders decomposition technique (Benders [1962]),a series of master and sub-problems are solved iteratively. The master problem is arelaxation of the original formulation obtained by removing some constraints, andis solved to obtain an initial solution. A sub-problem that typically represents thedual, or pricing, of the relaxed constraints is solved to generate a new (Benders)cut if either the whole problem has not converged to within an MIP optimality gaptolerance or when the current master problem solution is violated. This constraintis added to the master problem, and is then re-solved. This process is continueduntil the master problem and sub-problem converge to an optimal solution or withinan MIP optimality gap tolerance. Since the constraints are relaxed and iterativelyadded into the master problem, Benders decomposition technique is also known asa relaxation strategy.

We extend the Benders decomposition (that involves two stages) to implement a


Nested Benders Decomposition (NBD) based methodology comprising of three stages.Louveaux [1980] first performed an outer linearization built on the two-stage L-shaped method or the Benders decomposition method for multistage quadratic prob-lems. Birge [1985] extended the two-stage method in the linear case. The approachalso appears in Pereira and Pinto [1985]. Almost entire literature reported for NBDmethod is devoted to multistage stochastic case. We apply this approach, however,to a deterministic problem, specifically, to solve Model BLP, that enables solutionof a real-life-sized problem. The novelty of our work is that we deal with an MIPprogram in the sub-problem and generates logic-based Benders optimality cuts assuch. This variation will be described later.

Now, we describe the Nested Benders decomposition to solve Model BLP. The firststage is referred to as Model BLP-MP. It concerns selection of BePs, SSLs andfields, and assignment of chosen fields to SSLs. For a given solution to Model BLP-MP, the remaining model is separable over individual BePs, and essentially becomesa Capacitated Vehicle Routing Problem (CVRP). We chose to solve it by furtherdecomposing into two stages. Thus, for a given solution to Model BLP-MP, and lth

BeP, the second stage is referred to as Model BLP-SPl. It concerns assignment ofSSLs to tours (Note: A tour is a single set of equipment system assigned to a routeamong multiple SSLs pre-designated to it).


Given a solution to Model BLP-SPl, the third stage is constructed, referred to asModel BLP-SSPl. It is essentially a collection of multiple yet separable TravelingSalesman Problems (TSPs), with a variant that no tour has a fixed depot. Eachequipment system is routed through a set of SSLs predesignated to it in a tour thatis repeated twice (since each SSL is visited twice by the same equipment system dueto bi-annual harvesting season). Model BLP-MP is formulated as follows.

BLP-MP:

Minimize∑l∈L

mp costl +∑l∈L

z spl (2.32)

subject to:

mp costl = 2cS∑j∈Jl

zj +∑j∈Jl

∑i∈NIj

fijlyij + cR∑j∈Jl

∑i∈NIj

haiyij

+ Yflbepl, ∀l ∈ L (2.33)

(Emax)sl ≥ bepl, ∀l ∈ L (2.34)

sl ∈ Z+, ∀l ∈ L (2.35)

{(2.2)− (2.10), (2.24)− (2.26)}

z spl captures the cost of Model BLP-SPl in Model BLP-MP. Constraints (2.33)equates mp costl to be the sum total of SSLs’ equipment setup cost, biomass trans-portation cost from fields to BeP (in-field hauling from a field to SSl and trailer/truckfrom SSLs to BeP subsequently), SSLs’ usage/development cost, and biomass trans-portation cost from BeP to refinery via rail on the right hand side respectively forlth BeP.

Model BLP-SPl is formulated as follows.

BLP-SPl(: ql = 1):

Minimize sp costl + z sspl (2.36)

subject to:

sp costl = e0

∑t∈Tl

z0t (2.37)∑t∈Tl

zjt = zj, ∀j ∈ Jl ← αj (2.38)


∑t∈Tl

z0t ≥ sl ← βl (2.39)

zjt ≤ z0t, ∀j ∈ Jl, t ∈ Tl (2.40)∑t∈Tl

yijt = yij, ∀j ∈ Jl, i ∈ NIj ← γij (2.41)

yijt ≤ zjt, ∀j ∈ Jl, i ∈ NIj, t ∈ Tl (2.42)∑j∈Jl

∑i∈NIj

haiyijt ≤ Emax, ∀t ∈ Tl ← δt (2.43)z sspl ≥ . . .

...

≡ OSPl ← νlφ, ∀l ∈ L, φ = 1, . . . , |OSP

l | (2.44)

z0t ≥ 0, ∀t ∈ Tl (2.45)

zjt ≥ 0, ∀j ∈ Jl, t ∈ Tl (2.46)

yij ≥ 0, ∀j ∈ Jl, i ∈ NIj (2.47)

z0t ∈ {0, 1}, ∀t ∈ Tl (2.48)

zjt ∈ {0, 1}, ∀j ∈ Jl, t ∈ Tl (2.49)

yijt ∈ {0, 1}, ∀j ∈ Jl, i ∈ NIj, t ∈ Tl (2.50)

z sspl captures the cost of Model BLP-SSPl in Model BLP-SPl. Constraints (2.34),(2.35) and (2.39) together enforce

∑t∈Tl z0t ≥ dbepl/Emaxe,∀l ∈ L. For practical

purposes, the size of constraints (2.42) presents heavy computational burden. Assuch, we drop them from Model BLP-SPl in favor of their aggregated version asgiven in constraints (2.51).∑

i∈NIj

yijt ≤ |NIj|zjt, ∀j ∈ Jl, t ∈ Tl (2.51)

We add symmetry breaking constraints to Model BLP-SPl in the form of (2.52)-(2.58), which enforce that the maximum index among all SSLs assigned to a tourfollow an increasing order with tour indices. This results in faster convergence be-tween SP and SSP stages.

bjt ≤ 1, ∀j ∈ Jl, t ∈ Tl ← εjt (2.52)

vt ≥ jzjt, ∀j ∈ Jl, t ∈ Tl (2.53)

vt− ≤ vt, ∀t ∈ Tl\T 1l (2.54)


vt ≤ jzjt + (maxj∈Jl{j})(1− bjt), ∀j ∈ Jl, t ∈ Tl ← ζjt (2.55)∑j∈Jl

bjt ≥ 1, ∀t ∈ Tl ← ηt (2.56)

bjt ≥ 0, ∀j ∈ Jl, t ∈ Tl (2.57)

bjt ∈ {0, 1}, ∀j ∈ Jl, t ∈ Tl (2.58)

Note that constraints (2.52) bound the variables, bjt, from above, and hold anymeaning only when Model BLP-SPl is solved as relaxed and the dual values at theseare non-zero.

In view of symmetry breaking constraints (2.52)-(2.58) used in Model BLP-SPl, bjtcan be used in place of wjt in deciding makeshift depot for a tour. Note that, thiswill also eliminate use of symmetry breaking constraints (2.20) for wjt, since choiceof bjt being unity is already implied uniquely by constraints (2.52)-(2.58) for eachtour.

We now present Model BLP-SSPl, wherein constraints (2.18)-(2.23) and (2.31) arerewritten as constraints (2.62)-(2.64).

BLP-SSPl(: ql = 1):

Minimize 2∑j∈Jl

∑k∈Jl,k 6=j

∑t∈Tl

cjkxjkt (2.59)

subject to:∑j∈Jl,j 6=k

xjkt = zkt, ∀k ∈ Jl, t ∈ Tl ← θkt (2.60)∑j∈Jl,j 6=k

xkjt = zkt, ∀k ∈ Jl, t ∈ Tl ← ιkt (2.61)

ujt − ukt + |Jl|xjkt + (|Jl| − 2)xkjt

≤ |Jl| − 1 + |Jl|(bjt + bkt), ∀ ∈ Jl, k ∈ Jl, k 6= j, t ∈ Tl ← κjkt(2.62)

ujt ≥ zjt − (bjt + z0t − 1), ∀j ∈ Jl, t ∈ Tl ← λjt (2.63)

ujt ≤∑k∈Jl

zkt − z0t, ∀j ∈ Jl, t ∈ Tl ← µjt (2.64)

xjkt ≥ 0, ∀j ∈ Jl, k ∈ Jl, k 6= j, t ∈ Tl (2.65)

xjkt ∈ {0, 1}, ∀j ∈ Jl, k ∈ Jl, k 6= j, t ∈ Tl (2.66)


ujt ≥ 0, ∀j ∈ Jl, t ∈ Tl (2.67)

Optimality cut from BLP-SSPl to BLP-SPl: For a given solution of ModelsBLP-SPl and BLP-SSPl, if the SP-SSP optimality gap is not within the set optimality

tolerance, τ sp, i.e.

∣∣∣∣g(BLP-SSPl)−z ssplz sspl

∣∣∣∣ > τ sp, we add the following φth set of optimality

cuts to optimality cuts’ set, OSPl (Note that, the dual solution is obtained by solving

Model BLP-SSP′

l).

z sspl ≥ LBφ

=

F(z,b|zφ,bφ

), which represents L.B. obtained by solving ∀(z,b), except current

the dual of BLP-SSP′

l, solution, (zφ,bφ).integer solution value, g (BLP-SSPl), at (zφ,bφ).

= F(z,b|zφ,bφ

)+(g (BLP-SSPl)− g

(BLP-SSP

′

l

))rsplφ (2.68)

=∑k∈Jl

∑t∈Tl

(θkt + ιkt)zkt +∑j∈Jl

∑t∈Tl

λjtzjt +∑t∈Tl

(∑j∈Jl

µjt

)(∑k∈Jl

zkt

)−∑j∈Jl

∑t∈Tl

(λjt + µjt)z0t + |Jl|∑j∈Jl

∑k∈Jl,k 6=j

∑t∈Tl

κjkt(bjt + bkt)−∑j∈Jl

∑t∈Tl

λjtbjt

+ (|Jl| − 1)∑j∈Jl

∑k∈Jl,k 6=j

∑t∈Tl

κjkt +∑j∈Jl

∑t∈Tl

λjt︸︷︷︸c1splφ

+(g (BLP-SSPl)− g

(BLP-SSP

′

l

))rsplφ ← νlφ (2.69)

psplφ =

( ∑j∈Jl∪{0}t∈Tl

(j,t)∈Zspl,1

zjt +∑j∈Jlt∈Tl

(j,t)∈Bl,1

bjt

)−( ∑j∈Jl∪{0}t∈Tl

(j,t)∈Zspl,0

zjt +∑j∈Jlt∈Tl

(j,t)∈Bl,0

bjt

)(2.70)

psplφ ≤ rsplφ +

(|Zsp

l,1|+ |Bl,1| − 1

)︸︷︷︸

c2splφ

← ξlφ (2.71)

psplφ ≥(|Zsp

l,1|+ |Bl,1| −M splφ

)︸︷︷︸

c3splφ

+M splφ r

splφ ← πlφ (2.72)

rsplφ ≥ 0 (2.73)


rsplφ ≤ 1 ← ρlφ (2.74)

rsplφ ∈ {0, 1} (2.75)

Proposition 2.1. Constraints (2.69)-(2.75) are valid optimality cuts to Model BLP-SPl

and therefore lead to convergence between the solutions of Models BLP-SPl andBLP-SSPl.

Proof. For all (z,b), the expression z sspl ≥ F(z,b|zφ,bφ

)represents the standard

Benders optimality cut which is obtained by solving the dual of Model BLP-SSP′

l,i.e. the dual of relaxed Model BLP-SSPl, and thus is a valid optimality cut. Also,g (BLP-SSPl) is the exact integer solution value corresponding to BLP-SSPl, whichis a valid lower bound on z sspl, but only for the current solution, (zφ,bφ).

Next, we show now that the cut given by Equation (2.68) reduces to, z sspl ≥g (BLP-SSPl), at the current solution, (zφ,bφ), corresponding to the φth iteration,and z sspl ≥ F

(z,b|zφ,bφ

), for all {(z,b)}\(zφ,bφ). Note that, at current solution,

(zφ,bφ), F(z,b|zφ,bφ

)= F

(zφ,bφ

)= g

(BLP-SSP

′

l

), therefore, g (BLP-SSPl) can

be rewritten as, g (BLP-SSPl) = F(zφ,bφ

)+(g (BLP-SSPl)− g

(BLP-SSP

′

l

)). We

now only need to show that rsplφ takes a value 1 for (z,b) = (zφ,bφ), and 0, for all

(z,b), except (zφ,bφ), in order to prove the validity of the proposed cut. This isaccomplished by enforcing the additional Constraints (2.70)-(2.75). Here, Z

sp

l,1 and

Zsp

l,0 represent indices of the zjt variables that were equal to one and zero, respectively,

in the current solution of BLP-SPl. Similarly, Bl,1 and Bl,0 represent indices of thebjt variables that were equal to one and zero, respectively, in the current solution ofBLP-SPl. An appropriate value for scalar, M sp

lφ , is |Zsp

l,1|+ |Bl,1|+ |Zsp

l,0|+ |Bl,0|. Notethat, rsplφ takes value one if and only if a solution corresponding to Model BLP-SPl

repeats (more specifically those variables that are input to Model BLP-SSPl, i.e.,(z,b)).

Since the number of solution pairs, (z,b) are finite due to the binary nature ofvariables, the application of the proposed cuts is guaranteed to lead to convergencebetween solution values of Models BLP-SPl and BLP-SSPl.

Note that, Model BLP-SSPl will always be feasible for any given solution of ModelBLP-SPl. Therefore, we do not need to consider sending the feasibility cuts back toSP stage.

Optimality cut from BLP-SPl to BLP-MP: For a given solution of Models


BLP-MP and BLP-SPl, if the MP-SP optimality gap is not within set optimality

tolerance, τmp, i.e.

∣∣∣∣g(BLP-SPl)−z splz spl

∣∣∣∣ > τmp, we add the following χth set of optimality

cuts to Model BLP-MP (Note that, the dual solution is obtained by solving ModelBLP-SP

′

l).

z spl ≥ LBχ

=

F (z,y|zχ,yχ), which represents L.B. obtained by solving ∀(z,y), except current

the dual of BLP-SP′

l, solution, (zχ,yχ).integer solution value, g (BLP-SPl), at (zχ,yχ).

= F (z,y|zχ,yχ) +(g (BLP-SPl)− g

(BLP-SP

′

l

))rmplχ (2.76)

=∑j∈Jl

αjzj +∑j∈Jl

∑i∈NIj

γijyij + βlsl + Emax∑t∈Tl

δt +∑j∈Jl

∑t∈Tl

εjt

+ maxj∈Jl{j}∑j∈Jl

∑t∈Tl

ζjt +∑t∈Tl

ηt +

|OSPl |∑φ=1

(c1splφνlφ + c2splφξlφ + c3splφπlφ + ρlφ

)+(g (BLP-SPl)− g

(BLP-SP

′

l

))rmplχ (2.77)

pmplχ =

( ∑j∈Jlj∈Zmpl,1

zj +∑j∈Jli∈NIj

(i,j)∈Y l,1

yij

)−( ∑

j∈Jlj∈Zmpl,0

zj +∑j∈Jli∈NIj

(i,j)∈Y l,0

yij

)(2.78)

pmplχ ≤ rmplχ +

(|Zmp

l,1 |+ |Y l,1| − 1

)(2.79)

pmplχ ≥(|Zmp

l,1 |+ |Y l,1|)−Mmp

lχ (1− rmplχ ) (2.80)

rmplχ ∈ {0, 1} (2.81)

Proposition 2.2. Constraints (2.77)-(2.81) are valid optimality cuts to Model BLP-MP and therefore lead to convergence between solutions of Models BLP-MP andBLP-SPl.

Proof. For all (z,y), the expression z spl ≥ F (z,y|zχ,yχ) represents the standardBenders optimality cut which is obtained by solving the dual of Model BLP-SP

′

l,i.e. the dual of relaxed Model BLP-SPl, and thus is a valid optimality cut. Also,g (BLP-SPl) is the exact integer solution value corresponding to BLP-SPl, which isa valid lower bound on z spl, but only for the current solution, (zχ,yχ).


Next, we show now that the cut given by Equation (2.76) reduces to, z spl ≥g (BLP-SPl), at the current solution, (zχ,yχ), corresponding to the χth iteration,and z spl ≥ F (z,y|zχ,yχ), for all {(z,y)} \ (zχ,yχ). Note that, at current solution,

(zχ,yχ), F (z,y|zχ,yχ) = F (zχ,yχ) = g(

BLP-SP′

l

), therefore, g (BLP-SPl) can be

rewritten as, g (BLP-SPl) = F (zχ,yχ) +(g (BLP-SPl)− g

(BLP-SP

′

l

)). We now

only need to show that rmplχ takes a value 1 for (z,y) = (zχ,yχ), and 0 for all (z,y)except (zχ,yχ) in order to prove the validity of the proposed cut. This is accom-plished by enforcing the additional Constraints (2.78)-(2.81). Here, Z

mp

l,1 and Zmp

l,0

represent indices of the zj variables that were equal to one and zero, respectively,in the current solution of BLP-MP. Similarly, Y l,1 and Y l,0 represent indices of theyij variables that were equal to one and zero, respectively, in the current solutionof BLP-MP. An appropriate value for scalar, Mmp

lχ , is |Zmp

l,1 |+ |Y l,1|+ |Zmp

l,0 |+ |Y l,0|.Note that, rmplχ takes value one if and only if a solution corresponding to Model BLP-MP repeats (more specifically those variables that are input to Model BLP-SPl, i.e.,(z,y)).

Since the number of solution pairs, (z,y) are finite due to the binary nature ofvariables, the application of the proposed cuts is guaranteed to lead to convergencebetween solution values of Models BLP-MP and BLP-SPl.

Note that, Model BLP-SPl will always be feasible for any given solution of ModelBLP-MP. Therefore, we do not need to consider sending the feasibility cuts back toMP stage.

In the remaining section, we would like to point out the difference between our versionof the optimality cut(s) from that developed by Laporte and Louveaux [1993] in theirseminal work. Laporte and Louveaux [1993] derived optimality cuts for problemswherein the second-stage variables are integer/binary, with the first stage containingbinary decision variables only, a setting that is similar to that in our case (the inputsolution to the second-stage and the third-stage integer sub-problem is a set of binarydecision variables of the first-stage and second-stage master problem, respectively).

First, we introduce the relevant notation and equations necessary to represent theoptimality cut (referred to as ‘Improved optimality cut‘) given in Proposition 6 ofLaporte and Louveaux [1993], which we reproduce as follows.

θ ≥ a

∑i∈Sr

xi −∑i∈Sr

xi

+ θr − a|Sr|, (2.82)


where, x denotes a general solution of the Stage 1 master problem, which serves asinput to the second stage sub-problems. Sr and Sr are the sets of variable indicesfor which xi = 1 and xi = 0, respectively, in the current solution, xr, at the rth

iteration. Here θ, in general, is an approximation to Q(x) (i.e., the expected valueof the objective function over the second-stage sub-problems with integer/binaryvariables) for a given x, whereas, θr denotes the true value of the second stage,Q(xr), at the current solution. Further, let Nr(s) define the set of solutions thatconstitute ‘s-neighbors’ (s varies from 1 to |Sr|+|Sr|), such that

∑i∈Sr xi−

∑i∈Sr xi =

|Sr| − s. Note that, they consider x to be a multi-dimensional binary variable; assuch, s = 0 ⇐⇒ x = xr. Let, λr(s) denote a lower bound on minx∈Nr(s){Q(x)},and L denote an overall lower bound on Q(x). The value of a is given by, a =max{θr − λr(1), (θr − L)/2}.

Following the definition of λr(s) and L, their selection is arbitrary and is usuallyproblem dependent. Understandably, the authors have not prescribed any rigorousmethodology for computing them uniquely. However, they do prescribe certain stepsthat could be used to compute these lower bound estimates in a general case. Beforewe describe how to compute these lower bound estimates, we must point out that theauthors considered a stochastic integer program with complete recourse, whereas, wedeal with a deterministic case in our work. For a fair comparison, therefore, we willfix the variable representing the probability distribution to be uni-variate, therebyconsidering only a single sub-problem at Stage 2 for every master problem solutionin their work. We reproduce a lower bound on Q(x) under such an assumption(following Equation (17) in their work) as follows.

L = minx,λ{λ|x ∈ X and (λ, x) follows Equation (2.84) for k = 1, . . . , s}, (2.83)

for any finite value s ≥ 1. Here, X is the domain of all binary first-stage vector ofdecision variables, i.e., X = {x|Ax = b, x = {0, 1}n1}.

λ ≥ R(x) = R(xk) + ∂R(xk)(x− xk), (2.84)

where, R(x) = miny{qy|Wy = h− Tx, 0 ≤ y ≤ 1}, represents the continuous relax-ation of Q(x), and ∂R(xk) is a sub-gradient of R at xk. Note that, the computation ofL in Equation (2.83) can either be performed in advance (whenever we can representQ(x) or R(x) analytically as a function of x, for all x ∈ X, which is only possiblein problems with a special structure in second stage; however, such an analytic rep-resentation of second stage is not possible in our case), or by using the information


available up until rth iteration in the scheme of Benders decomposition. Therefore,for a general case, we obtain L as follows.

L = minx∈X{λ|λ ≥ R(xk) + ∂R(xk)(x− xk),∀k = 1, . . . , r}. (2.85)

Similarly, we compute λr(s) as follows.

λr(s) = minx∈Nr(s)⊂X

{λ|λ ≥ R(xk) + ∂R(xk)(x− xk),∀k = 1, . . . , r}. (2.86)

Clearly, we have, λr(s) ≥ L,∀s = 0, 1, . . . , |Sr| + |Sr|. Even though, we have de-scribed a general procedure to obtain λr(s) for all s values, it is to be noted thatthe optimality cut proposed by the authors in the rth iteration (as given by Equa-tion (2.82)) uses only λr(1). Now, we will decompose this cut into different casesdepending upon the value of x as follows.

θ ≥

θr at x = xr,θr − a = λr(1)↓ ∀x ∈ Nr(1) ⊂ X,θr − 2a = L↓ ∀x ∈ Nr(2) ⊂ X,θr − s · a = L

′< L ∀x ∈ Nr(s) ⊂ X, s ≥ 3.

(2.87)

The symbol ↓ as an exponent denotes that the actual value is less than or equalto the base value (nonetheless, the cut remains valid). We will now draw compar-ison between Cut (2.87) and Cut (2.68) that we propose, (i.e. the optimality cutfrom BLP-SSPl to BLP-SPl; note that comparison to Cut (2.76) follows in a similarmanner). We rewrite Cut (2.68) for the φth iteration as follows.

zssp ≥{gzφ(BLP-SSP) at z = zφ,

gz(BLP-SSP′) = F (zφ,s|zφ) ∀z ∈ Nφ(s) ⊂ Z, s ≥ 1.

(2.88)

In combination with all the cuts added from iteration k = 1, . . . , φ, the above takesthe following form.

zssp ≥

max{gzφ(BLP-SSP), max

k=1,...,φ−1{F (zφ|zk)}} at z = zφ,

= gzφ(BLP-SSP)maxk=1,...,φ

{F (z|zk) + δ+ · rspφ } ∀z ∈ Nφ(s) ⊂ Z, s ≥ 1,

≥ LBφ(s)

(2.89)


where LBφ(s) is obtained in a similar manner as λr(s) in Equation (2.86) as follows.

LBφ(s) = minz∈Nφ(s)⊂Z

{λ|λ ≥ F (z|zk),∀k = 1, . . . , φ}. (2.90)

Since F (z|zk) gives the value of continuous relaxation solution to the third stage(obtained using dual-based Benders cut corresponding to zk solution of second stageat some iteration, k < φ), it is exactly similar to using R(x) = R(xk)+∂R(xk)(x−xk)in the work of Laporte and Louveaux [1993], and therefore, LBφ(s) = λr(s) ≥ L, forall, s ≥ 1. Therefore, if we branch out Cut (2.89) for different values of s, we can seethat, the lower bound on zssp (given by the right hand side of the cut) has the sameexact value to that of the lower bound on θ in Cut (2.87), for s = 0, 1, whereas, it islarger for s ≥ 2.

Therfore, apart from the fact, that we have developed a multi-cut version of op-timality cuts compared to the single integer cut version proposed by Laporte andLouveaux [1993], we have shown now that our cut offers a tighter lower bound onthe value of an integer sub-problem in the context of Benders decomposition scheme.

2.4 Results and Discussion

In this section, we present computational results on the use of proposed Nested Ben-ders decomposition method. To assess the effectiveness of our modeling and decom-position scheme, we use the largest available GIS data for this region, i.e. a catchmentarea of 48-Km around Gretna, VA. We refer to the Nested Benders decomposition ofthe original Model BLP as Model BLP+NBD in our results, which is tested against acombination of varying parameters such as switchgrass production scenarios, equip-ment system, and the exact/heuristic method. Modeling for the constituent parts,namely, BLP-MP, BLP-SPl and BLP-SSPl, and the associated work-flow is doneusing C++ Concert technology API of CPLEX R©(12.6). CPLEX R©was run in ‘deter-ministic’ mode with a maximum of 32 threads in parallel. All numerical tests wereexecuted on Intel R©Xeon R©Processor E5-2687W, using 8GB DDR3 memory.

2.4.1 Case of a single BeP

In the first part of this section, we restrict our attention to only a single BeP at thecenter of Gretna, VA, i.e. |L| = 1. Also, we do not consider the rail transportation


cost for this analysis. Comparisons are conducted for all six scenarios, described inTable 2.1, with the minimum amount of biomass required, BePmin, fixed at 42,752,45,176, 47,601, 51,700, 55,799 and 68,847 ha, respectively, for the scenarios. Thisamounts to utilizing roughly 90% of the available biomass from the region for allthe scenarios considered. We report the scenario used, program run time in secs,total cost in $ (i.e., the objective function value of Model BLP+NBD, together withthe additional cost of densification that is necessary at the BeP for the two racksystems), and the cost of production for a unit Mg of biomass utilized at BeP inColumns 2, 3, 4, and 5, respectively, of Table 2.4. Columns 6, 7, and 8 list thenumber of production fields, SSLs, and equipment sets used, respectively, in the finalsolution at convergence within the set optimality criterion. Columns 9 and 10 listthe average distance in Km from a field to its associated SSL and from an SSL tothe BeP, respectively.

For lth BeP (currently, we have, l = 1), the average distance from the fields to SSLis calculated as follows. ∑

j∈Jl

∑i∈NIj dijhaiyij∑

j∈Jl

∑i∈NIj haiyij

The average distance from the SSLs to the BeP (this is referred to as ‘mass-distanceparameter’ in Resop study) is calculated as follows.∑

j∈Jl djl

(∑i∈NIj haiyij

)∑

j∈Jl

∑i∈NIj haiyij

Based on the data aggregation criterion described earlier, we have a data of 1,000production fields with all of them serving as potential SSL locations for this region.An implementation of Model BLP+NBD for these parameters is hereby referred toas ‘Optimal solution’ (Table 2.4). In order to make a comparison to the Resop study,we devise a heuristic approach to the problem, wherein, we only use those SSLs thatmatch the same locations as used in Resop study, i.e. we fix the zj to be unity forthe exact 199 SSLs used in their study. We refer to this solution as ‘Resop-solution’.Further, we devise another heuristic method, wherein we only fix the number ofSSLs to be used a priori, which is the same as used in Resop study, i.e. 199, but nottheir exact locations. We refer to such a solution as ‘Optimal-Fixed’. The followingconstraint is added to Model BLP-MP to obtain this solution.∑

j∈J

zj = 199


Note that, we will use Model BLP+NBD only for implementing both the heuristicmethods, i.e. ‘Resop-solution’ as well as ‘Optimal-Fixed’.

We now present the results of Model BLP+NBD for the three different equipmentsystems, namely, ‘rear-loading”, ‘side-loading’ and ‘densification’ under scenario 1,with the optimal setting, i.e. using all 1,000 potential SSLs. We present the resultsfor the same in the first three rows of Table 2.4. The ‘side-loading’ rack systemperforms the best among all three loading/unloading systems in terms of total cost.Even though the ‘densification’ system is able to reduce the average distance of travelfrom a field to an SSL, its use can not be justified because of overall largest cost. Notethat such results for average distance to SSL are similar to that obtained by Juddet al. [2012]. They report the total cost of $11.82M and unit cost of 24.53 $/Mg forthe single scenario they considered with the ‘side-loading’ rack system for the 48-Kmanalysis in Gretna region. However, it is noteworthy that they considered a muchhigher yield per hectare of production area (15 Mg/ha). They utilized 32,110 ha ofland out of 43,434 ha considered available, which amounts to roughly 74% utilizationof the available biomass. In view of these figures, even though it is expected thattheir approach would result in shorter average travel distances and thereby totalcost and unit cost, still, our results outperform theirs, as we obtain the total cost of$10.23M and unit cost of 21.37 $/Mg.

Since the objective value for ‘side-loading’ rack system is best among all three equip-ment systems, the remaining experiments are conducted with this equipment systemonly.

In the next half of Table 2.4, comparison is made among the different solution meth-ods, namely, ‘Optimal’, ‘Optimal-Fixed’ and ’Resop-solution’ for all six differentscenarios. The total cost increases with each scenario for all the solution methods,however the trend in unit cost is not clear. Still, based on the results of ‘Optimal’method, Scenario 5 is found to be the most favorable with the lowest unit cost ofproduction of 21.04 $/Mg. As expected, both the total and unit production costs ob-tained using ‘Optimal-Fixed’ solution method is lower than that for ‘Resop-solution’,but higher than that for the ‘Optimal’ method, for all the six scenarios considered.

The average distance from the SSLs to BeP decreases continuously from Scenario 1to Scenario 5 from 42.2 to 40.6 Km and increases slightly to 41.4 Km for Scenario 6,thereby showing that most of the cropland and pastureland are located close to thecenter of the city and the scrubland and grasslands are concentrated on the outskirt.For comparison, the average distance reported in Resop study varies from 44.8 to42.7 Km. This shows that the optimal solution based on cost analysis is better able


to allocate fields to the SSLs than a heuristic rule does as in case of Resop study.

2.4.2 Case of multiple BePs

We now consider the biomass logistics problem in its entirety as stated earlier, i.e.we have two potential BeP locations, the first one at Beford, VA and the secondat Gretna, VA. Note that, we now include the rail transportation costs as well.We consider the most conservative scenario for this analysis, i.e. Scenario 1 andevaluate the trend in the results of Model BLP+NBD for varying the amount ofbiomass required, i.e. the overall demand at the refinery, R. It is assumed thatif a BeP is functional at either of these two locations, it will be operated at theminimum operational capacity of 478,822.40 Mg (=42,752 ha * 11.2 Mg/ha). Themaximum amount of biomass processed at either of the locations is based on thescenario availability, which is 526,411.20 Mg (=47,001 ha * 11.2 Mg/ha) for scenario1 considered here.

Table 2.5 lists experimental results for overall demand at the refinery. Understand-ably, two BePs are used for the values of R that exceed the maximum availability inone region, i.e. R = 957,644.80, 1,005,233.60, and 1,052,822.40 Mg. The unit costof production increases from 44.69 $/Mg to 46.03 $/Mg as the biomass requirementat the refinery enforces the logistics system in a single BeP’s region to utilize 100%of the available biomass, i.e. from R = 478,822.40 Mg to R = 526,411.20 Mg. Asimilar trend is seen even when two BePs are used. For R = 957,644.80 Mg, both theBePs only process the amount equal to their minimum capacity, i.e. 478,822.40 Mg.As R is increased to 1,052,822.40 Mg, both the BeP regions operate so as to utilizethe maximum available biomass in their respective regions. For R = 1,005,233.60Mg, the total demand at refinery is between the minimum capacity of both BePscombined and the total available biomass from both the regions. As per the optimalsolution, the amount that is processed by either of the BePs lies somewhere betweentheir minimum and maximum processing capacity. For the first BeP that is closerto the refinery, it is 511,660.80 Mg, which is more than 493,572.80 Mg for the secondBeP. The average distance to refinery is 329.897 Km.

For the conservative yield scenario that was considered, i.e. Scenario 1, the maximumamount of ethanol that can be produced is 78.1 million gallons (MG) by using twoBePs at Gretna, VA and Bedford, VA. This amounts to unit cost of productionat $0.64/gallon. Note that the cost of conversion from bio-crude to ethanol at therefinery is not included in this analysis.


Table 2.4: Experimental results for Gretna region: 48-Km (without rail transporta-tion).

Scenario Run time (sec)Total cost

$/ MgNumber of Avg. dist. (km)

($ in millions) Prod. fields SSLs Equip. set Field to SSL SSL to BePComparison of equipment systems

(Model BLP + NBD; Optimal solution)Rear-loading Rack

12,920.00 11.50 24.01 806 358 12 0.804 42.170

Side-loading Rack 3,081.00 10.23 21.37 807 357 12 0.807 42.151Densification 3,817.00 13.44 28.07 790 359 11 0.744 42.414

Comparison of heuristic and optimal solutions(Model BLP + NBD; Side-loading Rack System)

Optimal1

3,081.00 10.23 21.37 807 357 12 0.807 42.151Optimal-Fixed 1,694.00 10.32 21.55 842 199 12 1.931 41.691Resop-solution 1,598.00 10.77 22.50 952 199 12 3.440 42.121

Optimal2


Optimal3


Optimal4


Optimal5


Optimal6


Table 2.5: Experimental results for overall demand at refinery with two potentialBeP regions - Gretna, VA and Bedford, VA.

R (Mg)Run time Total cost

$/ MgRail cost

BePs usedAvg. dist. (Km)

BePBiomass Number of Average distance (km)

(sec) ($ in millions) ($ in millions) BeP to refinery (Mg) Prod. fields SSLs Equip. set Field to SSL SSL to BeP478,822.40 3,081.00 21.40 44.69 11.17 1 297.000 1 478,822.40 807 357 12 0.807 42.151526,411.20 2,601.00 24.23 46.03 12.28 1 297.000 1 526,411.20 962 398 13 0.928 43.617957,644.80 3,215.00 44.31 46.27 23.84 2 330.500 1 478,822.40 807 357 12 0.807 42.151

2 478,822.40 807 357 12 0.807 42.1511,005,233.60 4,904.00 50.00 47.75 25.00 2 329.897 1 511,660.80 895 385 13 0.856 43.200

2 493,572.80 844 368 12 0.833 42.6121,052,822.40 2,848.00 50.13 47.61 26.21 2 330.500 1 526,411.20 962 398 13 0.928 43.617

2 526,411.20 962 398 13 0.928 43.617


2.5 Concluding Remarks

In this chapter, we presented a biomass logistics system for the production of ethanolfrom seasonal harvest of switchgrass in Upper Southeast of U.S.A. We developeda mathematical model to capture all the features that we have considered. Wedecomposed this model into multiple smaller problems based on Nested Bendersdecomposition. The novelty of this decomposition scheme is that the sub-problemscan be solved as integer programs rather than as their LP relaxation. We proposedoptimality cuts that can capture solution value at an integer solution for the sub-problem(s). We also showed validity of these cuts and that they lead to convergenceof our decomposition scheme. We first obtained optimal solution for a single bio-energy plant (BeP) operation (without consideration of rail transportation cost)under a single harvest scenario and drew comparisons among three different typesof equipment systems used. We conclude that ‘side-loading’ rack system gives thesmallest total cost, and use this system in all the remaining test cases studied. Theoptimal solution corresponding to the use of this equipment system is obtained in3,081.00 secs, having a total cost of $10.23M and a unit cost of $21.37/Mg. Then, wemade comparison between the performance of our decomposition methodology withthat of a heuristic approach to locate SSLs in this region, for six harvest scenariosas reported in Resop study. We conclude that our approach gives a lower totalcost, ranging from $10.23M to $16.38M, as against from $10.77M to $17.96M forthe heuristic method. Also, the average distance of transportation from a field to itsassigned SSL is significantly lower for our method, for all the scenarios, although it isslightly higher in case of average distance from an SSL to BeP. The maximum averagefield to SSL distance is obtained as 0.819 Km, whereas the maximum average distancefrom the SSLs to BeP obtained is 42.151 Km for our model. We then presented resultsfor the complete problem statement by incorporating rail transportation cost fromBePs to refinery and consider logistics operations at two locations. We set up fivetest cases with increasing amount of total biomass intake required at the refinery.The average run time of the model over these test cases is 3,330.00 secs, whereas themaximum run time is 4,904.00 secs. The maximum amount of ethanol that can beproduced by using two BePs, one at Gretna, VA and the other at Bedford, VA is78.1 million gallons (MG), incurring a total cost of $50.13M. This amounts to a unitproduction cost for ethanol to be $0.64/gallon.

Chapter 3

Single Lot, 1+m Hybrid FlowShop Lot-streaming Problem

3.1 Introduction

In this chapter, we consider the problem of scheduling a single lot over a hybridflow shop. Depending upon the application, the lot is said to either have a size Uor consisting of U identical items. The hybrid flow shop that we consider has twostages with a single machine at Stage 1 and m parallel and identical machines atStage 2. We implement the process of lot-streaming, thereby allowing the lot tobe split into multiple sub-lots for simultaneous processing over parallel machines.Upon processing over the Stage 1 machine, each sublot incurs a fixed removal timeat Stage 1 before its transfer to one of the machines at Stage 2. We determine anoptimal schedule for the objective of minimizing makespan. As such, we obtain thenumber of sublots, sublot sizes, and sublots’ allocation over the machines at Stage 2in such a schedule. We consider two cases for sublot sizes - continuous and discrete(integer) sublot sizes. We designate this problem as 1 + m Two-stage Hybrid FlowShop Lot-streaming Problem (1 +m TSHFS-LSP).

The Hybrid Flow Shops (HFS) have important applications in flexible manufactur-ing systems that work on the principle of agile manufacturing. (Devor et al. [1997])mention the view shared by industrial executives that competitive advantage in thefuture lays in strategies promoting speed to market and adherence to changing cus-tomer demands. Agile manufacturing is a collective expression for all such strategies

38

Sanchit Singh Chapter 3. 1 +m Hybrid Flow Shop Problem 39

geared towards thriving in a continuously changing production environment, andtherefore it is becoming a system of increasing importance now-a-days because of itsability to effectively respond to changing customers’ demand. Flexible manufacturingsystems have attracted a significant amount of attention of the research communityin the past. The HFSs are the physical entities that define the core of such sys-tems. They have been employed for producing discrete products in the electronics,furniture and steel industries (Tang et al. [2005]), and also, that have been usedin continuous processing industries such as textile (Jungwattanakit et al. [2008]),food-processing (Yaurima-Basaldua et al. [2018]), and chemical and pharmaceutical(Gholami et al. [2009]). Some specific examples of HFSs include: assembly of printedcircuit boards (PCB) (Wittrock [1988],Jin et al. [2002]), where the lots of differentsizes of PCBs move sequentially through a three stage HFS, with each stage havingseveral insertion machines in parallel to insert electronic components on PCBs; andthe scheduling problem encountered at the blade line for Pratt and Whitney Inc. (Li[1997]), where various part-types are grouped into part families or batches at Stage1 for processing on the machines at Stage 2. Carpov et al. [2012] identify the use of1 +m HFS in scheduling of a program on a parallel-task computer, where there is asingle machine for memory access at Stage 1 that performs sequential data loadingfrom the external memory. Subsequently, the program is split into tasks that areperformed concurrently on parallel CPUs at Stage 2.

An extensive overview of research on the HFSs has been presented in Ribas et al.[2010] and Ruiz and Vazquez-Rodrıguez [2010]. The scheduling of jobs or lots ona two-stage HFS (with a single machine at Stage 1) with varied constraints andobjectives has been addressed by Gupta [1988], Sriskandarajah and Sethi [1989],Kusiak [1989], Gupta and Tunc [1991], Gupta and Tunc [1994], Hoogeveen et al.[1996] and Carpov et al. [2012]. However, their work does not consider lot-streaming.Gupta [1988] has shown that the two-stage HFS scheduling problem is NP-hard fora given number of jobs with the objective of makespan minimization. We claim thatthe 1 + m TSHFS-LSP is a NP hard problem as well, since it is a generalization ofthe problem studied by Gupta [1988]. If we fix the number and sizes of sublots in1 + m TSHFS-LSP, resulting sublots have similar role as jobs requiring schedulingover the two-stage HFS. Although, the 1 + m TSHFS-LSP is NP hard in a generalsense, yet we will show later that there exists a special case such that this problempresents itself with a pseudo-polynomial time solution.

Even though lot-streaming promises benefits in terms of minimization of completiontime-based measures in flow shop configurations and its variants (including HFS),its use in the solution of the HFS-based problems is limited in literature, due to


the added modeling and computational complexity. Notably, Tsubone et al. [1996],Zhang et al. [2003], Zhang et al. [2005], Liu [2008] and Cheng et al. [2015] are the onlyresearch studies in this direction. Tsubone et al. [1996] studied the impact of usinglot-streaming in a 1 + m hybrid flow shop on makespan, total flow time, maximumwork-in-process, and capacity utilization, by using a simulation model. Zhang et al.[2005] considered the m + 1 HFS problem for the objective of minimizing meancompletion times of the sublots. They made the choice of using equal sublot sizesand presented two heuristics together with specifying a lower bound on the solutionvalue. Liu [2008] studied the m + 1 HFS problem for the objective of minimizingmakespan by treating sublots sizes to be continuous. They proved a property of anoptimal solution, in which sublots can be allocated on Stage 1 machines following a‘rotation’ rule. Even though they specified sublots’ allocation decisions, the sublotsizes were obtained only using a linear program for a given number of sublots. Aheuristic method was presented to determine the number of sublots having assumedequal sizes for all sublots. Cheng et al. [2015] made an important contributiontowards reducing the complexity of the algorithm for obtaining the optimal scheduleof a special case of 1 + m HFS problem (where they studied m = 2). They notonly determined an optimal number of sublots, but also their sizes for both thecontinuous and discrete sublot sizes. Their key contribution lies in the developmentof closed-form expressions for continuous sublot sizes when the number of sublots isspecified. In this chapter, we have generalized their work to 1 + m two-stage HFS(where m ≥ 2), and designate this problem as 1+m TSHFS-LSP. Moreover, we havedeveloped an efficient methods to determine an optimal number as well as sizes ofsublots for both the cases of sublot sizes, i.e., continuous and discrete.

Organization of the chapter. The remainder of this chapter is organized asfollows. §3.2 lists the notation used along with model formulation for the 1 + mTSHFS-LSP. §3.3 addresses the case of continuous sublot sizes. Here, we obtain theclosed-form expressions for sublot sizes when the number of sublots, n is specified.We also present an algorithm to determine an optimal value of n in this section. Thecase of discrete sublot sizes is addressed in §3.4, wherein a branch-and-bound-basedalgorithm is developed to obtain the number, sizes and allocation of sublots on themachines at Stage 2 in an optimal schedule. The performance of this algorithm iscompared with that of directly solving the proposed model formulation for the 1+mTSHFS-LSP by CPLEX R©for which the results are presented in §3.5. Concludingremarks are then contained in §4.5.

Keeping in view that we are studying a HFS, it is assumed that m > 1 for thematerial presented in §3.3 and §3.4. For the case when m = 1 (i.e., we have a


two-machine flow shop) with multiple sublots, we refer to Vickson [1995].

3.2 Model Formulation


Parameters:

U Lot size (u is used as an index for lot size, 1 ≤ u ≤ U).t Removal time for a sublot at the Stage 1 machine incurred before transferring

it to any machine at Stage 2 (note that the unit processing time of the machineat Stage 1 is fixed at unity; other parameter values, p and t, and variablesmeasuring time are scaled accordingly).

p Unit processing time of machines at Stage 2.m Number of machines available at Stage 2 (j and k are used as either indices

or number of machine(s) at Stage 2, 1 ≤ j, k ≤ m).n Maximum number of sublots allowed, for both continuous and integer-sized

sublots (i and n are used as either indices or number of sublot(s), 1 ≤ i, n ≤ n).ρ (p+ 1)/p.

Variables:

M(u, n, k) Makepan value of an optimal schedule, for a given u, n and k, wherethe lot of size u is split into n sublots and processed on precisely kmachines at Stage 2.

M(u) Makepan value of an optimal schedule for a given lot size u. Thenumber of sublots in such a schedule is designated as n(u), andthese sublots are assigned to k(u) machines at Stage 2 (1 ≤ k(u) ≤m, 1 ≤ n(u) ≤ n).

M(u, no) Makepan value of an optimal schedule, for a given u and no. An op-timal number of sublots in such a schedule is designated as n(u, no),

and these sublots are assigned to k(u, no) machines at Stage 2

(1 ≤ k(u, no) ≤ m, 1 ≤ n(u, no) ≤ no ≤ n).


si(u, n, k) Sublot size for the ith sublot (sublots are numbered in order of theirsequence at Stage 1 machine) in an optimal schedule, where the lotof size u is split into n sublots and processed on precisely k machinesat Stage 2. We also use sublots numbered in the reverse order inwhich case si(u, n, k) represents size of the ith sublot from the endof a schedule at Stage 1, i.e. si(u, n, k) = sn−i+1(u, n, k).

C1,i(u, n, k) Completion time of the ith sublot at Stage 1 in an optimal sched-ule, when the lot of size u is split into n sublots and processed onprecisely k machines at Stage 2.

C2,i,j(u, n, k) Completion time of the ith sublot on the jth machine at Stage 2 inan optimal schedule when the lot of size u is split into n sublots andprocessed on precisely k machines at Stage 2 (1 ≤ j ≤ k).

We now present an MIP model for the 1 +m TSHFS-LSP, and designate it as modelTSHFS-LSP. We also define three binary variables: yi = 1, if sublot i is used, and= 0, otherwise, ∀i = 2, . . . n; xi,j = 1, if sublot i is assigned to machine j at Stage 2,and = 0, otherwise, ∀i = 1, . . . n,∀j = 1, . . .m; and hj = 1, if jth machine is used forscheduling any sublot at Stage 2, and = 0, otherwise, ∀j = 1, . . .m.

Minimize: M+ εm∑j=1

hj

M≥ C2,n,j ∀j = 1, . . .m (3.1)

C1,1 = s1 + t (3.2)

C1,i ≥ C1,i−1 + si + tyi ∀i = 2, . . . n (3.3)

C2,i,j ≥ C1,i + psi − θ(1− xi,j) ∀i = 1, . . . n, j = 1, . . .m (3.4)

C2,i,j ≥ C2,i−1,j + psi − θ(1− xi,j) ∀i = 2, . . . n, j = 1, . . .m (3.5)

C2,i,j ≥ C2,i−1,j ∀i = 2, . . . n, j = 1, . . .m (3.6)m∑j=1

xi,j = 1 ∀i = 1, . . . n (3.7)

n∑i=1

si = U (3.8)

si ≤ Uyi ∀i = 1, . . . n (3.9)

yi ≤ yi−1 ∀i = 2, . . . n (3.10)


n∑i=1

xi,j ≤ nhj ∀j = 1, . . .m (3.11)

n∑i=1

xi,j ≤n∑i=1

xi,j−1 ∀j = 2, . . .m (3.12)

si ≥ 0

C1,i ≥ 0, yi ∈ {0, 1}, hj ∈ {0, 1}C2,i,j ≥ 0, xi,j ∈ {0, 1}

∀i = 1, . . . n, j = 1, . . .m (3.13)

The objective is to minimize makespan of the schedule, along with a small penaltyproportional to total machines used at Stage 2 (in order to reduce the number ofmachines used as much as possible without compromising on the makespan value,we use a small enough value of ε; here, we use ε = 0.01). Constraints (3.1) enforcethe makespan value to be at-least as large as the last sublot’s completion time onevery machine at Stage 2. Note that, it is not necessary for the last sublot (orany other sublot) to have non-zero size for its completion time to be defined. Therelationships among sublots’ completion times are captured by Constraints (3.2)and (3.3) over Stage 1 machine, and by constraints (3.4), (3.5), and (3.6) over themachines at Stage 2. Constraints (3.7) assert unique assignment for each sublotover second stage. Constraint (3.8) ensures all sublot sizes to add up to the lot sizeU . Constraints (3.9) ensure that the sublot size is zero when a sublot is not used.Constraints (3.10) strengthen the model by adding symmetry-breaking conditions.Constraints (3.11) ensure that no sublot is assigned to a machine when that machineis not used. Constraints (3.12) strengthen the model by adding another symmetry-breaking condition that enforces the number of sublots allocated to the machinesto be in the non-increasing order of their indices. Constraints (3.13) capture thedomains of the variables. The standard value of θ for use in constraints (3.4) and(3.5) is taken as pU .

3.3 Solution Methodology For Continuous Sublot

Sizes

In this section, we determine an optimal schedule for the 1 +m TSHFS-LSP for thecase when the sublot sizes can take on continuous values. It is further divided into twosub-sections. §3.3.1 addresses determination of sublot sizes si(U, n,m)(or si(U, n,m)),∀i =


1, . . . n, allocation of sublots to the Stage 2 machines, and makespan valueM(U, n,m)when the number of sublots is fixed to n(≤ n). Some key results are also derived foruse in the determination of an optimal schedule when the number of sublots is notspecified, the case that is presented in §3.3.2.

3.3.1 Determination of optimal schedule when the numberof sublots is fixed

Theorem 3.1. For a given lot of size U , number of machines, m (≥ 2), and numberof sublots, n,

1. if n ≤ m, the optimal continuous sublot sizes can be obtained by only utilizing nnumber of machines, and assigning each sublot to a different machine on Stage2. The sublot sizes are thus obtained by solving the following expressions.

si = ρsi−1 + t/p, ∀i = 2, . . . , n (3.14)n∑i=1

si = U (3.15)

2. if n ≥ m, the optimal continuous sublot sizes are obtained by solving the fol-lowing expressions.

si = ρsi−1 + t/p, ∀i = 2, . . . ,m (3.16)

si = (si−1 + . . .+ si−m)/p+mt/p, ∀i = m+ 1, . . . , n (3.17)n∑i=1

si = U (3.18)

Sublots are assigned to machines at Stage 2 following the ‘alternate assign-ment’ rule, i.e., sublots s1, sm+1, s2m+1, . . . are assigned to machine 1, sublotss2, sm+2, s2m+2, . . . are assigned to machine 2, and so on.

Note that si is used as a shorthand for si(U, n,m) here. Also, the proof of Theorem3.1 and those of other results in the sequel are presented in the Appendix section, inorder not to detract the reader from the main ideas presented here.

Remark 3.1. For the case, n ≤ m, M(U, n,m) is equivalent to M(U, n, n); simi-larly, si(U, n,m) ≡ si(U, n, n),∀i = 1, . . . , n.


Determination of s1 (when n ≤ m). The value of s1 is obtained by straight-forwardsubstitutions using Equations (3.14) and (3.15) as follows.

s1 =U(ρ− 1)− t(ρn − nρ+ n− 1)

ρn − 1(3.19)

si = ρi−1s1 + t(ρi−1 − 1), ∀i = 2, . . . , n (3.20)

Determination of s1 (when n ≥ m). In the matrix notation, Equations (3.16)-(3.18) can be represented in the form, Ax = b, where x denotes a vector of unknownsublot sizes of dimension n. As such, various established exact methods can be usedto obtain x, or equivalently, si, ∀i = 1, . . . , n. These include ‘Gaussian Elimination’,which is known to have arithmetic complexity of O(n3). At best, the inverse op-eration takes O(n2.376) order of complexity using Coppersmith-Winograd algorithm(Coppersmith and Winograd [1990]) which involves performing ‘LU decomposition’of matrix A. Our aim is to further improve upon the time complexity of obtainingsi,∀i = 1, . . . , n, by utilizing the inherent structure present in (3.16)-(3.18).By using Theorem 3.1 and the results from linear algebra pertaining to recurrencesequences, the value of s1 is obtained by the expression given in (3.21) when p 6= mand by (3.22) when p = m (refer to Appendix B).

s1 =U −

((ωn − e− nc) + t

(∑m−1k=2 (βn,k − bk)(ρk−1 − 1)

))∑m−1k=1 (βn,k − bk)ρk−1

(3.21)

s1 =U −

((ω′n − ne

′ − n(n+ 1)c′/2) + t

(∑m−1k=2 (β

′

n,k − nb′

k)(ρk−1 − 1)

))∑m−1k=1 (β

′n,k − nb

′k)ρ

k−1(3.22)

Among the scalars, of particular interest are βn,l and ωn, ∀l = 1, . . . ,m − 1, whenp 6= m, and β

′

n,l and ω′n, ∀l = 1, . . . ,m − 1, when p = m. These are obtained as

follows. [βn,1 βn,2 . . . βn,m−1

]T= (Bm,m−1)T (Rm,m)−1rnm (3.23)

ωn = wm(Rm,m)−1rnm (3.24)[β′n,1 β

′n,2 . . . β

′n,m−1

]T= (B

′

m,m−1)T

(Rm,m)−1rnm, and (3.25)

ω′

n = w′

m(Rm,m)−1rnm, (3.26)

where Bm,m−1,Rm,m and B′m,m−1 are scalar matrices, whereas rnm,wm and w

′m are

scalar vectors. Note that the calculations in Equations (3.23)-(3.26) involve compu-tation of the inverse of matrix Rm,m, which has an order of complexity, O(m3) in


general. Also, it takes O(m log n) time to obtain rnm, thus requiring O(m3 +mn) timeto obtain s1 alone. Finally, we can determine sizes of the remaining n − 1 sublotsin no more than O(n) time, after having computed the value of s1, as shown next.Overall, the complete schedule can be obtained in polynomial time, O(m3 + mn).Since, n >> m typically, our approach offers a significant reduction in time com-plexity over conventional approach as discussed earlier, i.e., compare O(m3 + mn)v/s O(n3), respectively.

The rest of the sublot sizes, si,∀i = 2, . . . ,m, follow from the s1 value, and they canbe obtained using (3.16) as follows.

si = ρi−1s1 + t(ρi−1 − 1) ∀i = 2, . . . ,m (3.27)

The value of sm+1 can be obtained by using (3.17). For i = m+2, . . . , n, si is obtainedas follows.

si = ρsi−1 − si−m−1/p, ∀i = m+ 2, . . . , n (3.28)

Equations (3.16)-(3.18) do not guarantee attainment of non-negative values for allsublot sizes. The next result provides necessary and sufficient conditions that ensurethe schedule obtained using Theorem 3.1 to be feasible, i.e. all the sublot sizes arenon-negative.

Corollary 3.1. If the last sublot size is non-negative, then all other sublot sizes arestrictly positive, i.e. s1 ≥ 0 =⇒ si > 0,∀i = 2, . . . , n, and the complete schedule isa feasible one.

Corollary 3.2. The schedule given by the sublots is critical (i.e., no sublot waitsfor processing over the machines, once it starts processing on the machine at Stage1, and there is no idle time on that machine), and the completion times of the lastsublots on Stage 2 machines are the same. In such a case, the makespan value,M(U, n,m) is given by,

M(U, n,m) = U + nt+ ps1(U, n,m) (3.29)

Next, we present some results (Theorem 3.2-Theorem 3.4) in §3.3.2, that are notonly beneficial in understanding the application of lot-streaming in hybrid flow shopscheduling, but also will be used in determining an optimal schedule when the numberof sublots is not specified.

Theorem 3.2. For a given number of machines, m (≥ 2) and number of sublots, n,si(U, n,m),∀i = 1, . . . n, increase (decrease) with increment (decrement) in lot size,U .


Theorem 3.3. For a given lot of size, U , and number of machines, m (≥ 2),

1. s1(U, n,m) decreases monotonically with increment in the number of sublots,n.

2. There exists an integer, ni, s.t. ∀n > ni, the solution given by Theorem 3.1 isinfeasible, i.e., s1(U, n,m) is negative (we designate by nf the smallest of allsuch ni values and n.)

Corollary 3.3. For a given U , no and m (≥ 2),

1. if s1(U, no,m) < 0, then a solution given by Theorem 3.1 is not feasible ∀n ≥no.

2. if s1(U, no,m) ≥ 0, then a solution given by Theorem 3.1 is feasible ∀n ≤ no.

Remark 3.2. Definition of nf together with Corollary 3.3 imply that s1(U, n,m) ≥0,∀n ≤ nf and that if nf 6= n, then s1(U, n,m) < 0, ∀n > nf .

We now present an algorithm to obtain nf , based on the binary search method.

Algorithm 3.1. Determination of nf , for a given U and m (≥ 2).

1. Initialize nl = 2, and nu = n, at the start.

2. If s1(U, nu,m) < 0, go to Step 3, else if s1(U, nu,m) ≥ 0, let nf = nu and stop.

3. If s1(U, nl,m) ≥ 0, go to Step 4, else if s1(U, nl,m) < 0, let nf = nl − 1 andstop.

4. Let nz = b(nl+nu)/2c. If s1(U, nz,m) < 0, then nu = nz−1, else, nl = nz +1.Return to Step 2.

Theorem 3.4. For a given U , n (≤ nf ) and m (≥ 2), there exists an integer, ns,s.t. the makespan for the lot of size U obtained using number of sublots larger thanns, where ns = n + bps1(U, n,m)/tc, does not improve over the makespan obtainedusing n sublots.


3.3.2 Determination of optimal schedule when the numberof sublots is not specified

By Remark 3.1, we have the expression for optimal makespan value,

M(U) = min1 ≤ n ≤ n1 ≤ k ≤ m

M(U, n, k) = min1 ≤ n ≤ n

1 ≤ k ≤ min(n,m)

M(U, n, k) (3.30)

Note that, for 1 ≤ m1 < m2 ≤ n, M(U, n,m2) ≤ M(U, n,m1), since we can alwaystransfer m2 − m1 sublots from the optimal schedule on m1 machines to each oneof the extra m2 − m1 machines at Stage 2 that become available when using m2

machines in total. Therefore, (3.30) reduces to the following.

M(U) = min1 ≤ n ≤ n

M (U, n,min (n,m)) (3.31)

For a given set of problem parameters, U, n,m, t, and p, we next present an algorithmto determine optimal makespan value, M(U), following (3.31), which leverages uponthe results given in Remark 3.2 and Theorem 3.4 in order to reduce the search spacew.r.t. n. Even though, overwhelming number of results show that the optimalmakespan value for a specified number of sublots, n, is a convex function of n (inthe discrete sense), we only state this as a conjecture at this point, and not use itin the design of the algorithm. As such, in the worst-case scenario, we perform alinear search from 1 to n to determine optimal n. We designate the algorithm asLinear Search Algorithm (LSA). We also determine, M(U, n),∀1 ≤ n ≤ n, withinthis algorithm, that will be used later in Heuristic 3.1. Algorithm LSA has a pseudo-polynomial time complexity of order O(m3 + mn) (note that the parameter, n, isusually set as a linear function of U). The steps of LSA are as follows.

Algorithm 3.2. LSA: Determination of best makespan value (continuous sublotsizes).

1. Let M(U)←M(U, 1, 1)← U+t+pU , n(U)← 1, and m(U)← 1. M(U, 1)←M(U, 1, 1). Let k ← 2.

2. If k < m, continue, else go to Step 5.

3. Let n← k. If n ≤ nf (U,m), continue, else go to Step 7.


4. Obtain M(U, n, k) by (3.29). If M(U) > M(U, n, k), then let M(U) ←M(U, n, k), n(U) ← n, and m(U) ← k. Let M(U, n) ← min{M(U, n −1),M(U, n, k)}. Let k ← k + 1. Return to Step 2.

5. Let n ← m. If n ≤ nf (U,m), let ¯n ← nf (U,m), and continue, else go to Step7.

6. Obtain M(U, n,m) by (3.29). If M(U) > M(U, n,m), then let M(U) ←M(U, n,m), n(U) ← n, and m(U) ← m. Let M(U, n) ← min{M(U, n −1),M(U, n,m)}. Let ¯n ← min(¯n, ns(U, n,m)). Let n ← n + 1. If n ≤ ¯n, thenrepeat Step 6, else go to Step 7.

7. If n ≤ n, let M(U, n′)← M(U, n− 1),∀n′ = n, . . . n. Stop.

3.4 Case of Discrete Sublot Sizes

In this section, we present a branch-and-bound-based heuristic method to solve 1+mTSHFS-LSP when sublot sizes are discrete. However, for comparative purposes, wealso solve the MIP model 1 +m TSHFS-LSP directly using state-of-the-art commer-cial solver, for a given U , n, m, t and p, after imposing the integrality restrictionson sublot sizes. Corresponding to the best MIP solution, we report the makespanvalue, M(U) =M∗− ε

∑mj=1 h

∗j , number of sublots, n(U) =

∑ni=1 y

∗i , and number of

machines required, m(U) =∑m

j=1 h∗j , in the results section later on.

3.4.1 Branch-and-bound-based method

For a given set of problem parameters, U, n,m, t, and p, we now present a heuristicmethod, which we designate as Branch-and-Bound-based Heuristic (B&BH). Notethat, we call it a heuristic method because the maximum numbers of sublots canbe restricted to be smaller than n in this method. It is a constructive procedure,wherein a tree structure evolves starting from a single root node. At any time duringthe procedure, the tree structure may contain multiple nodes with each node in theproposed branch-and-bound scheme representing a partial schedule for 1+m TSHFS-LSP (note that we use the word ‘partial’ in a general sense, which encompasses anycomplete schedule as well). We define the following additional notation to representa partial schedule corresponding to a node denoted by subscript ‘a’.


ua Sum of the items, 1 ≤ ua ≤ U .

na Number of sublots, 1 ≤ na ≤ ¯n.

sa A vector of dimension, na, representing discrete sizes of sublots in the order ofsequence of sublots over Stage 1 machine.

Xa A binary matrix of dimensions, na×m, representing assignment of each sublotto one of the machines at Stage 2.

ca A vector of dimension, m, representing completion times of sublots on Stage 2machines (with only sublots that are part of this partial schedule).

Mlba A lower bound on the makespan value over all those complete schedules that

can be formed after extending this partial schedule.

We designate the node representing the best incumbent schedule (also a completefeasible schedule) by B. The members constituting this node are: nb, sb,Xb, cb, andMb (makespan). We use the notation N.m to represent m as a member of some noden, and the notation v[i] to represent ith element of some vector, v. For example,B.sb[k]← 1 means that the size of the kth sublot for node B is initialized to 1.

The steps of B&BH are as follows.

Heuristic 3.1. B&BH: Determination of best makespan value (integer sublot sizes).

1. Initialize node B with any schedule that is both feasible and complete, in thesense that it has all integer-sized sublots and follows (3.2)-(3.13). We choseto populate B with a schedule made up of a single sublot having size U . Also,initialize a priority queue with a node corresponding to zero allocated sublots,i.e. its members are initialized as follows. ua ← 0, na ← 0, sa and Xa are nullvector and matrix, respectively, ca[k] = 0,∀k = 1, . . . ,m. We also restrict themaximum number of sublots to ¯n, calculated as, ¯n← min(n, n(U) + dU/100e),where n(U) is obtained using Algorithm 3.2.

2. If the queue is not empty, continue, else go to Step 6.

3. Remove the node from the top of the queue, call it P. If (Mb − P.Mlba )/Mb ∗

100 > τ (where τ is the set tolerance for optimality gap in percentage), thencontinue, else go to Step 6.

4. Depending upon the number of sublots currently in P, we create either a singleor multiple child nodes from the parent node. If P.na = ¯n− 1, we define a set,


I = {U − P.ua}, otherwise, I = {1, 2, . . . U − P.ua}. A child node is created inthe following manner (where each node is denoted by Qi,∀i ∈ I).

(a) Qi.ua ← P.ua + i, and Qi.na ← P.na + 1.

(b) Qi.sa[j]← P.sa[j], ∀j = 1, . . . ,P.na, and Qi.sa[Qi.na]← i, i.e. other thanthe last sublot added which has size i, first P.na sublots have the same sizein node Qi as in its parent node P.

(c) The assignment of sublots at Stage 2 machines remains the same for nodeQi as is for P, for all the sublots except that of (Qi.na)

th one. The (Qi.na)th

sublot is assigned to machine (denoted by k′i), which has the smallest com-

pletion time, i.e., k′i = arg mink=1,...m{P.ca[k]}.

(d) The completion time vector member ca remains the same as in the par-ent node, except for machine k

′i. We have, Qi.ca[k

′i] ← max(Qi.ua +

(Qi.na)t,P.ca[k′i]) + pi.

(e) Case 1: Qi.ua 6= U .

i. We use two ways of computing lower bound, Mlba known as sublot-

based or machine-based bounds. For the sublot-based bound, we haveknowledge of the allocated sublots and those that remain un-allocated,i.e. U − Qi.ua (we call this size of un-allocated lot as Qi.uu). Notethat, we can split this un-allocated lot further into a maximum of¯n − Qi.na sublots. This is an important distinction that helps us toimprove the quality of lower bound. The allocated sublots finish theirprocessing at Stage 1 at time Ta ← Qi.ua + (Qi.na)t. If we relaxthe sublot sizes to be continuous for the remaining un-allocated lotof size Qi.uu, and consider it to start processing at time T = 0 atStage 1, then M(Qi.uu) provides the optimal makespan value. Thus,

M(Qi.uu) gives the least time bound ahead of time Ta, for the un-allocated lot uu, which is lower than if this un-allocated lot were bothrestricted to have discrete sublot sizes, and also scheduled so as toprevent any overlap with the sublots that have already been allocatedat the Stage 2 machines. Note that, while obtaining M(Qi.uu), wemight also get, n(Qi.uu) > ¯n−Qi.na, which would violate the feasibilitycondition regarding the maximum number of sublots. This would havelead to an even smaller bound on the makespan value. Therefore, weuse M(Qi.uu, ¯n−Qi.na) instead of M(Qi.uu) to obtain a better lowerbound on an extrapolated complete schedule for the child node, i.e.Mlb1

a ← Ta+M(Qi.uu,min(Qi.uu, ¯n−Qi.na)). For the machine-based


bound, we use the maximum of completion times of all the machines atStage 2, with the current knowledge about the assignment of allocatedsublots to the machines at Stage 2, i.e. Mlb2

a ← max1≤k≤m(Qi.ca[k]).Thus overall, we have,

Qi.Mlba ← max(Mlb1

a ,Mlb2a )

ii. Insert Qi into the priority queue. (Note that the priority queue issorted based on the lower bound value of nodes, with the node withleast value of lower bound placed at the top of the queue. In case ofa tie, a node with a smaller un-alloacted lot is given a higher prior-ity). We use a heap-based priority queue to make the insertion andremoval operations (in Step 3) computationally effective (Sedgewickand Wayne [2015]).

Case 2: Qi.ua = U .

i. Since the schedule is complete, the lower bound is the same as the truemakespan for the schdule, i.e.

Qi.Mlba ← max

1≤k≤m(Qi.ca[k])

ii. If, either Qi.Mlba <Mb, or Qi.Mlb

a = Mb,Qi.na < nb , then updateB with the schedule represented by Qi, i.e. nb ← Qi.na, sb ← Qi.sa,Xb ← Qi.Xa, cb ← Qi.ca, and Mb ← Qi.Mlb

a .

5. Return to Step 2.

6. We now have the desired solution for 1+m TSHFS-LSP (within the optimalitygap of τ%), for a lot of size U , having integer-sized sublots, given by the best

node B, i.e. M(U) = Mb. Stop.

3.5 Computational Investigation

The purpose of this section is to evaluate the performance of B&BH by comparingagainst the direct solution of MIP model 1+m TSHFS-LSP using CPLEX R©. B&BHwas implemented and solved entirely using C++ on Xcode (v10.0), whereas theMIP model was solved using C++ Concert library of CPLEX R©(v12.8) in multi-threaded mode. All tests were conducted on a 2.6 GHz Intel Core i5 processor, witha maximum available RAM of 8GB.


We present six tables of comparative results (Tables 3.3-3.8), one each for a uniquepair of U and t values. In all of these tables, the maximum number of sublots,n, is fixed at half of the lot size, U , which itself takes the value of either 100 or1000. The value of sublot-attached removal time, t is fixed at either 0.20, 1.00, or5.00 secs, whereas the value of p is varied. A tolerance value of τ = 0.1% is set foroptimality gap for both the methods. Both the methods are terminated either whenthe allowable CPU time limit of 1800.00 secs is reached, or when the lower boundon the makespan value and the best incumbent solution’s makespan value fall withinthe tolerance value set for optimality gap.

First, we make remarks on the general trends observed for the 1 + m TSHFS-LSPbased on the results presented in Tables 3.3-3.8: (1) For a given U and m, thecomputational difficulty (time taken to achieve optimality within the prescribed tol-erance value) increases for both the methods either with an increment in the valueof p for a fixed t value, or with a decrement in the value of t for a fixed p value.This is attributed to increment in the value of the ratio p/t, which, in turn, has adirect correlation with the number of sublots, n(U), corresponding to the optimalschedule, and the number of machines, m(U), used for allocating these sublots at

Stage 2, m. (2) For a given U and m, makespan value, M(U), corresponding tothe optimal schedule increases for both the methods either with an increment in thevalue of p for a fixed t value, or with an increment in the value of t for a fixed p value,as expected. It also decreases with increment in the number of machines availableat Stage 2, m, for a given U , p and t.

Regarding comparative performances of the two methods: (1) B&BH method is ableto achieve optimality in 83 out of 90 total cases tested. In the remaining 7 cases,it is forced to terminate with the maximum optimality gap of 1.20% only from itslower bound value. Even so, in such cases, it reports a better makespan value thanthe direct solution method. (2) The direct solution method is not able to achieveoptimality in 50 out of 90 total cases tested, with the maximum optimality gap ashigh as 99.89%. Note that, the direct solution method struggles with the lower boundvalue and is unable to ramp it up during the course of test run at higher values of anycombination of U , m, p, or 1/t in general. Still, in 20 such cases, the makespan valuematches to that using the B&BH method, and in only 8 of these cases, it reports animprovement of 0.13% on an average over B&BH in terms of the makespan value (notethat even though the B&BH method achieved optimality within the set tolerance of0.1% in all the 8 cases, the lower bound reported for B&BH method is slightly higherthan the makespan observed for the direct solution method in 6 out of these 8 cases,


which is due to the fact that B&BH is actually a heuristic method, therefore the lowerbounds observed for the two methods are not directly comparable). However, in theremaining 22 cases, B&BH method reports a smaller makespan value compared withthat obtained by the direct solution method by an average of 3.84% and a maximumas high as 15.12%. In fact, the standard deviation in the improvement is 4.67% whichpoints to a strong positive skewness in the distribution of makespan improvementpercentage for these 22 cases. (3) For all the 90 test cases, B&BH method reports animprovement in makespan value of 0.90% on an average compared with that obtainedby the direct solution method. Even though, this improvement of the makespan valueis not sufficiently large, the difference in time required before termination by boththe methods is a clear indication of the superiority of the B&BH method, where thedirect solution method requires close to 1095.00 secs compared to 315.00 secs for theB&BH method on an average.

Table 3.3: Results for 1 + m TSHFS-LSP in the case of discrete sublot sizes (U =100, t = 0.20 secs).

Direct solution by CPLEX R© B&BH % drop with B&BH

row m p time (sec) M(U) n(U) m(U) gap% (L.B.) time (sec) M(U) n(U) m(U) gap% (L.B.) time makespan1 2 0.20 0.98 101.00 4 1 0.09 (100.91) 1.12 101.00 4 2 0.02 (100.98) -14.29 0.002 2 0.60 1.48 101.80 6 2 0.09 (101.71) 1.44 101.80 6 2 0.03 (101.77) 2.70 0.003 2 1.00 15.51 102.80 9 2 0.09 (102.71) 1.54 102.80 9 2 0.06 (102.74) 90.07 0.004 2 3.00 20.00* 152.40* 26 2 32.55 (102.79) 22.50 152.40 21 2 0.02 (152.37) 98.75 0.005 2 5.00 15.00* 252.40* 32 2 59.20 (102.98) 33.19 252.40 12 2 0.08 (252.2) 98.16 0.006 5 0.20 11.01 101.00 4 1 0.00 (101) 3.64 101.20 3 2 0.01 (101.19) 66.94 -0.207 5 0.60 48.00* 101.80* 6 2 0.19 (101.60) 3.98 102.00 5 5 0.05 (101.95) 99.78 -0.208 5 1.00 2.30* 102.40* 7 4 1.17 (101.20) 21.06 102.60 8 5 0.09 (102.51) 98.83 -0.209 5 3.00 700.00* 106.80* 17 5 5.62 (100.80) 73.57 106.60 16 5 0.09 (106.49) 95.91 0.19

10 5 5.00 1748.00* 116.00* 48 5 12.76 (101.20) 254.02 115.20 37 5 0.04 (115.15) 85.89 0.6911 10 0.20 174.00 101.00 4 1 0.09 (100.91) 15.58 101.20 3 3 0.09 (101.11) 91.05 -0.2012 10 0.60 16.00* 101.80* 6 2 1.17 (100.60) 19.09 102.00 5 5 0.07 (101.93) 98.94 -0.2013 10 1.00 60.00* 102.40* 7 4 1.75 (100.60) 66.00 102.60 7 7 0.07 (102.53) 96.33 -0.2014 10 3.00 1600.00* 106.00* 12 7 4.91 (100.80) 300.27 105.80 13 10 0.08 (105.72) 83.32 0.1915 10 5.00 1632.00* 109.20* 21 10 7.33 (101.20) 612.62 109.00 20 10 0.06 (108.93) 65.97 0.18

* For these cases, (i) the test run was forced to terminate at 1800.00 secs, (ii) the optimality gap at termination was > 0.1%, (iii) the makespan value, n(U),and m(U) reported correspond to the best incumbent solution, along with the time when it was observed during run, and (iv) the optimality gap (%) and thelower bound value reported correspond to termination point.Among both the methods, the lower values of best makespan and time before termination are marked in bold.




row m p time (sec) M(U) n(U) m(U) gap% (L.B.) time (sec) M(U) n(U) m(U) gap% (L.B.) time makespan1 2 0.20 0.46 103.40 3 2 0.00 (103.40) 0.31 103.40 3 2 0.02 (103.38) 32.61 0.002 2 0.60 0.75 105.60 5 2 0.00 (105.60) 0.52 105.60 5 2 0.04 (105.56) 30.67 0.003 2 1.00 0.71 108.00 7 2 0.00 (108.00) 0.77 108.00 6 2 0.03 (107.97) -8.45 0.004 2 3.00 11.00* 157.00* 42 2 23.44 (120.20) 6.20 157.00 15 2 0.06 (156.91) 99.66 0.005 2 5.00 1.50* 254.00* 12 2 49.43 (128.45) 12.23 254.00 12 2 0.09 (253.75) 99.32 0.006 5 0.20 5.38 103.40 3 2 0.08 (103.32) 1.88 103.40 3 3 0.09 (103.31) 65.06 0.007 5 0.60 80.21 105.60 5 2 0.09 (105.50) 1.93 105.60 5 5 0.05 (105.55) 97.59 0.008 5 1.00 281.52 107.00 6 3 0.08 (106.91) 8.60 108.00 5 5 0.03 (107.97) 96.95 -0.939 5 3.00 56.00* 117.00* 12 5 6.89 (108.00) 22.09 116.00 12 5 0.04 (115.95) 98.77 0.85

10 5 5.00 1610.00* 131.00* 18 5 17.56 (108.00) 49.55 128.00 22 5 0.09 (127.88) 97.25 2.2911 10 0.20 17.09 103.40 3 2 0.00 (103.40) 6.81 103.40 3 3 0.02 (103.38) 60.15 0.0012 10 0.60 6.00* 105.60* 5 2 1.51 (104.01) 8.88 105.60 5 5 0.03 (105.57) 99.51 0.0013 10 1.00 190.00* 107.00* 6 3 3.74 (103.00) 33.29 107.00 6 6 0.08 (106.91) 98.15 0.0014 10 3.00 308.00* 115.00* 11 7 8.69 (105.00) 148.72 115.00 11 10 0.06 (114.93) 91.74 0.0015 10 5.00 1670.00* 126.00* 21 10 16.67 (105.00) 246.71 123.00 15 10 0.03 (122.96) 86.29 2.38

* For these cases, (i) the test run was forced to terminate at 1800.00 secs, (ii) the optimality gap at termination was > 0.1%, (iii) the makespan value, n(U),and m(U) reported correspond to the best incumbent solution, along with the time when it was observed during run, and (iv) the optimality gap (%) andthe lower bound value reported correspond to termination point.Among both the methods, the lower values of best makespan and time before termination are marked in bold.



row m p time (sec) M(U) n(U) m(U) gap% (L.B.) time (sec) M(U) n(U) m(U) gap% (L.B.) time makespan1 2 0.20 0.30 112.00 2 2 0.00 (112.00) 0.20 112.20 2 2 0.05 (112.14) 33.33 -0.182 2 0.60 0.49 119.00 3 2 0.00 (119.00) 0.31 119.00 3 2 0.09 (118.88) 36.73 0.003 2 1.00 0.54 125.00 4 2 0.00 (125.00) 0.53 125.00 4 2 0.09 (124.88) 1.85 0.004 2 3.00 50.67 173.00 14 2 0.09 (172.84) 3.78 173.00 11 2 0.03 (172.95) 92.54 0.005 2 5.00 12.00* 264.00* 21 2 3.98 (253.50) 8.07 264.00 11 2 0.06 (263.84) 99.55 0.006 5 0.20 6.32 112.20 2 2 0.00 (112.20) 1.20 112.20 2 2 0.06 (112.13) 81.01 0.007 5 0.60 4.79 118.60 3 3 0.01 (118.59) 1.46 118.60 3 3 0.09 (118.49) 69.52 0.008 5 1.00 7.54 123.00 4 4 0.00 (123.00) 3.78 123.00 4 4 0.09 (122.88) 49.87 0.009 5 3.00 408.13 143.00 8 5 0.09 (142.87) 18.26 144.00 7 5 0.05 (143.93) 95.53 -0.70

10 5 5.00 70.00* 164.00* 11 5 14.11 (140.00) 31.22 163.00 11 5 0.09 (162.85) 98.27 0.6111 10 0.20 27.54 112.20 2 2 0.00 (112.20) 3.98 112.20 2 2 0.02 (112.18) 85.55 0.0012 10 0.60 13.73 118.60 3 3 0.00 (118.60) 5.90 118.60 3 3 0.07 (118.52) 57.03 0.0013 10 1.00 54.67 123.00 4 4 0.03 (122.96) 20.01 123.00 4 4 0.09 (122.88) 63.40 0.0014 10 3.00 15.00* 143.00* 8 5 9.79 (129.00) 102.00 143.00 8 8 0.06 (142.91) 94.33 0.0015 10 5.00 243.00* 156.00* 10 10 16.67 (130.00) 163.52 156.00 10 10 0.04 (155.94) 90.92 0.00

* For these cases, (i) the test run was forced to terminate at 1800.00 secs, (ii) the optimality gap at termination was > 0.1%, (iii) the makespan value, n(U),and m(U) reported correspond to the best incumbent solution, along with the time when it was observed during run, and (iv) the optimality gap (%) andthe lower bound value reported correspond to termination point.Among both the methods, the lower values of best makespan and time before termination are marked in bold.




row m p time (sec) M(U) n(U) m(U) gap% (L.B.) time (sec) M(U) n(U) m(U) gap% (L.B.) time makespan1 2 0.20 8.80 1001.20 5 2 0.06 (1000.6) 18.44 1001.60 4 2 0.07 (1000.9) -109.55 -0.042 2 0.60 89.00 1002.40 9 2 0.09 (1001.5) 23.25 1002.80 8 2 0.03 (1002.5) 73.88 -0.043 2 1.00 783.00 1003.80 14 2 0.09 (1002.80) 39.75 1003.60 13 2 0.04 (1003.2) 94.92 0.024 2 3.00 927.00* 1502.40* 90 2 33.16 (1004.20) 311.50 1504.00 22 2 0.02 (1503.7) 82.69 -0.115 2 5.00 142.00* 2502.40* 73 2 59.93 (1001.58) 602.21 2502.40 17 2 0.09 (2500.15) 66.54 0.006 5 0.20 176.00 1001.40 6 2 0.09 (1000.50) 91.20 1001.60 4 4 0.03 (1001.3) 48.18 -0.027 5 0.60 509.00* 1002.20* 8 4 0.14 (1000.81) 134.67 1002.20 8 5 0.02 (1002) 92.52 0.008 5 1.00 645.00* 1003.20* 11 5 0.25 (1000.64) 282.66 1004.20 9 5 0.03 (1003.9) 84.30 -0.109 5 3.00 1101.00* 1013.80* 53 5 75.78 (245.56) 1378.09 1011.80 25 5 0.09 (1010.79) 23.44 0.20

10 5 5.00 1789.00* 1124.40* 79 5 99.13 (9.81) 1440.00* 1051.20* 83 5 0.75 (1043.32) 0.00 6.5111 10 0.20 830.00 1001.20 5 3 0.08 (1000.40) 332.30 1001.60 4 4 0.05 (1001.1) 59.96 -0.0412 10 0.60 1123.00* 1002.60* 10 3 0.22 (1000.40) 443.87 1002.60 7 7 0.09 (1001.6) 75.34 0.0013 10 1.00 869.00* 1004.20* 16 10 0.38 (1000.40) 1489.34 1003.40 10 10 0.09 (1002.4) 17.26 0.0814 10 3.00 897.00* 1151.60* 346 6 99.89 (1.22) 1207.00* 1010.40* 18 10 0.78 (1002.52) 0.00 12.2615 10 5.00 436.00* 1050.20* 101 10 99.18 (8.60) 1335.00* 1013.60* 33 10 1.20 (1001.44) 0.00 3.49

* For these cases, (i) the test run was forced to terminate at 1800.00 secs, (ii) the optimality gap at termination was > 0.1%, (iii) the makespan value, n(U), andm(U) reported correspond to the best incumbent solution, along with the time when it was observed during run, and (iv) the optimality gap (%) and the lowerbound value reported correspond to termination point.Among both the methods, the lower values of best makespan and time before termination are marked in bold.



row m p time (sec) M(U) n(U) m(U) gap% (L.B.) time (sec) M(U) n(U) m(U) gap% (L.B.) time makespan1 2 0.20 6.40 1004.80 4 2 0.09 (1003.90) 12.57 1004.80 4 2 0.01 (1004.7) -96.41 0.002 2 0.60 13.00 1008.20 7 2 0.09 (1007.29) 22.34 1008.20 7 2 0.09 (1007.19) -71.85 0.003 2 1.00 87.00 1013.00 12 2 0.09 (1012.09) 27.21 1013.00 10 2 0.09 (1012.09) 68.72 0.004 2 3.00 1675.00* 1567.00* 29 2 33.05 (1009.00) 190.48 1507.00 23 2 0.09 (1505.64) 89.42 3.835 2 5.00 196.00* 2504.00* 41 2 59.66 (1010.00) 394.85 2504.00 17 2 0.01 (2503.75) 78.06 0.006 5 0.20 225.00 1004.60 4 3 0.08 (1003.80) 62.38 1004.80 4 4 0.06 (1004.2) 72.28 -0.027 5 0.60 1359.00 1007.60 7 4 0.09 (1006.69) 88.58 1008.40 6 5 0.07 (1007.69) 93.48 -0.088 5 1.00 698.00* 1011.00* 10 4 0.79 (1003.00) 202.02 1011.00 9 5 0.09 (1009.99) 88.78 0.009 5 3.00 1740.00* 1041.00* 24 5 99.33 (7.00) 1012.75 1028.00 22 5 0.06 (1027.38) 43.74 1.25

10 5 5.00 1661.00* 1283.00* 258 5 71.26 (368.76) 1603.20 1089.00 53 5 0.04 (1088.56) 10.93 15.1211 10 0.20 578.00* 1004.60* 4 4 0.26 (1002.00) 203.38 1004.80 4 4 0.05 (1004.3) 88.70 -0.0212 10 0.60 470.00* 1007.60* 7 7 0.56 (1002.00) 287.98 1007.80 7 7 0.07 (1007.09) 84.00 -0.0213 10 1.00 1716.00* 1011.00* 9 8 0.89 (1002.00) 1200.76 1010.00 9 9 0.07 (1009.29) 33.29 0.1014 10 3.00 923.00* 1031.00* 24 10 2.43 (1005.00) 1026.00* 1023.00* 17 10 0.52 (1017.68) 0.00 0.7815 10 5.00 1780.00* 1144.00* 47 8 90.08 (113.54) 887.00* 1041.00* 26 10 0.61 (1034.65) 0.00 9.00





row m p time (sec) M(U) n(U) m(U) gap% (L.B.) time (sec) M(U) n(U) m(U) gap% (L.B.) time makespan1 2 0.20 16.70 1019.00 3 2 0.00 (1019.00) 7.21 1019.20 3 2 0.07 (1018.49) 56.83 -0.022 2 0.60 23.00 1030.20 6 2 0.09 (1029.27) 9.34 1031.80 6 2 0.03 (1031.49) 59.39 -0.163 2 1.00 71.00 1045.00 8 2 0.09 (1044.06) 16.23 1045.00 8 2 0.09 (1044.06) 77.14 0.004 2 3.00 610.00* 1525.00* 35 2 30.09 (1066.03) 156.88 1525.00 23 2 0.05 (1524.24) 91.28 0.005 2 5.00 74.00* 2514.00* 15 2 57.64 (1065.00) 235.49 2514.00 14 2 0.05 (2512.74) 86.92 0.006 5 0.20 165.00 1018.80 3 3 0.00 (1018.80) 34.22 1019.40 3 3 0.04 (1018.99) 79.26 -0.067 5 0.60 225.00 1029.80 5 5 0.09 (1028.87) 44.90 1029.80 5 5 0.01 (1029.7) 80.04 0.008 5 1.00 1735.00 1039.00 7 4 0.09 (1038.06) 112.66 1039.00 7 5 0.06 (1038.38) 93.51 0.009 5 3.00 1645.00* 1088.00* 17 5 5.33 (1030.00) 590.12 1088.00 15 5 0.09 (1087.02) 67.22 0.00

10 5 5.00 1756.00* 1215.00* 27 5 15.44 (1027.37) 966.09 1184.00 33 5 0.09 (1182.93) 46.33 2.5511 10 0.20 1638.00 1018.80 3 3 0.08 (1017.98) 120.38 1019.40 3 3 0.09 (1018.38) 92.65 -0.0612 10 0.60 896.00* 1029.80* 6 6 1.92 (1010.00) 177.37 1029.80 5 5 0.09 (1028.77) 90.15 0.0013 10 1.00 1562.00* 1039.00* 7 7 2.31 (1015.00) 502.50 1039.00 7 7 0.01 (1038.9) 72.08 0.0014 10 3.00 987.00* 1174.00* 34 5 99.53 (5.48) 201.00* 1077.00* 14 10 0.24 (1074.42) 0.00 8.2615 10 5.00 1792.00* 1291.00* 28 6 99.59 (6.04) 1007.82* 1115.00* 20 10 0.36 (1110.99) 0.00 13.63



In this chapter, we addressed scheduling of a lot over a two-stage HFS with 1 + mconfiguration using the lot-streaming concept, for the objective of minimizing themakespan. A HFS configuration is encountered in a variety of practical situationssuch as continuous processing industries, flexible manufacturing systems and parallelcomputing frameworks. We have generalized the work of Cheng et al. [2015] to a1+m HFS. The novelty of our work is in obtaining an optimal schedule in polynomialtime, O(m3 + mn), where n is the specified number of sublots the lot is to be splitinto for scheduling over m machines at Stage 2, for the case when sublot sizes arerelaxed to be continuous. A branch-and-bound-based method is also developed forthe case when sublot sizes are discrete that relies on the use of a tight lower boundon makespan. Its efficacy is revealed after testing its performance against that of thedirect solution of a mixed-integer formulation of 1 + m TSHFS-LSP by CPLEX R©.The former is able to obtain solutions within 0.10% optimality gap in 83 out of90 instances, and require an average of 315.00 secs before termination, whereas thelatter obtains solutions within 0.10% optimality gap in only 40 out of 90 instances,and require as much as 1095.00 seconds on an average before termination. Theresults clearly show the superiority of our developed method over the direct solutionapproach.

Chapter 4

A New Production MethodologyTowards Achieving MassCustomization

4.1 Introduction

The trends such as interdependence of global economies, off-shoring of contracts,breakdown of monopolies, last mile delivery, and rise in social awareness owing tospread of internet and mobile technology, have all created dynamic global markets,which ultimately benefit the customer. Today’s customers enjoy far greater choicein customizing a product to their needs and preference as compared to what waspossible decades ago. This has affected how the companies design their supply chainsin order to offer customer-driven products and services. Also, the products arecharacterized by a short product life cycle, especially in high-technology sector. Thishas pushed the manufacturers to adopt a new approach called mass customization,which emphasizes on the shift from conventional mass production techniques in favorof flexible and rapid response methods (Qiao et al. [2002], Kumar and Allada [2007],Qu et al. [2011]), with the end goal to provide their customers as much variety aspossible without adding the burden of higher costs and extended delivery times.The work done by Park and Simpson [2008] and Zhang et al. [2008] point to theoperational benefits of mass customization such as lower costs and smaller lead times.

Platform Product Design (PPD) and Platform Architecture are two approaches that

58

Sanchit Singh Chapter 4. Towards Achieving Mass Customization 59

set us closer to achieving a mass customization system (Wheelwright and Clark[1992], Meyer and Lehnerd [1997], Robertson and Ulrich [1998], and Salvador et al.[2002]). Using these approaches, manufacturers have been able to produce a vari-ety of end-products for the market place, while at the same time, reduce varietyof constituent components and raw materials required during manufacturing pro-cess. Leading manufacturers such as Black and Decker and Hewlett-Packard haveimplemented PPD approach to rationalize their assembly lines (Meyer and Lehnerd[1997]). Volkswagen used Platform Architecture approach to reduce the developmentand production costs (Wilhelm [1997]). A review of various aspects of the PPD ap-proach has been presented by Krishnan and Ulrich [2001], Simpson [2004], Jose andTollenaere [2005], and Allada et al. [2006].

A platform is described as a foundation assembly comprising of a set of componentsshared by the variants within a product family (Ben-Arieh et al. [2009]). Differentproducts in the product family can be produced from a platform by adding and re-moving only a few of the parts on the assembly/disassembly lines. Ben-Arieh et al.[2009] address the problem of determining an optimal number as well as the con-figuration of multiple platforms for a given demand of products for the objective ofminimizing the production cost owing to the mass assembly of platforms togetherwith the addition and removal of individual components from the platforms. Theymake an essential contribution to the development of the theory of products’ plat-forms towards achieving a successful mass customization system. However, theirwork has few shortcomings: (1) there is no consideration given to products’ Bill-of-materials (BOM) or the notion of sub-assemblies which might lead to an inconsis-tency in the physical assembly of parts across different products, (2) the resultantplatforms upon optimization might not turn out to be feasible engineering-wise, asno such restrictions or necessary constraints are imposed in their study either a priorior post-optimization.

Zhang et al. [2010] list four cost-effective strategies that can be used in conjunctionwith the PPD approach: (1) commonality (standardization and sharing of compo-nents across products without compromising the variety), (2) modularity (creationof pre-configured modular design options catering to specific business and marketneeds), (3) postponement or delayed product differentiation (design of product struc-tures with proliferation of parts’ variety limited to the top levels of assembly; seeLee and Tang [1997]), and, (4) scalability (design of products’ platforms that canbe reconfigured with ease according to changeable set of products’ parameters; seeSimpson et al. [2001]).

In our work, we focus on effectively using the strategy of commonality. Some of


the positive benefits of commonality are simplified planning and scheduling withlower setup and holding costs (Berry et al. [1992], Collier [1981]), economy of scale(Gerchak and Henig [1989]), and reduction of vendor lead time uncertainty (Bentonand Krajewski [1990]). In this chapter, we strive to benefit from the commonalityaspect of parts that might exist in the assembly hierarchy across different products.Similar to most of the research on the topic of commonality, Collier [1981] too assumesa simple form of commonality among products without giving attention to high-levelassembly structures. We broaden the scope by considering the commonality of sub-assemblies across products from the viewpoint of reducing overall production cost.

Problem Statement: Given a set of products with stochastic demand, each of whichconstitutes sub-assemblies that are common across products, and a set of machinesfor performing assembly operations, determine an optimal schedule for in-house pro-duction of sub-assemblies, in order to minimize the sum of production cost, loss dueto excess production and cost of delay in products’ order fulfillment.

We call this problem as Stochastic Demand Assembly Job Shop Scheduling Problem(SD-AJSSP). Assembly Job Shop Scheduling Problem (AJSSP) is a strongly NP-hard problem even with the deterministic demand. Most of the current research usesdispatching rules to solve it to reduce its complexity (Maxwell and Mehra [1968],Fry et al. [1989], and Philipoom et al. [1991]). Stochastic demand adds to the com-plexity of the AJSSP. On the operational level, we propose to handle this by havingproduction in two stages, one prior to products’ demand realization and the otherpost that. As such, we propose a new production methodology, Assemble-To-Orderwith Commonality of Sub-assemblies and Timing Aspect (ATO-CST) to solve theSD-AJSSP. We claim that ATO-CST is particularly suitable to achieve desired goalsof a mass customization system by offering operational benefits such as lower produc-tion costs, delayed product differentiation and thereby reduced losses due to excessproduction, and smaller lead times pertaining to products’ order fulfillment, in theface of stochastic demand. ATO-CST is flexible enough to accommodate productionin an assembly job shop environment in a two-stage manner and it leverages thepresence of commonality of sub-assemblies across products. Moreover, we solve it asan optimization model.

In both stages of production, different assemblies (products) or sub-assemblies areproduced or assembled in-house as per their BOM. Each contiguous assembly opera-tion incurs a setup cost and time and unit production/assembly cost and time. Thesub-assemblies can either be produced/assembled and then stocked up as inventory(in anticipation of stochastic demand) at Stage 1, or assembled to order after prod-ucts’ demand is realized at Stage 2. We assume that the demand for all the products


is realized at the onset of Stage 2. If a unit of sub-assembly produced in excess is notutilized towards overall demand fulfillment, it incurs a unit inventory cost or loss.For every product, there is a cost of delay which is proportional to its lead time, i.e.,the time between getting an order (i.e., when the demand occurs) and its delivery(i.e., time of completion for the order at an assembly machines).

To begin with, in §4.2, we address a relaxed version of the problem in which the timerequired to assemble sub-assemblies over the machines is not taken into account. Assuch, this amounts to determining only optimal quantities of sub-assemblies producedat either of the two stages of production. We evaluate the performance of some of thecommonly used production methodologies, namely Make-To-Stock (MTS), Make-To-Order (MTO) and Assemble-To-Order (ATO) for the objective of minimizingtotal production cost and inventory losses. We develop a new methodology for thiscase, Assemble-To-Order with Commonality of Sub-assemblies (ATO-CS), which isnot only superior to MTS, MTO, and ATO in the prsence of the problem featuresmentioned above, but also, is expected to outperform them if the timing aspect ofthe production is also considered.

Then, in §4.3, we address the given problem by incorporating the timing aspectof production, thereby considering a full-fledged assembly job shop. As such, wedetermine the optimal quantity of sub-assemblies produced at both the stages ofproduction and the production schedule over the machines. The objective is tominimize the sum of production cost, loss due to excess inventory, and cost of delayin products’ order fulfillment (which is proportional to the lead time of a product’sorder delivery for all the products). We extend the ATO-CS for this case and call it asAssemble-To-Order with Commonality of Sub-assemblies and Timing Aspect (ATO-CST). Its performance is evaluated against those of MTS, MTO, and ATO. We alsodevelop an algorithm that can effectively solve a mathematical model formulation ofATO-CST for large-sized problem instances.

We make the following assumptions for SD-AJSSP:

1. Demand for products is stochastic and defined by a set of probable scenarios.The production must meet demand in all the scenarios.

2. Each product’s BOM can be represented as a tree structure with sub-assembliesas nodes and directed edges capturing assembly precedence relationship be-tween a parent sub-assembly and its constituent child sub-assemblies. Similarly,each sub-assembly has a unique tree structure representation. The dependencyrelationship and time and cost of assembly operations for a sub-assembly are


independent of the product it constitutes.

3. Each contiguous assembly operation for a sub-assembly incurs a fixed sub-assembly dependent setup cost independent of units of sub-assemblies put to-gether. There is also a sub-assembly dependent unit assembly cost.

4. When the timing aspect of production is taken into consideration, we onlyconsider processing over the assembly machines at Stage 2. During Stage 1production, we assume that there is infinite time to schedule production overthe resources. This distinction makes sense due to the importance of deliveringthe products having the shortest possible lead times only once their demandoccurs.

4.2 Without Timing Aspect

Now, we describe two fundamental methods used for the in-house production of sub-assemblies and products. We explain their implicit use in commonly-used productionmethodologies such as MTO, MTS, ATO, and our proposed methodology, ATO-CS.

1. Method 1 (Sub-assembly-specific): All the units of a sub-assembly, whichmay belong to one or more products, are assembled contiguously on its assemblymachine.

2. Method 2 (Product-specific): All the sub-assembly units corresponding toa product (after accounting for the multiplicity of a sub-assembly within thatproduct) are assembled contiguously on its assembly machine, one productat-a-time.

For the case of stochastic demand, Method 1 production is said to take place duringStage 1, prior to demand realization. The sub-assemblies are then stored as inventoryuntil they are required to sustain production post products’ demand realization,which is carried out as per Method 2 and is said to take place during Stage 2.

Also note that, MTO, MTS, and ATO only make sense in the context of stochasticdemand. In such a case: (1) MTS follows purely Method 1 production, where allthe sub-assemblies are produced at Stage 1 in anticipation of the worst-case demandscenario and subsequently stored as inventory, (2) MTO follows purely Method 2production, where all the sub-assemblies are produced at Stage 2, having known exact


demand, and (3) ATO follows Method 2 production for the top level assemblies inproducts’ BOM only, whereas the rest of the sub-assemblies are produced by Method1 in anticipation of the worst-case demand scenario.

ATO-CS is a cost-based method and is implemented in a two-stage manner. Itcombines features of both Method 1 and Method 2. In the first stage, assembly ofcertain sub-assemblies (mostly those that have a higher degree of commonality acrossproducts) is carried out as per Method 1. In the second stage, all the remaining sub-assemblies are put together as per Method 2, i.e., specific to one product at-a-time.For the case of stochastic demand, ATO-CS is compared with MTS, MTO, and ATO,whereas for the case of deterministic demand, we will compare the performance ofATO-CS directly with that of Method 1 and Method 2.

In this section, we do not consider the timing aspect of production when developingATO-CS. Further, we represent ATO-CS by two mathematical models - one for thecase of deterministic demand and the other for stochastic demand. Even thoughthe problem inherently assumes the stochastic nature of products’ demand, we relaxthis, to begin with (along with timing aspect), and assume that the demand isdeterministic. We do so only to develop our proposed methodology from its mostessential elements and to present a simple illustration of the complex decision-makingintertwined within ATO-CS.


Next, we present a representation of ATO-CS as a mathematical model, for the caseof deterministic demand.


K Set of products (k is used as an index).

J Set of sub-assemblies (j, l is used as an index).

Jk Set of sub-assemblies forming product k. Conversely, J j representsthe set of products for which sub-assembly j is a part of their BOMs.

rk Assembly (it will be refered to as a sub-assembly as well) at theroot of product k.

dk Demand (deterministic) for product k.

Rj Set of sub-assemblies that are immediate children (i.e. put togetherfor the formation) of sub-assembly j.


afixj Fixed setup cost of a contiguous assembly operation resulting in oneor more unit(s) of sub-assembly j.

aunitj Cost of assembling a single unit of sub-assembly j from its con-stituent sub-assemblies and/or base components.

mkj Number of times sub-assembly j appears within product k (m isused as an index).

<k, j,m> A triplet representing a uniquely labeled node, nkjm (as shown inFigure 4.3). m denotes the order of occurrence of sub-assembly jfrom top to bottom and left to right fashion within the BOM ofproduct k. The set of all such triplets is denoted by N .

Rkjm Set of triplets (nodes) that are immediate children of node nkjm.

Pkjm Parent node of nkjm.

Decision variables:

w1j , f

ass,1j = Number of units of jth sub-assembly produced/assembled at Stage

1 (in a single setup) and the associated production/assembly cost,respectively.

w2kj, f

ass,2kj = Number of jth sub-assembly to be produced/assembled at Stage 2

(in a single setup) for product k, and the associated cost, respec-tively.

y1j =

{1, if w1

j > 0,

0, otherwise.

y2kj =

{1, if w2

kj > 0,

0, otherwise.

In the context of nkjm, we have,

w2kjm = Number of units that are assembled at Stage 2.

w1kjm = Number of units that are assembled at Stage 1, such that all of them

are fully utilized towards ultimate fulfillment of products’ demand.It is a sum of two components, w1,2

kjm and w1,2kjm.

w1,1kjm = A portion of the total number of units of node, nkjm, that are as-

sembled at Stage 1, and procured from inventory towards direct ful-fillment of assembly of its parent node, Pkjm, at Stage 2 (together

with w2kjm units). As such, w1,1

kjm + w2kjm = w2

Pkjm.


w1,2kjm = A portion of the total number of units of node, nkjm, that are as-

sembled at Stage 1, and used towards direct fulfillment of assemblyof its parent node, Pkjm, at Stage 1. As such, w1,2

kjm = w1Pkjm

.

We now present a Mixed Integer Programming (MIP) formulation of ATO-CS forthe case when the products’ demand is deterministic.

Model ATO-CS-D:

Minimize:∑j∈J

fass,1j +∑k∈K

∑j∈Jk

f 2kj (4.1)

w1,1k,rk,1

+ w2k,rk,1

= dk, ∀k ∈ K (4.2)

w1,1klm + w2

klm = w2kjm, ∀<k, j,m> ∈ N, ∀<k, l,m> ∈ Rkjm

(4.3)

w1,2klm = w1

kjm, ∀<k, j,m> ∈ N, ∀<k, l,m> ∈ Rkjm

(4.4)

w1,1kjm + w1,2

kjm = w1kjm, ∀<k, j,m> ∈ N

(4.5)∑k∈Jj

mkj∑m=1

w1kjm = w1

j , ∀j ∈ J (4.6)

mkj∑m=1

w2kjm = w2

kj, ∀k ∈ K, j ∈ Jk (4.7)

fass,1j = aunitj w1j + afixj y1

j , ∀j ∈ J (4.8)

w1j ≤ wubj y

1j , ∀j ∈ J (4.9)

f 2kj = aunitj w2

kj + afixj y2kj, ∀k ∈ K, j ∈ Jk

(4.10)

w2kj ≤ wubkjy

2kj, ∀k ∈ K, j ∈ Jk

(4.11)

w1kjm, w

2kjm, w

1,1kjm, w

1,2kjm, w

1j , w

2kj ∈ Z+, ∀j ∈ J,∀k ∈ J j,∀m = 1, . . . mkj

(4.12)

y1j , y

2kj ∈ {0, 1}, ∀j ∈ J,∀k ∈ J j

(4.13)


Constraints (4.2) relate the demand for each product, k, to the units of correspondingsub-assembly, rk, being either procured directly from inventory (w1,1

k,rk,1) or assembled

during Stage 2 (w2k,rk,1

). Constraints (4.3) ensure that for w2k,j,m units of sub-assembly

j, that are assembled during Stage 2, the equivalent units of its constituent sub-assembly, l (where nklm is immediate child of nkjm) are available either procureddirectly from inventory (w1,1

k,l,m) or assembled during Stage 2 (w2k,l,m). Constraints

(4.4) enforce relationship between sub-assembly, l, that is used in the assembly ofother sub-assemblies, j (such that nklm is a child of nkjm, directly or indirectly) duringStage 1, by recursively equating w1,2

klm to w1kjm (where nklm is the immediate child

of nkjm). Note that w1,2klm is not procured from inventory for assembly during Stage

2. Constraints (4.5) provide total number of units of sub-assembly j (in context ofnkjm), w1

kjm, that are to be produced as per Method 1. Note that during Stage 1(Method 1), assembly is done sub-assembly wise and not specific to each individualnode nkjm. As such, we sum over all such nodes corresponding to sub-assembly jand get w1

j (constraints (4.6)), the quantity that is produced contiguously duringStage 1. Constraints (4.7) calculate w2

kj, the quantity that is produced contiguouslyduring Stage 2, for each sub-assembly j specific to product k of which it is partin accordance with BOM. Constraints (4.8) and (4.9) relate the cost of assemblingsub-assembly j at Stage 1. Similarly, constraints (4.10) and (4.11) relates the costof assembly of sub-assembly j specific to product k during Stage 2. The objectivefunction combines the cost of production during Stage 1 and Stage 2 over all sub-assemblies and products.

The values of parameters, wubj and wubkj in constraints (4.9) and (4.11) can be fixedas following.

wubj =∑k∈Jj

dkmkj,

wubkj = dkmkj.

Figure 4.1 illustrates some of the decision variables used, w1kjm, w

2,1kjm, w

2,2kjm and w2

kjm

for the two products in Example 4.1 in §4.2.2. The illustration assumes the BOM andgraph representation for the products as given in Figures 4.2 and 4.3, respectively.The products’ demands (deterministic) are assumed to be, d1 = 10 and d2 = 16,and only a partial solution feasible to Model ATO-CS-D is provided for some of thenodes corresponding to sub-assemblies 5, 2, and 1 for Product 1 and 7, 6, 2, and 1for Product 2.


2 8

1

𝑤 "#$%,' =𝑤 ',%,'

%,' =11

𝑤 "#$%,% =𝑤 ',%,'

%,% =2

T=0Method2

Method1j

5

7

6

2

1

5

52

2 5

5

3

2

8

6 4

5

3

8

82

821

1

84

2 𝑤 "#$% =𝑤 ',%,'

% =11+2=13

𝑤 "#$' =𝑤 ',%,'

' =3

k=1 (𝐷%= 10) k=2 (𝐷'= 16)

𝑤 "#$%,'

K,j,m Node<k,j,m>

𝑤 "#$'

𝑤 "#$%,%

2,7,1

2,1,12,6,1

2,2,1

2,1,2

𝑤 "#$%,% +𝑤 "#$

%,' = 𝑤 "#$%

1,5,1

1,2,1

1,1,1

Figure 4.1: An illustration of the implementation of ATO-CS-D.


For the case when the products’ demands are stochastic, ATO-CS is formulated asa two-stage stochastic program with recourse. The stochastic demand is modeled asa set of demand scenarios with known occurrence probabilities.

We define the following additional notation for the case of stochastic demand.

S Set of demand scenarios (s is used as an index).ps Probability of occurance of scenario s.dks Demand of product k under scenario s.hj Inventory holding cost (or loss) per unit of sub-assembly j that is

not used to meet products’ demand.

Stage 1 decision variables:

w1j , f

ass,1j = Number of jth sub-assembly to be produced by Method 1, and the

associated cost, respectively.

y1j =

{1, if w1

j > 0,

0, otherwise.


w2kjs, f

ass,2kjs = Number of jth sub-assembly to be produced by Method 2 for product

k, and the associated cost, respectively, under scenario s.

y2kjs =

{1, if w2

kjs > 0,

0, otherwise.

w1kjms, w

2kjms

w1,1kjms, w

1,2kjms

}= The variables hold the same meaning in the context of nkjm as in the

deterministic case, except with an additional index for each scenarios.

ujs, finvjs (≡ hjujs) = Number of jth sub-assembly that do not get utilized and

the associated cost, respectively, under scenario s.

We now present an MIP formulation of ATO-CS for the case when the products’demand is stochastic.

Model ATO-CS-S:

Minimize:∑j∈J

fass,1j +∑s∈S

ps

(∑k∈K

∑j∈Jk

fass,2kjs +∑j∈J

hjujs

)(4.14)


fass,1j = aunitj w1j + afixj y1

j , ∀j ∈ J (4.15)

w1j ≤ wubj y

1j , ∀j ∈ J (4.16)∑

k∈Jj

mkj∑m=1

w1kjms + ujs = w1

j , ∀j ∈ J, s ∈ S (4.17)

w1,1k,rk,1,s

+ w2k,rk,1,s

≥ dks, ∀k ∈ K, s ∈ S (4.18)

w1,1klms + w2

klms = w2kjms, ∀<k, j,m> ∈ N, ∀<k, l,m> ∈ Rkjm, s ∈ S

(4.19)

w1,2klms = w1

kjms, ∀<k, j,m> ∈ N, ∀<k, l,m> ∈ Rkjm, s ∈ S(4.20)

w1,1kjms + w1,2

kjms = w1kjms, ∀<k, j,m> ∈ N, s ∈ S (4.21)

mkj∑m=1

w2kjms = w2

kjs, ∀k ∈ K, j ∈ Jk, s ∈ S (4.22)

fass,2kjs = aunitj w2kjs + afixj y2

kjs,∀k ∈ K, j ∈ Jk, s ∈ S (4.23)

w2kjs ≤ wubkjsy

2kjs, ∀k ∈ K, j ∈ Jk, s ∈ S (4.24)

w1j , ujs, w

2kjs, w

1kjms,

w1,1kjms, w

1,2kjms, w

2kjms

}∈ Z+, ∀j ∈ J,∀k ∈ J j,∀m = 1, . . . mkj,∀s ∈ S

(4.25)

y1j , y

2kjs ∈ {0, 1}, ∀j ∈ J,∀k ∈ J j,∀s ∈ S (4.26)

Constraints (4.15) and (4.16) are exactly the same as Constraints (4.8) and (4.9).Constraints (4.17) are similar to Constraints (4.6) under each scenario, s, with anappropriate adjustment owing to the stochastic nature of demand due to which someof the Stage 1 production might go un-utilized. Hence we use an extra term ujs tocapture the units of un-utilized inventory. Constraints (4.18), (4.19), (4.20), (4.21),(4.22), (4.23) and (4.24) are similar to Constraints (4.2), (4.3), (4.4), (4.5), (4.7),(4.10) and (4.11) under each scenario s. The objective function combines the cost ofproduction during Stage 1 and the expected production cost plus un-utlized inventorylosses during Stage 2 over all sub-assemblies and products.

The values of parameters, wubj and wubkjs in constraints (4.16) and (4.24) can be fixedas following.

wubj = maxs∈S

{∑k∈Jj

dksmkj

},


wubkjs = dksmkj.

4.2.2 A Numerical Example

Example 4.1. We present a small example consisting of two products and seven sub-assemblies for an assembly job shop configuration in the presence of commonality ofsub-assemblies across products (as is evident in the BOM of the products shown inFigure 4.2). The data for this example is provided in Table 4.1 and Table 4.2. Figure4.3 illustrates the BOMs of the products using a tree graph, wherein each node hasa unique label of the form <k, j,m>, where k and j are indices for the product andsub-assembly, respectively, and m is the multiplicity index of the sub-assembly in thatproduct.

Through this example, we will show the implementation of ATO-CS and other method-ologies for both the cases when demand is deterministic and stochastic.

5

23 4

1

(a) Product 1.

7

6 1

2

1

(b) Product 2.

Figure 4.2: Example 4.1-BOM of products.

Table 4.1: Products Data.

Product (k) Root sub-assembly (rk)1 52 7


Table 4.2: Composition of sub-assemblies.

Sub-assembly (j) Set of children (Rj) Setup cost (afixj ) Unit assembly cost (aunitj )

1 φ 10 12 {1} 10 13 φ 5 14 φ 5 15 {2,3,4} 5 26 {2} 5 17 {1,6} 5 1

1, 5, 1

1, 2, 11, 3, 1 1, 4, 1

1, 1, 1

(a) Product 1

2, 7, 1

2, 6, 1 2, 1, 1

2, 2, 1

2, 1, 2

(b) Product 2

Figure 4.3: Example 4.1-Unique representation of each sub-assembly (node) in theBOM of products.


SOLUTION:Deterministic demand. We present the cost of production by three methodologies,namely Method 1 (Table 4.3), Method 2 (Table 4.4) and ATO-CS (Table 4.5) whenthe demands for the products are, d1 = 10 and d2 = 10.

Table 4.3: Example 1: Cost of production by Method 1 (sub-assembly-specific) fordeterministic demand (d1 = 10, d2 = 10).

Sub-assembly (quantity) Cost1 (30) 10+1x30=403 (10) 5+1x10=154 (10) 5+1x10=152 (20) 10+1x20=306 (10) 5+1x10=157 (10) 5+1x10=155 (10) 5+2x10=25

Total cost: 155

Table 4.4: Example 1-Cost of production by Method 2 (product-specific) for deter-ministic demand (d1 = 10, d2 = 10).

Product Sub-assembly (quantity) Cost1 1 (10) 10+1x10=20

3 (10) 5+1x10=154 (10) 5+1x10=152 (10) 10+1x10=205 (10) 5+2x10=25

2 1 (20) 10+1x20=302 (10) 10+1x10=206 (10) 5+1x10=157 (10) 5+1x10=15

Total cost: 175


Table 4.5: Example 1: Cost of production by ATO-CS for deterministic demand(d1 = 10, d2 = 10).

Sub-assemblyQuantity produced by

Method 1 (cost) ≡ w1j (f

ass,1j )

Quantity produced by Method 2,w2kj, specific to product, k and

its cost, f 2kj ≡ k, w2

kj(f2kj)

1 30 (40) φ, 0 (0)3 0 (0) 1, 10 (15)4 0 (0) 1, 10 (15)2 20 (30) φ, 0 (0)6 0 (0) 2, 10 (15)7 0 (0) 2, 10 (15)5 0 (0) 1, 10 (25)

Total cost:70 85

155

Stochastic demand. We present the cost of production by four methodologies,namely MTS (Table 4.6), MTO (Table 4.7), ATO (Table 4.8) and ATO-CS (Ta-ble 4.9) when the demand for each product follows a uniform distribution. Letd1 ∼ Unif(9, 11) and d2 ∼ Unif(9, 11). We consider a total of nine scenarios of prod-ucts’ demands. For a fair comparison among different methodologies, it is assumedthat the demand for both the products has to be met under all scenarios. Underthis premise, it is understood that the appropriate amount of sub-assemblies needto be assembled and stocked up as inventory prior to the actual demand realization.We assume a uniform unit inventory holding cost (or loss) across sub-assemblies, i.e.hj = h,∀j ∈ J .


Table 4.6: Example 1: Cost of production and inventory losses in MTS system forstochastic demand (d1 ∼ Unif(9, 11), d2 ∼ Unif(9, 11)).

Sub-assemblyQuantity producedby Method 1 (cost)

Scenarios represented by (d1, d2)(9,9) (9,10) (9,11) (10,9) (10,10) (10,11) (11,9) (11,10) (11,11)

Inventory cost of sub-assemblies unused1 33 (43) 6h 4h 2h 5h 3h 1h 4h 2h 03 11 (16) 2h 2h 2h 1h 1h 1h 0 0 04 11 (16) 2h 2h 2h 1h 1h 1h 0 0 02 22 (32) 4h 3h 2h 3h 2h 1h 2h 1h 06 11 (16) 2h 1h 0 2h 1h 0 2h 1h 07 11 (16) 2h 1h 0 2h 1h 0 2h 1h 05 11 (27) 2h 2h 2h 1h 1h 1h 0 0 0

Total cost: 166 20h 15h 10h 15h 10h 5h 10h 5h 0Expected cost: 166+1/9×90h=166+10h

Table 4.7: Example 1: Cost of production and inventory losses in MTO system forstochastic demand (d1 ∼ Unif(9, 11), d2 ∼ Unif(9, 11)).

Product Sub-assemblyScenarios represented by (d1, d2)

(9,9) (9,10) (9,11) (10,9) (10,10) (10,11) (11,9) (11,10) (11,11)Quantity produced by Method 2 (cost)

1

1 9 (19) 9 (19) 9 (19) 10 (20) 10 (20) 10 (20) 11 (21) 11 (21) 11 (21)3 9 (14) 9 (14) 9 (14) 10 (15) 10 (15) 10 (15) 11 (16) 11 (16) 11 (16)4 9 (14) 9 (14) 9 (14) 10 (15) 10 (15) 10 (15) 11 (16) 11 (16) 11 (16)2 9 (19) 9 (19) 9 (19) 10 (20) 10 (20) 10 (20) 11 (21) 11 (21) 11 (21)5 9 (23) 9 (23) 9 (23) 10 (25) 10 (25) 10 (25) 11 (27) 11 (27) 11 (27)

2

1 18 (28) 20 (30) 22 (32) 18 (28) 20 (30) 22 (32) 18 (28) 20 (30) 22 (32)2 9 (19) 10 (20) 11 (21) 9 (19) 10 (20) 11 (21) 9 (19) 10 (20) 11 (21)6 9 (14) 10 (15) 11 (16) 9 (14) 10 (15) 11 (16) 9 (14) 10 (15) 11 (16)7 9 (14) 10 (15) 11 (16) 9 (14) 10 (15) 11 (16) 9 (14) 10 (15) 11 (16)

Total cost: 164 169 174 170 175 180 176 181 186Expected cost: 1/9×1575=175

Under MTS policy, we must produce the sub-assemblies and stock them as inventoryin anticipation of the worst-case scenario, since by definition, this methodology doesnot allow for production post demand realization. Column 2 of Table 4.6 shows thequantity of sub-assemblies assembled before demand realization. Columns 3-11 showthe cost of inventory for the sub-assemblies left unused after demand realization.Under the assumption that all the sub-assemblies be stocked up for the worst-casescenario, MTS is expected to deliver the shortest lead times (theoretically zero) forproducts if the timing aspect of the scheduling is to be considered. However, this


advantage is countered by higher expected cost through inventory losses.

Under MTO policy, all the sub-assemblies start their assembly operation only afterthe products’ demand is realized. All the sub-assemblies specific to a product areassembled first before starting with the next product. The cost of production is givenin Columns 3-11 of Table 4.7. Note that MTO does not result in inventory lossessince the production is carried out after exact demand is realized. When the unitinventory holding cost or loss is significant, MTO is expected to perform the best interms of the total of production and inventory cost. However, this gain is offset bylarge expected lead times for products, if the timing aspect of the scheduling is tobe considered.

Table 4.8: Example 1: Cost of production and inventory losses in ATO system forstochastic demand (d1 ∼ Unif(9, 11), d2 ∼ Unif(9, 11)).

Sub-assemblyQuantity producedby Method 1 (cost)

Scenarios represented by (d1, d2)(9,9) (9,10) (9,11) (10,9) (10,10) (10,11) (11,9) (11,10) (11,11)

Inventory cost of sub-assemblies unused1 33 (43) 6h 4h 2h 5h 3h 1h 4h 2h 03 11 (16) 2h 2h 2h 1h 1h 1h 0 0 04 11 (16) 2h 2h 2h 1h 1h 1h 0 0 02 22 (32) 4h 3h 2h 3h 2h 1h 2h 1h 06 11 (16) 2h 1h 0 2h 1h 0 2h 1h 0

Quantity produced by Method 2 (cost)7 0 (0) 9 (14) 10 (15) 11 (16) 9 (14) 10 (15) 11 (16) 9 (14) 10 (15) 11 (16)5 0 (0) 9 (23) 9 (23) 9 (23) 10 (25) 10 (25) 10 (25) 11 (27) 11 (27) 11 (27)

Total cost: 123 37+16h 38+12h 39+8h 39+12h 40+8h 41+4h 41+8h 42+4h 43Expected cost: 123+1/9×(360+72h)=163+8h

ATO is similar to MTS in that all but top level assemblies in products’ BOM areassembled before demand realization during Stage 1. The sub-assemblies are subse-quently stocked up as inventory. The assembly of the final products is carried outfor an exact amount after demand realization. Column 2 of Table 4.8 shows thequantity of sub-assemblies assembled before demand realization. Columns 3-11 showthe cost of inventory for the sub-assemblies left unused after demand realization forall the sub-assemblies except the ones corresponding to the products. Note that thesub-assemblies numbered five and seven are assembled only after demand realizationfor Product 1 and 2, respectively, where their production costs are given in Columns3-11.


Table 4.9: Example 1: Cost of production and inventory losses in ATO-CS systemfor stochastic demand (d1 ∼ Unif(9, 11), d2 ∼ Unif(9, 11)).

Production by Method 1 Scenarios represented by (d1, d2)

Sub-assembly (j) Quantity (w1j ) Cost (fass,1j )

(9,9) (9,10) (9,11) (10,9) (10,10) (10,11) (11,9) (11,10) (11,11)Inventory cost of sub-assemblies unused

2 20 60 2h h 0 1 0 0 0 0 0

Product (k) Sub-assembly (j) Multiplicity (m) Quantity produced by Method 2 (cost) ≡ w2kjm(f 2

kjm)

1

1 1 0 (0) 0 (0) 0 (0) 0 (0) 0 (0) 0 (0) 0 (0) 1 (11) 1 (11)3 1 9 (14) 9 (14) 9 (14) 10 (15) 10 (15) 10 (15) 11 (16) 11 (16) 11 (16)4 1 9 (14) 9 (14) 9 (14) 10 (15) 10 (15) 10 (15) 11 (16) 11 (16) 11 (16)2 1 0 (0) 0 (0) 0 (0) 0 (0) 0 (0) 0 (0) 0 (0) 1 (11) 1 (11)5 1 9 (23) 9 (23) 9 (23) 10 (25) 10 (25) 10 (25) 11 (27) 11 (27) 11 (27)

21

1 9 (19) 10 (20) 11 (21) 9 (19) 10 (20) 11 (21) 9 (19) 10 (20) 11 (21)2 0 (0) 0 (0) 0 (0) 0 (0) 0 (0) 1 (11) 0 (0) 0 (0) 1 (11)

2 1 0 (0) 0 (0) 0 (0) 0 (0) 0 (0) 1 (11) 0 (0) 0 (0) 1 (11)6 1 9 (14) 10 (15) 11 (16) 9 (14) 10 (15) 11 (16) 9 (14) 10 (15) 11 (16)7 1 9 (14) 10 (15) 11 (16) 9 (14) 10 (15) 11 (16) 9 (14) 10 (15) 11 (16)

Total cost: 60 98+2h 101+h 104 102+h 106 130 105 131 156Expected cost: 60+1/9×(1033+4h)=174.78+0.44h

We now present in Table 4.9 a solution following the ATO-CS methodology. We donot claim this solution to be an optimal one from the viewpoint of model ATO-CS-S. However, a comparison will characterize the superiority of ATO-CS methodologyover others. Method 1 production takes place before demand realization, and assuch we let 20 units of sub-assembly numbered two to be assembled in advance andkept as inventory. Under different scenarios, these units are used for the assembly ofdifferent products. For example, when the demand scenario (10, 9) occurs (Column7), we can effectively use 10 units of Sub-assembly 2 for the fulfillment of the demandfor Product 1, and another 9 units for Product 2. A single unit of the sub-assemblygoes un-utilized and contributes towards the inventory loss amounting to h. Method2 production takes place after products’ demand realization, and as such when aparticular demand scenario occurs, the remaining and least necessary sub-assembliesare assembled, specific to one product at-a-time.

4.2.3 Numerical Example - Analysis of Results

So far, we have considered a production environment that resembles an assemblyjob shop, even though the timing aspect of scheduling has not yet been taken intoaccount. Two cases of products’ demand have been studied - deterministic andstochastic (with demand scenarios for products). The products share certain sub-assemblies in their BOM. The objective is to minimize production cost and inventory


0 1 2 3 4 5

140

180

220

h

Cos

t

MTOMTSATOATO−CS−hybrid

Figure 4.4: Example 1: Comparison of expected cost (production and inventory loss)for MTO, MTS, ATO and ATO-CS.

loss (only for the stochastic case). We have proposed a new production methodol-ogy, ATO-CS, and formulated two integer programs, model ATO-CS-D and modelATO-CS-S, corresponding to the two cases of demand, deterministic and stochastic,respectively.

Deterministic demand. ATO-CS has been compared directly with productionsolution following Method 1 and Method 2. As expected, cost of production forMethod 1 is smaller than that for Method 2 since all the units of a sub-assembly areassembled contiguously thus reducing cost by the ‘economy of scale’ because of thepresence of setup cost. Model ATO-CS-D is solved to optimality for the data givenin this example, and it can be seen from the solution presented in Table 4.5 that itonly differs from Method 1 solution in that it operates in a two-stage manner ratherthan as a single-stage production. Nonetheless, all the units of a single sub-assemblyare assembled contiguously resulting in the total cost equal to that for Method 1.The advantage of using ATO-CS over Method 1 production can only be seen for thecase when the demand is stochastic.

Stochastic demand. The performance of ATO-CS has been compared with threecommonly practiced production methodologies, MTO, MTS, and ATO. The results


for all the four methodologies are presented in Figure 4.4 for different values of h.The cost incurred by ATO is lesser than that by MTS. ATO is better than ATO-CSonly marginally for small values of h but is otherwise outperformed by ATO-CS forhigher values of h. As for MTO, it results in the least cost for higher values of h,even though ATO-CS has only a slightly higher cost. However, this advantage ofMTO over ATO-CS will no longer occur if the timing aspect of scheduling is to betaken into account. As such, MTO is expected to result in the largest lead times forproducts.

Even though, the solution for ATO-CS presented in Example 4.1 is not necessarilyoptimal for all values of h, the results do show that ATO-CS is a superior productionmethodology for a broad range of h.

The preliminary results for ATO-CS are satisfactory. Therefore, we develop it furtherin §4.3 by incorporating an additional key feature of the problem, i.e., the assemblytime of sub-assemblies over the machines. As such, we aim to solve a full-fledged as-sembly job shop problem. To do so, we adopt an extension of ATO-CS (the proposednew methodology is referred to as ATO-CST). We first present the mathematical for-mulation of the problem, and then we design an algorithm that can effectively solvethis mathematical model for large-sized data-set. In §4.4, we present results forATO-CST to showcase its superiority over MTS, MTO, and ATO. We then presentresults comparing the performance of this algorithm with that of a direct solution ofthe mathematical model using CPLEX R©.

4.3 With Timing Aspect

In this section, we introduce the timing aspect of production over the constrainedresources. As discussed previously in §4.1, we assume that setup and processingtimes are incurred for assembly operations only during Stage 2. We consider acost proportional to the lead time of a product’s order delivery in addition to theproduction cost and inventory loss. We believe that this additional feature is anintegral part of mass customization and holds critical importance to the businessesfor whom the earliest possible delivery of products to their customers is of primeimportance.

In a nutshell, we claim that our proposed methodology, ATO-CST, is suited toachieving the goals of mass customization by essentially implementing the followingtwo ideas: (1) common sub-assemblies that make a variety of products are assembled


in large quantities in advance prior to products’ demand realization, thus taking ad-vantage of ’economy of scale’ and commonality of sub-assemblies in products’ BOM.In turn, this achieves lower production costs overall as well as expedited deliveryof final assemblies/products once the real demand arises, (2) sub-assemblies or finalassemblies that belong to the upper levels of products’ BOM are assembled only afterproducts’ demands are realized. This results in fewer losses in excess inventory builtin the anticipation of stochastic demand. It is also referred to as ’delayed productdifferentiation’ in the literature on the subject of mass customization.


Next, we present the notation that we use for the model formulation of ATO-CST.Note that, for convenience, we have repeated some notation here that was presentedin §4.2.1 earlier.

K Set of products (k is used as an index).J Set of sub-assemblies (j, l is used as an index).M Set of machines (i is used as an index).S Set of demand scenarios (s is used as an index).Jk Set of sub-assemblies forming product k. Conversely, J j represents

the set of products for which sub-assembly j is a part of their BOMs.Additionally, we also define J0 = {0}, which will be used when defininga dummy node for resolving sequence order of assembly of differentsub-assemblies over a same machine.

Li Set of sub-assemblies assembled on machine i. Conversely, Lj repre-sents the assembly machine of jth sub-assembly.

Li Set of all pairs <kj>, such that k ∈ K, j ∈ Jk, j ∈ Li, i.e. sub-assembly j forming product k requires assembly on machine i at Stage2. Such a pair, <kj> is synonymous to a job which requires processingtime, pkjs on its assembly machine, i = Lj during Stage 2, underscenario s . Note that, pkjs is not know a priori and is a decisionvariable.

rk Assembly (it will be referred to as a sub-assembly as well) at the rootof product k.

Rj Set of sub-assemblies that are immediate children (i.e. put togetherfor the formation) of sub-assembly j.

afixj , bfixj Fixed setup cost and time incurred for a contiguous assembly operationresulting in one or more unit(s) of sub-assembly j, respectively.


aunitj , bunitj Unit cost and time incurred for assembling sub-assembly j from itsconstituent sub-assemblies and/or base components, respectively.

mkj Number of times sub-assembly j appears within product k (m is usedas an index).

<kjm> A triplet representing a uniquely labeled node, nkjm (as shown in Fig-ure 4.3). m denotes the order of occurrence of sub-assembly j fromtop to bottom and left to right fashion within the BOM of product k.The set of all such nodes is denoted by N .

Rkjm Set of child nodes of nkjm.Pkjm Parent node of nkjm.ps Probability of occurrence of scenario s.dks Demand of product k under scenario s.ek Unit delay cost for delivery of product k.hj Unit cost of un-utilized sub-assembly j in the inventory.


w1j , f

ass,1j Number of units of jth sub-assembly produced/assembled at Stage 1

(in a single setup) and the associated production/assembly cost, re-spectively.

y1j =

{1, if w1

j > 0,

0, otherwise.


w2kjs Number of units of jth sub-assembly produced/assembled at Stage 2

(in a single setup) for product k, under scenario s.

pkjs, fass,2kjs Time and cost for assembly of job, <kj>, under scenario s, respec-

tively.

y2kjs =

{1, if w2

kjs > 0,

0, otherwise.

tskjs, tekjs Start and end time of job, <kj>, under scenario s, respectively. Note

that, t = 0 marks the time when all the products’ demand is realizedat the onset of Stage 2. An upper bound parameter for start and endprocessing times can be obtained for a given machine and scenario,and is denoted by pubis , where i = Lj.


xk1j1k2j2is =

1, if w2

k1j1sunits corresponding to job, <k1j1>, are assem-

bled immediately before w2k2j2s

units corresponding to job,

<k2j2>, on their common assembly machine, i = Lj1 = Lj2 ,

0, otherwise.

αk1j1k2j2is =

1, if w2

k1j1sunits corresponding to job, <k1j1>, are assembled

before w2k2j2s

units corresponding to job, <k2j2>, on their

common assembly machine, i = Lj1 = Lj2 ,

0, otherwise.

fdelks Total cost of delay of product k, under scenario s (proportional tolead time of its delivery).

ujs, finvjs Number of jth sub-assembly that do not get utilized, and the associ-

ated inventory loss, under scenario s.

In the context of node, nkjm, we have,

w2kjms Number of units that are assembled at Stage 2, under scenario s.

w1kjms Number of units that are assembled at Stage 1, such that all of them

are fully utilized towards ultimate fulfillment of products’ demand, underscenario s. It is a sum of two components, w1,1

kjms and w1,2kjms.

w1,1kjms A portion of the total number of units of node, nkjm, that are assembled at

Stage 1, and procured from inventory towards direct fulfillment of assemblyof its parent node, Pkjm, at Stage 2, under scenario s (together with w2

kjms

units). As such, w1,1kjms + w2

kjms = w2Pkjms

.

w1,2kjms A portion of the total number of units of node, nkjm, that are assembled at

Stage 1, and used towards direct fulfillment of the assembly of its parentnode, Pkjm, at Stage 1, under scenario s. As such, w1,2

kjms = w1Pkjms

.

Note that, if (·) is an optimization program, then V(·) represents its optimal value,F(·) is the set of feasible solutions and (·) is the same problem with integralityconditions relaxed.

We now formulate ATO-CST as a linear deterministic equivalent of a two-stagestochastic program with recourse.

Model F:

Minimize:∑j∈J

fass,1j +∑s∈S

ps

(∑k∈K

∑j∈Jk

fass,2kjs +∑j∈J

f invjs +∑k∈K

fdelks

)(4.27)


w1j ≤ wubj y

1j , ∀j ∈ J (4.28)

fass,1j = afixj y1j + aunitj w1

j , ∀j ∈ J (4.29)

w1,1k,rk,1,s

+ w2k,rk,1,s

≥ dk,s, ∀k ∈ K, s ∈ S (4.30)

w1,1klm2s

+ w2klm2s

≥ w2kjm1s

, ∀nkjm1 ∈ N, nklm2 ∈ Rkjm1 , s ∈ S (4.31)

w1,2klm2s

= w1kjm1s

, ∀nkjm1 ∈ N, nklm2 ∈ Rkjm1 , s ∈ S (4.32)

w1,1kjms + w1,2

kjms = w1kjms, ∀nkjm ∈ N, s ∈ S (4.33)∑

k∈Jj

mkj∑m=1

w1kjms + ujs = w1

j , ∀j ∈ J, s ∈ S (4.34)

f invjs = hjujs, ∀j ∈ J, s ∈ S (4.35)mkj∑m=1

w2kjms = w2

kjs, ∀k ∈ K, j ∈ Jk, s ∈ S (4.36)

w2kjs ≤ wubkjsy

2kjs, ∀k ∈ K, j ∈ Jk, s ∈ S (4.37)

fass,2kjs = afixj y2kjs + aunitj w2

kjs, ∀k ∈ K, j ∈ Jk, s ∈ S (4.38)

pkjs = bfixj y2kjs + bunitj w2

kjs, ∀k ∈ K, j ∈ Jk, s ∈ S (4.39)

tekjs − tskjs = pkjs, ∀k ∈ K, j ∈ Jk, s ∈ S (4.40)

tskjs ≥ tekls, ∀k ∈ K, j ∈ Jk, l ∈ Rj, s ∈ S (4.41)∑<k2j2>∈Li∪<0,0>,<k2j2>6=<k1j1>

xk2j2k1j1is = y2k1j1s

, ∀i ∈M,<k1j1> ∈ Li ∪<0, 0>, s ∈ S (4.42)

∑<k2j2>∈Li∪<0,0>,<k2j2>6=<k1j1>


, ∀i ∈M,<k1j1> ∈ Li ∪<0, 0>, s ∈ S (4.43)

tsk2j2s ≥ tek1j1s − pubis (1− xk1j1k2j2is), ∀i ∈M,<k1j1>,<k2j2> ∈ Li,

<k1j1> 6= <k2j2>, s ∈ S (4.44)

fdelks = ektekrks

, ∀k ∈ K, s ∈ S (4.45)

w1j , ujs, w

2kjs, w

1kjms,

w1,1kjms, w

1,2kjms, w

2kjms

}∈ Z+, ∀j ∈ J, k ∈ J j,m = 1, . . .mkj, s ∈ S (4.46)

y1j ∈ {0, 1}, ∀j ∈ J (4.47)

y2kjs ∈ {0, 1}, ∀j ∈ J ∪ {0}, k ∈ J j, s ∈ S (4.48)

tskjs, tekjs ≥ 0, ∀k ∈ K, j ∈ Jk, s ∈ S (4.49)


xk1j1k2j2is ∈ {0, 1}, ∀i ∈M, j1 ∈ Li ∪ {0}, k1 ∈ J j1 ,j2 ∈ Li ∪ {0}, j2 6= j1, k2 ∈ J j2 , s ∈ S (4.50)

Constraints (4.28) and (4.29) define the assembly cost of jth sub-assembly at Stage1. Constraints (4.30) relate the demand for each product, k, to the units of corre-sponding sub-assembly, rk, being either procured directly from inventory (i.e. w1,1

k,rk,1,s

units) or assembled at Stage 2 (i.e. w2k,rk,1,s

units), under each scenario s. Constraints(4.31) ensure that assembly of w2

k,j,m1,sunits of node, nkjm1 , at Stage 2 require at-

least equivalent units of its child node, nklm, either procured directly from inventory(i.e. w1,1

k,l,m2,sunits) or assembled at Stage 2 itself (i.e. w2

k,l,m2,sunits), under scenario

s. Constraints (4.32) relate w1,2klm2s

, i.e. the number of units corresponding to node,nklm2 , that are assembled at Stage 1 towards the fulfillment of the assembly of itsparent node, nkjm1 = Pklm2 at Stage 1 itself, under scenario s. Constraints 4.33enforce that w1

kjms is a sum of two parts, w1,1kjms and w1,2

kjms as per its definition. Notethat during Stage 1, a contiguous assembly operation is undertaken sub-assemblywise only and not individual node nkjm wise. All such units, w1

kjms, corresponding toa sub-assembly j are therefore put together and assembled in one contiguous fashionusing a single setup at Stage 1. Since production is carried out at Stage 1 beforethe demand is realized, some units, ujs, remain un-utilized in inventory, under eachscenario s. Constraints (4.34) define this relationship. Constraints (4.35) determinethe un-utilized units of sub-assembly j remaining in the inventory, under scenario s.Constraints (4.36) define w2

kjs, the number of units that are produced in a contigu-ous fashion, corresponding to sub-assembly j and product k at Stage 2, under eachscenario s. Constraints (4.37) and (4.38) capture the cost of assembling a job atStage 2 associated with wkjs units. Constraints (4.39) determine the processing timeof job <kj>, under scenario s. Constraints (4.40)-(4.44) capture with scheduling ofjobs on assembly machines during Stage 2. Constraints (4.40) bind the start andend time of completion of a job on its assembly machine, at Stage 2. Constraints(4.41) are the precedence constraints during Stage 2 production, and enforce thatjob <kl> is scheduled prior to job <kj>, where sub-assembly l is the immediatechild in the BOM of sub-assembly j, under each scenario s. By Constraints (4.36),w2kjm1s

,∀m1 = 1 . . . mkj and w2klm2s

,∀m2 = 1 . . . mkl are dictated by w2kjs and w2

kls

units assembled corresponding to job <kj> and <kl>, respectively, under scenarios. As such, Constraints (4.41) will in turn enforce that assembly of w2

kjms units cor-responding to node nkjm can only start after assembly of w2

klms units correspondingto node nklm is completed, where nklm ∈ Rkjm, under scenario s. For a given ma-chine i and scenario s, Constraints (4.42)-(4.43) are the degree constraints (similarto a TSP formulation). A dummy job <0, 0> is created for each machine and sce-


nario pair, similar to a dummy depot used for sequencing cities in a TSP. Note thatthe Sub-tour Elimination Constraints (SECs) are not required in lieu of constraints(4.44), which enforce that the assembly start time of job <k2j2> follows that of theassembly completion time of job <k1j1>, if the former follows immediately after thelatter in the sequence order, i.e. xk1j1k2j2is = 1. Constraints (4.45) capture the costof delay of product k, which is proportional to the lead time of its delivery, tekrks,under scenario s. Constraints (4.46)-(4.50) impose logical restrictions on the decisionvariables.

The objective function in (4.27) combines the total cost of production at Stage 1 andStage 2, excess or un-utilized inventory cost and the cost of delay in the delivery ofproducts.

Appropriate values of Big-M used in the above constraints can be determined asfollows.

wubj = maxs∈S

{∑k∈Jj

dksmkj

}, used in constraints (4.28),

wubkjs = dksmkj, used in constraints (4.37),

pubis ≥∑

<kj>∈Li

(bfixj + bunitj dksmkj

), used in constraints (4.44).

Note that, if pubis is fixed at the minimum value prescribed above, (F) might beinfeasible due to the presence of inter-machine precedence constraints. It is thereforerecommended to safely fix it at higher than the above minimum value by a factor oftwo or more when (F) is solved as an MIP program.

4.3.2 Solution Methodology

We now develop a decomposition scheme to effectively solve (F), and demonstrate itscomputational effectiveness in dealing with large-sized data-set. To begin with, wewill first discuss our motivation stemming from some of the nice structural propertiesof the model formulation.


Motivation

We rewrite the objective term and the block of constraints pertaining to the schedul-ing of jobs on assembly machines during Stage 2 production (excluding the prece-dence constraints given by (4.41)) as follows.

Model T:

Minimize:∑i∈M

∑s∈S

∑<krk>∈Li,rk∈Li

psektekrks

(4.51)

tekjs − tskjs = pkjs, ∀i ∈M,<kj> ∈ Li, s ∈ S (4.52)∑<k2j2>∈Li∪<0,0>,<k2j2> 6=<k1j1>


, ∀i ∈M,<k1j1> ∈ Li ∪<0, 0>, s ∈ S (4.53)

∑<k2j2>∈Li∪<0,0>,<k2j2> 6=<k1j1>


, ∀i ∈M,<k1j1> ∈ Li ∪<0, 0>, s ∈ S (4.54)


<k1j1> 6= <k2j2>, s ∈ S (4.55)

tskjs, tekjs ≥ 0, ∀i ∈M,<kj> ∈ Li, s ∈ S (4.56)

xk1j1k2j2is ∈ {0, 1}, ∀i ∈M,<k1j1>,<k2j2> ∈ Li ∪<0, 0>,<k1j1> 6= <k2j2>, s ∈ S(4.57)

Note that (T), is separable over each pair of assembly machine and scenario, (i, s),and is denoted by (Tis). Additionally, if the processing time for each job (i.e., pkjs)is given, (Tis) aims at scheduling a set of jobs (having fixed processing times) ona single machine, with the objective of minimizing the sum of weighted completiontime of jobs. This can be accomplished by a straightforward application of WeightedShortest Processing Time (WSPT) rule or Smith’s rule for every single machine andscenario.

We summarize the two salient observations made so far as follows.

1. Scheduling of jobs during Stage 1 production can be considered as multipleseparable tasks, one for each pair of an assembly machine and demand scenario,given by (Tis). This program can be solved in a trivial manner yielding anoptimal integer solution for the sequencing and timing variables.


2. Precedence constraints (4.41) are hard constraints and naturally present to usthe idea of relaxing them following ‘Lagrangian relaxation’ technique (we referthe reader to Fisher [1985] for an overview of this technique).

Use of Lagrangian Relaxation

We define Lagrangian relaxation of (F) relative to the precedence constraints and aconformable non-negative vector µ to be as follows.

Model FRµ :

Minimize:∑j∈J

fass,1j +∑s∈S

ps

(∑k∈K

∑j∈Jk

fass,2kjs +∑j∈J

f invjs +∑k∈K

fdelks

)+∑s∈S

∑k∈K

∑j∈Jk

∑l∈Rj

µkjls(tekls − tskjs) (4.58)

subject to: (4.28)-(4.40) and (4.42)-(4.50)

Lagrangian bounding principle. Note that for a choice of lagrangian multiplier,µ ≥ 0, the value of the ‘Lagrangian function’, L(µ) (= V(FRµ)), is a lower boundon the optimal objective function value V(F ) of the original problem.

For a given µ, the objective function (4.58) can be rewritten as follows for the purposeof decomposing it into two parts, one containing the production variables belongingto Stage 1 and Stage 2, and the other containing the sequencing and timing variablesbelonging to Stage 2 for each pair of assembly machine and scenario, (i, s).

∑j∈J

fass,1j +∑s∈S

ps

(∑k∈K

∑j∈Jk

fass,2kjs +∑j∈J

f invjs

)+∑s∈S

∑k∈K

∑j∈Jk

∑l∈Rj

pkjs

+∑s∈S

∑k∈K

psektekrks︸︷︷︸

i=Lrk

+∑s∈S

∑k∈K

∑j∈Jk

∑l∈Rj

(µkjlst

ekls︸︷︷︸

i=Ll

−µkjlstekjs︸︷︷︸i=Lj

)(4.59a)

=∑j∈J

fass,1j +∑s∈S

ps

(∑k∈K

∑j∈Jk

fass,2kjs +∑j∈J

f invjs

)+∑s∈S

∑k∈K

∑j∈Jk

∑l∈Rj

pkjs

+∑s∈S

∑i∈M

∑<kj>∈Fis

vkjstekjs −

∑s∈S

∑i∈M

∑<kj>∈Bis

|vkjs|tekjs (4.59b)


=∑j∈J

fass,1j +∑s∈S

ps

(∑k∈K

∑j∈Jk

fass,2kjs +∑j∈J

f invjs

)+∑s∈S

∑k∈K

∑j∈Jk

∑l∈Rj

pkjs

−∑s∈S

∑i∈M

∑<kj>∈Bis

v′

kjs(pubis + pkjs)

+∑s∈S

∑i∈M

∑<kj>∈Fis

vkjstekjs +

∑s∈S

∑i∈M

∑<kj>∈Bis

v′

kjste′

kjs (4.59c)

The braces used under the individual terms in Expression (4.59a) denotes the ma-chine index with which the corresponding term is associated. This was obtained fromExpression (4.58) by simply substituting tskjs with tekjs − pkjs (i.e. using Constraints(4.40) which are implicit in (FRµ). The expression (4.59b) is obtained by separatingthe last three terms in the expression (4.59a) for each of the (i, s) pairs. As such, eachof the variable terms, tekjs, is associated with a coefficient vkjs, that can have eithernon-negative or negative sign. Accordingly, we define two sets, Fis and Bis for eachof the (i, s) pair. Fis = {<kj> ∈ Li : vkjs ≥ 0} and Bis = {<kj> ∈ Li : vkjs < 0}.Note that, vkjs, Fis, and Bis are dependent on µ value. The expression (4.59c) is then

obtained by substituting for v′

kjs = −vkjs > 0, te′

kjs = pubis − (tekjs−pkjs),∀<kj> ∈ Bis.

Using the modified objective function given by the expression (4.59c), we can furtherrewrite (FRµ) as follows.

Model FRµ:

Minimize:∑j∈J

fass,1j +∑s∈S

ps

(∑k∈K

∑j∈Jk

fass,2kjs +∑j∈J

f invjs

)+∑s∈S

∑k∈K

∑j∈Jk

∑l∈Rj

pkjs

−∑s∈S

∑i∈M

∑<kj>∈Bis

v′

kjs(pubis + pkjs) +

∑s∈S

∑i∈M

(V(T µFis) + V(T µFis)

)(4.60a)

w1j ≤ wubj y

1j , ∀j ∈ J (4.60b)

fass,1j = afixj y1j + aunitj w1

j , ∀j ∈ J (4.60c)

w1,1k,rk,1,s

+ w2k,rk,1,s

≥ dk,s, ∀k ∈ K, s ∈ S (4.60d)

w1,1klm2s

+ w2klm2s

≥ w2kjm1s

, ∀nkjm1 ∈ N, nklm2 ∈ Rkjm1 , s ∈ S (4.60e)

w1,2klm2s

= w1kjm1s

, ∀nkjm1 ∈ N, nklm2 ∈ Rkjm1 , s ∈ S (4.60f)

w1,1kjms + w1,2

kjms = w1kjms, ∀nkjm ∈ N, s ∈ S (4.60g)


∑k∈Jj

mkj∑m=1

w1kjms + ujs = w1

j , ∀j ∈ J, s ∈ S (4.60h)

f invjs = hjujs, ∀j ∈ J, s ∈ S (4.60i)mkj∑m=1

w2kjms = w2

kjs, ∀k ∈ K, j ∈ Jk, s ∈ S (4.60j)

w2kjs ≤ wubkjsy

2kjs, ∀k ∈ K, j ∈ Jk, s ∈ S (4.60k)

fass,2kjs = afixj y2kjs + aunitj w2

kjs, ∀k ∈ K, j ∈ Jk, s ∈ S (4.60l)

pkjs = bfixj y2kjs + bunitj w2

kjs, ∀k ∈ K, j ∈ Jk, s ∈ S (4.60m)

w1j , ujs, w

2kjs, w

1kjms,

w1,1kjms, w

1,2kjms, w

2kjms

}∈ Z+, ∀j ∈ J, k ∈ J j,m = 1, . . .mkj, s ∈ S (4.60n)

y1j ∈ {0, 1}, ∀j ∈ J (4.60o)

y2kjs ∈ {0, 1}, ∀j ∈ J ∪ {0}, k ∈ J j, s ∈ S, (4.60p)

where, the inner minimization problems, TµFis

and TµBis

, are presented next.

Model TµFis

:

Minimize:∑

<kj>∈Fis

vkjstekjs (4.61)

tekjs − tskjs = pkjs, ∀<kj> ∈ Fis (4.62)∑<k2j2>∈Fis∪<0,0>,<k2j2>6=<k1j1>


, ∀<k1j1> ∈ Fis ∪<0, 0> (4.63)

∑<k2j2>∈Fis∪<0,0>,<k2j2>6=<k1j1>


, ∀<k1j1> ∈ Fis ∪<0, 0> (4.64)

tsk2j2s ≥ tek1j1s − pubFis

(1− xk1j1k2j2is), ∀<k1j1>,<k2j2> ∈ Fis, <k1j1> 6= <k2j2>

(4.65)

tskjs, tekjs ≥ 0, ∀<kj> ∈ Fis (4.66)

xk1j1k2j2is ∈ {0, 1}, ∀<k1j1>,<k2j2> ∈ Fis ∪<0, 0>,<k1j1> 6= <k2j2>(4.67)


Model TµBis

:

Minimize:∑

<kj>∈Bis

v′

kjste′

kjs (4.68)

te′

kjs − ts′

kjs = pkjs, ∀<kj> ∈ Bis (4.69)∑<k2j2>∈Bis∪<0,0>,<k2j2>6=<k1j1>


, ∀<k1j1> ∈ Bis ∪<0, 0> (4.70)

∑<k2j2>∈Bis∪<0,0>,<k2j2>6=<k1j1>


, ∀<k1j1> ∈ Bis ∪<0, 0> (4.71)

ts′

k2j2s≥ te

′

k1j1s− pubBis(1− xk1j1k2j2is), ∀<k1j1>,<k2j2> ∈ Bis, <k1j1> 6= <k2j2>

(4.72)

ts′

kjs, te′

kjs ≥ 0, ∀<kj> ∈ Bis (4.73)

xk1j1k2j2is ∈ {0, 1}, ∀<k1j1>,<k2j2> ∈ Bis ∪<0, 0>,<k1j1> 6= <k2j2>(4.74)

Let,ts′

kjs = pubis − (tskjs + pkjs), ∀<kj> ∈ Bis,

pubFis =∑

<kj>∈Fis


),

pubBis =∑

<kj>∈Bis


).

Note that either of the inner minimization problems, (TµFis

) or (TµBis

), is similarto (Tis)) discussed earlier, albeit with slightly different objective function and con-straints. More importantly, they both deal with scheduling a set of jobs (having fixedprocessing times) on a single machine, with the objective of minimizing the sum ofweighted completion time of jobs.

The potential usefulness of any relaxation of (F), and of a Lagrangian relaxation,(FRµ) in particular, is determined by how near its optimal value is to that of (F).To obtain the largest possible lower bound, we would need to solve the followingoptimization problem.

Model D:L∗ = max

µ≥0L(µ) = maxµ≥0V(FRµ),


which we refer to as the ‘Lagrangian dual’ problem associated with the originaloptimization problem (F). L(µ) is also referred to as ‘Lagrangian dual function.’

Weak duality. Note that the optimal solution L∗ of the Lagrangian dual problemserves as a lower bound on the value of an optimal solution of the original primalproblem, i.e., L∗ ≤ V(F ).

Duality gap. The optimal value of a corresponding Lagrangian dual (a maximiza-tion program) is equal to the optimal value of the original primal problem if thelatter is a linear program. However, no such guarantee holds for the case when theprimal problem is a mixed-integer program. If the two programs do not have optimalequal values, then a duality gap is said to exist.

We now focus our attention on solving the Lagrangian relaxation problem, (FRµ).To that end, we use Benders decomposition technique (Benders [1962]), in whichwe decompose (FRµ) into a master problem (with production related variables fromStage 1 and Stage 2) and sub-problem(s) with timing and sequence related variables.Even though the integer programs, (Tµ

Fis) and (Tµ

Bis), can be solved in polynomial

times by a straightforward application of WSPT (weighted shortest processing time)rule for a given value of processing time for each job (or <k, j>), this advantagecannot be leveraged when using the programs as sub-problems in Benders decompo-sition. Also, (Tµ

Fis) and (Tµ

Bis) produce weak LP bounds. Therefore, the optimality

cuts obtained for the ‘Reformulated Master Problem’ obtained by solving their LPrelaxations are not tight. In turn, this results in the value of the Lagrangian sub-problem (FRµ) being underestimated by use of such a reformulated master problemat every value of µ. Therefore, we now focus our attention towards developing asuitable LP formulation(s) for its use as the sub-problem(s) within Benders decom-position scheme, which can provide tight lower bound for the original schedulingsub-problem(s).

Use of Benders Decomposition

Working with Mixed Integer Linear Programs (MILP), a Benders decompositionscheme is usually designed in such a manner that the master problem is an MILP(that retains the hard or integer variables of the original problem), whereas, the sub-problem is either a linear program or the convex hull representation of the underlyingMILP.

In our case, each of the sub-problems, (TµFis

), (TµBis

), ∀i ∈ M, s ∈ S is a 1||∑vjtj


scheduling problem, i.e. the jobs (having fixed processing times) are to be scheduledon a single machine, with the objective of minimizing the sum of weighted completiontimes of jobs. It is formulated as following (containing only completion time variablesand no sequence-related variables).

Minimize:∑j∈J

vjtj

tj ≥ pj, ∀j ∈ Jtk ≥ tj + pk ∨ tj ≥ tk + pj, ∀j, k ∈ J, j 6= k, (4.75)

where, J is the set of jobs that are scheduled on the machine; pj, and vj are theprocessing time and weight associated with the completion time of job j ∈ J , suchthat, pj, vj ≥ 0,∀j ∈ J . The difficulty clearly arises from the disjunctive constraints(4.75). Queyranne [1993] has shown that the convex hull of feasible completion timevariables is completely described by the following LP formulation.∑

j∈A

pjtj ≥∑j,l∈A,j≤l

pjpl, ∀A ⊆ J (4.76)

The question now arises weather we can use the above formulation as a sub-problemin Benders decomposition. Even though this formulation is a complete convex hullrepresentation of the completion time variables, it is not directly suitable to obtainBenders optimality cuts upon solving its dual form. Besides the fact that the resul-tant optimality cut would contain quadratic terms in products of p variables, thereis yet another challenge, that the independent variables, p, are not in a linearly sep-arable form from the dependent variables, t. Also, we do not know of a methodto linearize the product of two continuous variables. As such, we look for otherformulations which can be solved yielding tight lower bound for this problem.

In order to solve problem, 1||∑vjtj, van den Akker [1994] discusses two types of

formulations, which are based on binary linear ordering variables, αjl (which are equalto 1, when job j precedes job l and equal to 0, otherwise). In the first formulation,T1, each of the disjunctive constraints (4.75) is modeled by two constraints thatcontain a Big M (which should be at least

∑j∈J p

ubj , where pubj represents an upper

bound on the processing time for job j). This formulation is given by:

Model T1:Minimize:

∑j∈J

vjtj (4.77a)


tj − tl ≥ pj −M(1− αlj), ∀j, l ∈ J, j 6= l (4.77b)

αjl + αlj = 1, ∀j, l ∈ J, j < l (4.77c)

αjl ∈ {0, 1}, ∀j, l ∈ J, j 6= l (4.77d)

tj ≥ pj, ∀j ∈ J, (4.77e)

The second formulation is given by:

Model T2:Minimize:

∑j∈J

vjtj (4.78a)

tj ≥ pj +∑l∈J,l 6=j

plαlj, j ∈ J (4.78b)

αjl + αlj = 1, ∀j, l ∈ J, j < l (4.78c)

αjl + αlk + αkj ≤ 2, j, l, k ∈ J, j 6= l 6= k (4.78d)

αjl ∈ {0, 1}, ∀j, l ∈ J, j 6= l (4.78e)

Formulation (T1) contains n2 constraints, whereas formulation (T2) contains n3

constraints, but this is a non-issue at this point, since we are only interested insolving the formulations as relaxed. The disadvantage with formulation (T1) is thatbig M makes the LP-relaxation very weak.

Constraints (4.78d) are also known as triangle inequalities, and they enforce thatthe order of jobs in the solution is acyclic. This is ensured by constraints (4.77b)in (T1). Nemhauser and Savelsbergh [1992] shows that the LP relaxation of (T2),even without including the triangle inequalities gives the same optimal value as theoriginal MIP formulation. Therefore, the triangle inequalities need not be included,if we are only interested in solving the LP relaxation of the formulation. Indeed, theLP relaxation of (T2) need not yield an integer feasible solution for both the comple-tion time and linear ordering variables. None-the-less the results in Nemhauser andSavelsbergh [1992] are crucial, and we proceed with the formulation (T2−) (i.e. (T2)minus the triangle inequalities) to develop a suitable sub-problem for the Bendersdecomposition.

Next, we take an approach motivated by the success of Reformulation Lineariza-tion Technique (RLT) in solving dfficult discrete optimization problems (Sherali andAdams [2013]). In essence, RLT takes a problem to a higher dimensional space for


the purpose of obtaining tight bounds. It does so by first reformulating the prob-lem using discrete binary factors. Thereupon, the problem is linearized either usingidempotent identity or use of continuous variables and binding inequalities. Our so-lution approach is slightly different in the sense that our goal here is not to develop aformulation with a tighter LP relaxation starting from a loose original formulation,but rather to obtain a linear formulation starting from an existing non-linear formu-lation. In doing so, we also intend to maintain a tight linear relaxation. As shown inNemhauser and Savelsbergh [1992], the original formulation (T2−) is already knownto be tightest possible w.r.t. its LP relaxation’s objective value. Our goal here isonly to reformulate and linearize it to the extent that we can obtain a new linearformulation (i.e. no quadratic or higher order terms involving variables, p), that stillis as much tight as possible w.r.t. to its LP relaxation.

First, we reformulate (T2−) such that the term, pα is homogenized across all con-straints. We multiply each of the constraint (4.78c) by pk, ∀k ∈ J (once again, thisis unlike RLT, wherein the reformulation step involves multiplication of constraintsby binary variable(s) as against continuous ones). In the linearization step, the termpkαjl is replaced by qkjl, together with incorporating the constraints to linearize thisrelationship using a Big M (in form of pubk , ∀k ∈ J). The resultant LP formulation isas follows (after relaxing the integrality restrictions on the linear ordering variables).

Model T2 :Minimize:

∑j∈J

vjtj (4.79a)

tj ≥ pj +∑l∈J,l 6=j

qllj, j ∈ J (4.79b)

qkjl + qklj = pk, ∀k, j, l ∈ J, j < l (4.79c)

qkjl ≤ pk, ∀k, j, l ∈ J, j 6= l (4.79d)

qkjl ≤ pkαjl, ∀k, j, l ∈ J, j 6= l (4.79e)

qkjl ≥ pk − pk(1− αjl), ∀k, j, l ∈ J, j 6= l (4.79f)

qkjl ≥ 0, ∀k, j, l ∈ J, j 6= l (4.79g)

αjl ∈ {0, 1}, ∀j, l ∈ J, j 6= l (4.79h)

In our preliminary investigation, we found that(T2)

yields higher objective value

for the weighted sum of completion times than (T1), quite comparable to the optimalvalue for a given p and s. Consequently, we rewrite (Tµ

Fis) in the same style as (T2−)


and(T2)

, as follows.

Model T2µFis :

Minimize:∑

<kj>∈Fis

vkjstekjs (4.80a)

tek1j1s ≥ pk1j1s +∑

<k2j2>∈Fis,<k2j2>6=<k1j1>

pk2j2sαk2j2k1j1is, ∀<k1j1> ∈ Fis (4.80b)

αk1j1k2j2is + αk2j2k1j1is = 1, ∀{<k1j1>,<k2j2>}|<k1j1>,<k2j2> ∈ Fis, <k1j1> 6= <k2j2>(4.80c)

αk1j1k2j2is ∈ {0, 1}, ∀<k1j1>,<k2j2> ∈ Fis, <k1j1> 6= <k2j2> (4.80d)

Model T2µ

Fis:

Minimize:∑

<kj>∈Fis

vkjstekjs (4.81a)

tek1j1s ≥ pk1j1s +∑

<k2j2>∈Fis,<k2j2> 6=<k1j1>

qk2j2k2j2k1j1is, ∀<k1j1> ∈ Fis (4.81b)

qk1j1k2j2k3j3is + qk1j1k3j3k2j2is = pk1j1s, ∀<k1j1> ∈ Fis, {<k2j2>,<k3j3>}|<k2j2>,<k3j3> ∈ Fis, <k2j2> 6= <k3j3> (4.81c)

qk1j1k2j2k3j3is ≤ pk1j1s, ∀<k1j1>,<k2j2>,<k3j3> ∈ Fis, <k2j2> 6= <k3j3>(4.81d)

qk1j1k2j2k3j3is ≤ pubk1j1sαk2j2k3j3is, ∀<k1j1>,<k2j2>,<k3j3> ∈ Fis, <k2j2> 6= <k3j3>

(4.81e)

qk1j1k2j2k3j3is ≥ pk1j1s − pubk1j1s(1− αk2j2k3j3is), ∀<k1j1>,<k2j2>,<k3j3> ∈ Fis,<k2j2> 6= <k3j3> (4.81f)

qk1j1k2j2k3j3is ≥ 0, ∀<k1j1>,<k2j2>,<k3j3> ∈ Fis, <k2j2> 6= <k3j3> (4.81g)

αk1j1k2j2is ∈ {0, 1}, ∀<k1j1>,<k2j2> ∈ Fis, <k1j1> 6= <k2j2>, (4.81h)

where, pubkjs = bfixj + bunitj dksmkj,∀i ∈ M,<kj> ∈ Fis, s ∈ S. The reformulation of(T2µBis) and (Tµ

Bis) follows in the similar fashion as follows.

Model T2µBis :

Minimize:∑

<kj>∈Bis

v′

kjste′

kjs (4.82a)


te′

k1j1s≥ pk1j1s +

∑<k2j2>∈Bis,

<k2j2>6=<k1j1>

pk2j2sαk2j2k1j1is, ∀<k1j1> ∈ Bis (4.82b)

αk1j1k2j2is + αk2j2k1j1is = 1, ∀{<k1j1>,<k2j2>}|<k1j1>,<k2j2> ∈ Bis, <k1j1> 6= <k2j2>(4.82c)

αk1j1k2j2is ∈ {0, 1}, ∀<k1j1>,<k2j2> ∈ Bis, <k1j1> 6= <k2j2> (4.82d)

Model T2µ

Bis:

Minimize:∑

<kj>∈Bis

v′

kjste′

kjs (4.83a)

te′

k1j1s≥ pk1j1s +

∑<k2j2>∈Bis,

<k2j2> 6=<k1j1>

qk2j2k2j2k1j1is, ∀<k1j1> ∈ Bis (4.83b)

qk1j1k2j2k3j3is + qk1j1k3j3k2j2is = pk1j1s, ∀<k1j1> ∈ Bis, {<k2j2>,<k3j3>}|<k2j2>,<k3j3> ∈ Bis, <k2j2> 6= <k3j3> (4.83c)

qk1j1k2j2k3j3is ≤ pk1j1s, ∀<k1j1>,<k2j2>,<k3j3> ∈ Bis, <k2j2> 6= <k3j3>(4.83d)

qk1j1k2j2k3j3is ≤ pubk1j1sαk2j2k3j3is, ∀<k1j1>,<k2j2>,<k3j3> ∈ Bis, <k2j2> 6= <k3j3>

(4.83e)

qk1j1k2j2k3j3is ≥ pk1j1s − pubk1j1s(1− αk2j2k3j3is), ∀<k1j1>,<k2j2>,<k3j3> ∈ Bis,

<k2j2> 6= <k3j3> (4.83f)

qk1j1k2j2k3j3is ≥ 0, ∀<k1j1>,<k2j2>,<k3j3> ∈ Bis, <k2j2> 6= <k3j3> (4.83g)

αk1j1k2j2is ∈ {0, 1}, ∀<k1j1>,<k2j2> ∈ Bis, <k1j1> 6= <k2j2>, (4.83h)

The Benders optimality cut derived upon solving(T2

µ

Fis

)and

(T2

µ

Bis

)is given as

follows.


zµFis ≥∑

<kj>∈Fis

βkjspkjs +∑

<k1j1>∈Fis,{<k2j2>,<k3j3>}|<k2j2>,<k3j3>∈Fis,<k3j3> 6=<k2j2>

(γk1j1k2j2k3j3is + δk1j1k2j2k3j3is + δk1j1k3j3k2j2is+

εk1j1k2j2k3j3is + εk1j1k3j3k2j2is) pk1j1s −∑

<k1j1>∈Fis,<k2j2>∈Fis,<k3j3>∈Fis,

<k3j3> 6=<k2j2>

εk1j1k2j2k3j3ispubk1j1s

, ∀i ∈M, s ∈ S

(4.84a)

zµBis ≥∑

<kj>∈Bis

β′

kjspkjs +∑

<k1j1>∈Bis,{<k2j2>,<k3j3>}|<k2j2>,<k3j3>∈Bis,<k3j3> 6=<k2j2>

(γ′

k1j1k2j2k3j3is+ δ

′

k1j1k2j2k3j3is+ δ

′

k1j1k3j3k2j2is+

ε′

k1j1k2j2k3j3is+ ε

′

k1j1k3j3k2j2is

)pk1j1s −

∑<k1j1>∈Bis,<k2j2>∈Bis,<k3j3>∈Bis,

<k3j3>6=<k2j2>

ε′

k1j1k2j2k3j3ispubk1j1s, ∀i ∈M, s ∈ S,

(4.84b)

where β, β′, γ, γ

′, δ, δ

′, ε and ε

′are the dual variables associated with Constraints

(4.81b), (4.83b), (4.81c), (4.83c), (4.81d), (4.83d), (4.81f) and (4.83f), respectively.

They represent the extreme points of the dual feasible region of(T2

µ

Fis

)and

(T2

µ

Bis

),

which is bounded, since the primal sub-problems are always feasible for given p values,obtained from the solution of the Reformulated Master Problem (RMP), (FRµ

RMP ),presented next.

Model FRµRMP :

Minimize:∑j∈J

fass,1j +∑s∈S

ps

(∑k∈K

∑j∈Jk

fass,2kjs +∑j∈J

f invjs

)+∑s∈S

∑k∈K

∑j∈Jk

∑l∈Rj

pkjs

−∑s∈S

∑i∈M

∑<kj>∈Bis

v′

kjs(pubis + pkjs) +

∑s∈S

∑i∈M

(zµFis + zµBis

)(4.85a)

subject to: (4.60c)-(4.60p), (4.84a) and (4.84b)


4.3.3 Algorithm To Solve ATO-CST

We express the Lagrangian dual function, L(µ) = V(FRµ) as L(µ) = min{cTx +µT (b − Ax)|x ∈ X ∩ Zn} = min{cT + µT (b − Ax)|x ∈ H(X ∩ Zn)}, where H(·)represents the convex hull. For the sake of simplicity, we assume that X is bounded,otherwise, we could artificially bound all the decision variables from above. Since Xis a polyhedral set in our problem, the minimization in the Lagrangian relaxationproblem occurs over the extreme points of the convex hull, H(X ∩ Zn), which arefinite and can be written as {x1, . . . xm}. Lagrangian dual function can then berewritten as,

L(µ) = mini=1,...m

{cTxi + µT (b− Axi)}.

Note that the Lagrangian dual function is the lower envelope of the set of hyper-planes, {cTxi+µT (b−Axi)}, ∀i = 1, . . .m. Therefore, L(µ) is a piece-wise linear anda concave function.

Generally, the number of constraints of such a program is exponential. Therefore,we use a gradient descent method to compute the maximum of Lagrangian functionas close as desired to the true Lagrangian dual optimal value.

We further propose to solve a variant of the Lagrangian dual problem, (D), whereinthe inner minimization problem, (FRµ) is replaced by (FRµ

RMP ). We call this variantof Lagrangian dual problem as (D

′) and the corresponding Lagrangian dual function

as L′(µ). Note that for a given value of µ, the optimal objective value of the RMP

estimates that of (FRµ) from below, i.e. ∀µ ≥ 0, Lµ = V(FRµ) ≥ V(FRµRMP ) =

L′(µ). Therefore, we have, V(F) ≥ V(D) = maxµ≥0 Lµ ≥ maxµ≥0 L

′(µ) = V(D

′).

Algorithm 4.1. Algorithm To Solve ATO-CST (Subgradient Method).

1. Set t = 0 and chose µ0 ≥ 0 (we chose µ0 = 0.2 to begin with). Set lower bound,LB = −∞.

2. Compute L′

µt by solving (FRµt

RMP ) in equation (4.85a), wherein the optimalitycuts given by Equations (4.84a) - (4.84b) are added in an iterative manner by

solving

(T2

µt

Fis

)and

(T2

µt

Bis

), until the values of zµ

t

Fisand zµ

t

Bisconverge to

values within a set tolerance gap to that of the objective value of

(T2

µt

Fis

)and


(T2

µt

Bis

), respectively.

3. If we write µt as a vector [µkjls],∀k ∈ K, j ∈ Jk, l ∈ Rj, s ∈ S, then thesubgradient gt of the function L

′is derived as [tekls − tskjs], ∀k ∈ K, j ∈ Jk, l ∈

Rj, s ∈ S (refer to the form of (FRµRMP ) in Equation (4.58)). Here, the values

of the timing variables are obtained from the solution of

(T2

µt

Fis

)and

(T2

µt

Bis

)at the end of RMP iterations in Step 2.

4. If gt = 0, then stop. In that case, the optimal solution to (D′) (as well as the

best lower bound, LB, to the original problem, (F)) is L′(µt) = V(FRµt

RMP ).Otherwise, update the value for the best lower bound as, LB = max{L′(µt), LB}.

5. Compute µt+1 = max{0, µt + θtgt}, where θt denotes the step size. By rightchoice of step size, θt (such that, θt ↓ 0,

∑t θ

t =∞), we can ensure convergenceto the optimal solution of the Lagrangian dual problem (Held et al. [1974]).

6. Increment t and go to Step 2.

Held et al. [1974] proposed the following formula for adapting the step size at the tth

of Algorithm 4.1 as follows.

θt = λtL′ − L′(µt)‖gt‖2 ,

where λt is a scalar satisfying 0 ≤ λt ≤ 2 and L′ is some upper bound on L′(µt). The

sequence λt is determined by setting λ0 = 2, and halving λt whenever L′(µt) fails to

increase in some fixed number of iterations, which we fix at 3.

Note that, there is no way of proving optimality in the subgradient method, unlessa value of L

′(µt) is obtained that is equal to the cost of a known primal feasible

solution. We terminate the method upon reaching a set time limit.

Since the solution obtained at each iteration of the subgradient method to (FRµt

RMP )may not be primal feasible to (F), we will next discuss how to obtain a primalfeasible solution. This is not only important for its practical implications, but alsoin providing a measure of evaluation for Algorithm 4.1.


Recovery Of Primal Feasible Solution

Next, we show how to recover a primal feasible solution to (F) at an iteration t of

Algorithm 4.1 for a given µt. In Step 2 of the algorithm, we solved (FRµt

RMP ). Westart off with this partial solution which is feasible to the original constraints (4.28)-(4.39) and (4.46) - (4.48) in (F). The partial solution contains the values for all theStage 1 and Stage 2 production variables and the processing times for various jobs onthe machines. Having known the values of pkjs, ykjs,∀k ∈ K, j ∈ Jk, s ∈ S, we solvethe scheduling problem(s) over the machines after re-introducing the precedenceconstraints between various jobs. As such, we solve |S| separable MIP problems,(RTs). and complete the recovery of primal feasible solution.

Model RTs:

Minimize:∑s∈S

ps∑k∈K

fdelks (4.86)

tekjs − tskjs = pkjs, ∀k ∈ K, j ∈ Jk (4.87)

tskjs ≥ tekls, ∀k ∈ K, j ∈ Jk, l ∈ Rj (4.88)∑<k2j2>∈Li∪<0,0>,<k2j2>6=<k1j1>


, ∀i ∈M,<k1j1> ∈ Li ∪<0, 0> (4.89)

∑<k2j2>∈Li∪<0,0>,<k2j2>6=<k1j1>


, ∀i ∈M,<k1j1> ∈ Li ∪<0, 0> (4.90)


<k1j1> 6= <k2j2> (4.91)

fdelks = ektekrks

, ∀k ∈ K, s ∈ S (4.92)

tskjs, tekjs ≥ 0, ∀k ∈ K, j ∈ Jk (4.93)

xk1j1k2j2is ∈ {0, 1}, ∀i ∈M, j1 ∈ Li ∪ {0}, k1 ∈ J j1 ,j2 ∈ Li ∪ {0}, j2 6= j1, k2 ∈ J j2 (4.94)

The total cost corresponding to the recovered solution provides an upper bound forthe subgradient method in Algorithm 4.1 as well as to the optimal solution to (F).Note that the key feature of this recovery procedure is that we are building upon

the current solution to Lagrangian relaxation. Essentially, we solved (FRµt

RMP ), and(RTs),∀s ∈ S, by direct application of an MIP solver. The recovery procedure isrelatively less strenuous in terms of time and memory use in comparison to direct


solution of (F). Nonetheless, it is sparingly used within the subgradient method.We do so only when L

′(µt) fails to increase in some fixed number of iterations, and

towards the end of the time limit set for Algorithm 4.1. We also record the solutionand the total cost corresponding to the best solution, i.e. which gives the least costamong all such recovered primal feasible solutions at different iterations within thealgorithm.

4.4 Results

In §4.2.3, we presented the results for Example 4.2.2. This example did not incor-porate the timing aspect of scheduling production. Therein, we presented a solutionfor each of the two methodologies ATO-CS-D and ATO-CS for the the case whenthe products’ demand is deterministic and stochastic, respectively. The preliminaryresults highlighted the advantages of ATO-CS over other commonly used productionmethodologies, MTO, MTS, and ATO in terms of minimization of production costsand inventory loss. In §4.3.1, we completed development of our proposed method-ology, ATO-CST, which is an extension of ATO-CS (having added the feature ofscheduling with timing constraints during production) and fully represents the com-plete problem, SD-AJSSP, studied in this chapter.

In this section, we will first proceed to showcase the superiority of ATO-CST overMTO, MTS, and ATO in terms of total cost. Then, we will demonstrate the effective-ness of Algorithm 4.1 in being able to handle large-sized data-sets for the completeproblem.

Superiority of ATO-CST over MTS, MTO, and ATO. We first present acomparison between MTS, MTO, and ATO analytically in Table 4.14, where wepoint out differences among them in terms of breakup of total cost studied, thatconsists of production cost, inventory cost or loss, and delay cost associated withproducts’ delivery. We further show how these three methodologies can be derivedas special cases of ATO-CST, where the base model for representing ATO-CST is(F). Therefore, ATO-CST is guaranteed to outperform MTS, MTO, and ATO interms of total cost.

We obtain optimal solution to ATO-CST as well as MTS, MTO, and ATO by directlysolving (F) (Table 4.14 shows how MTS, MTO, and ATO are derived from ATO-CST) . Table 4.15 presents the break up of various costs contributing to total costfor a range of parameters - fixed setup cost (afix), unit inventory loss cost (h), and


unit delay cost (e), which are assumed to be uniform across different sub-assembliesand products. These parameters reflect the proportion of costs in the breakup. Thenumber of products, i.e. |K|, number of sub-assemblies, i.e. |J |, number of machinesat Stage 2, i.e. |M |, and the number of scenarios, |S| is fixed at 3, 10, 2, and 6,respectively for all the results. For a fixed value of afix, higher values of h and e areexpected to result into higher cost due to losses in excess inventory and increaseddelay cost of products’ delivery to customers, respectively.

We observe the following trends from the results presented in Table 4.15 which con-form to the expected behavior following the analytic comparison made earlier andin keeping view of the definition of various methodologies concerned: (1) For anygiven combination of afix, h, and e, (a) MTO yields the highest production costs aswell as delay cost of product’s delivery, and (b) MTS yields the highest inventorycost or loss. (2) The delay cost is zero for MTS, whereas the inventory cost is zerofor MTO. (3) The total cost (sum of production cost at both the stages, inventorycost and the delay cost) is lowest for ATO-CST.

Table 4.14: Analytic comparison between MTS, MTO, and ATO and the procedureto represent them as a special case of ATO-CST.

MethodProduction

costInventorycost (loss)

Delay costof products’

delivery

Procedure to represent as aspecial case of ATO-CST

MTS - LargestSmallest

(zero)

No Stage 2 production, i.e.w2kjs = 0,∀k ∈ K, j ∈ Jk, s ∈

S

MTO LargestSmallest

(zero)Largest

No Stage 1 production, i.e.w1j = 0,∀j ∈ J

ATO - - -

No Stage 1 production ofproducts’ corresponding final

sub-assembly, i.e.w1rk

= 0,∀k ∈ K, and noStage 2 production of therest, i.e. w2

kjs = 0,∀k ∈K, j ∈ Jk, j 6= rk, s ∈ S

Performance of Algorithm 4.1 in solving a large-sized data-set.

We now make a comparison of the quality of the solution obtained for ATO-CST,between the direct solution of (F) as an MIP by CPLEX R©and the solution obtainedusing Algorithm 4.1, in the extreme cases when the size of (F) is large (owing tothe increased value of either of its parameters, |K|, |J |, |M | or |S|). As such, wepresent the results for the two methods in Table 4.16 and Table 4.17, respectively.


Tab

le4.

15:

Com

par

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ofM

TS,

MT

O,

AT

O,

and

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O-C

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.

he

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3.67

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6735

3.67

0.00

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380.

3328

3.00

97.3

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2.75

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283.

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330.

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360.6

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The test cases in Table 4.16 and Table 4.17 are characterized by the number of prod-ucts, i.e. |K|, number of sub-assemblies, i.e. |J |, number of machines at Stage 2,i.e. |M |, and the number of scenarios, |S|. Six test cases were generated as such(please refer to Appendix I for the complete data in these test cases). (1) For allthe test cases, the probability for each scenario, ps, is assumed to be equal, i.e.ps = 1/|S|, ∀s ∈ S. (2) For any of the test case, the distribution of the demand,dks, of a product, k, is assumed to be discrete uniform over the set of scenarios.The demand for each product, k, occurs uniformly over the interval, [ak, bk], wherethe values of ak and bk are fixed arbitrarily and differently for each product. (3)The configuration of products’ BOM is the same for test cases numbered 1, 2 and3 (i.e. <|K|, |J |, |M |, |S|> = <8, 40, 3, 9>,<8, 40, 5, 9>,<8, 40, 3, 15>) in both Ta-ble 4.16 and Table 4.17. As such, the values of structural parameters, rk, Rj, and

also the associated cost parameters, afixj , aunitj , bfixj , bunitj , ek, hj are the same for allthese test cases. The difference lies in the number of machines used at Stage 2,and the number of total scenarios considered. (4) The configuration of products’BOM is the same for the test cases numbered 4, 5 and 6 (i.e. <|K|, |J |, |M |, |S|>= <9, 45, 2, 9>,<9, 45, 5, 9>,<9, 45, 2, 15>) in both Table 4.16 and Table 4.17. Assuch, the values of structural parameters, rk, Rj, and also the associated cost param-

eters, afixj , aunitj , bfixj , bunitj , ek, hj are the same for all these test cases. The differencelies in the number of machines used at Stage 2, and the number of total scenariosconsidered. Also, these files have higher order of complexity than the first three.

We obtain the direct solution of (F) in Table 4.16 by using state-of-the-art com-mercial MIP solver CPLEX R©(12.6) running on a single thread, with a 2.6 GHzIntel R©Core i5 processor, and using 8GB DDR3 memory. We also deploy CPLEX R©underthe same platform setting as above to solve various MIP and LP models that aresolved in an iterative manner either called from within Algorithm 4.1 or during the

recovery procedure as described earlier (such as (FRµt

RMP ),

(T2

µt

Fis

),

(T2

µt

Bis

)and

(RTs)). For Step 2 of Algorithm 4.1, tolerance gap is set at 1% for the optimalitygap (in reference to Benders technique). In general, the tolerance gap is set at 1%,prior to solving any MIP directly. As such, direct solution approach will terminatewhen the gap between the best feasible solution value and the best discovered lowerbound (from LP relaxation) becomes lower than this tolerance gap. Algorithm 4.1terminates when µ becomes zero. We also terminate both these methods when themaximum time limit set at 3600.00 secs is reached.

For all test cases presented in Table 4.16 and Table 4.17, we observe that, (1) thevalue of the solution (i.e. ‘Total cost’) is lower for Algorithm 4.1 than that observed


for direct solution method, (2) the value of the ‘Lower bound’ for Algorithm 4.1 ishigher than that for direct solution method, therefore, (3) we obtain smaller gap(%) between upper bound (i.e. ‘Total cost’) and lower bound for Algorithm 4.1.Note that the upper bound reported in these results corresponds to the best primalfeasible solution to (F) for either of the methods, and the lower bound corresponds tothe LP relaxation value at the termination of MIP in case of direct solution method,and the largest Lagrangian dual function value in case of solution by Algorithm 4.1.Also note that, a smaller gap is indicative of higher confidence in concluding hownear a particular solution is to an optimal solution. Following these observations, weconclude that our proposed algorithm is quite effective in solving large-sized data-sets. In-fact, in some test cases, as we note in Table 4.17, a very small gap (for e.g.0.26% for test case numbered 2) is achieved. Also, note that, the maximum gapacross all test cases, is observed to be 10.26% for Algorithm 4.1 as against 36.18%by direct solution method.

Table 4.16: Results for direct solution by CPLEX R©.

Best feasible solution - cost LP relaxation|K|, |J |, |M |, |S| Complexity Time (secs) Total cost Prod I Prod II Prod total Inventory Delay gap (%) Lower bound8, 40, 3, 9 20979 binaries, 4027 generals 3600.00 3656.92 3052.35 282.80 3335.15 21.35 300.42 13.74 3215.068, 40, 5, 9 12780 binaries, 4027 generals 3600.00 3526.49 3043.95 283.62 3327.57 19.02 179.90 4.98 3359.088, 40, 3, 15 34965 binaries, 6685 generals 3600.00 4218.41 3119.60 482.88 3602.48 42.94 572.99 21.58 3469.769, 45, 2, 9 51786 binaries, 4779 generals 3600.00 4898.13 4206.30 188.88 4395.18 54.25 448.70 22.43 4000.899, 45, 5, 9 22113 binaries, 4779 generals 3600.00 4573.34 4017.30 305.43 4322.73 32.67 217.93 6.25 4304.249, 45, 2, 15 86310 binaries, 7935 generals 3600.00 5509.00 4084.56 437.17 4521.73 50.32 936.94 36.18 4045.43

Table 4.17: Results for Algorithm 4.1 (subgradient method).

Best feasible solution - cost Best Lagrangian dual|K|, |J |, |M |, |S| Total iterations Time (secs) Total cost Prod I Prod II Prod total Inventory Delay gap (%) Lower bound8, 40, 3, 9 86 3600.00 3419.44 3058.00 171.33 3229.33 27.44 162.67 5.86 3230.138, 40, 5, 9 87 3600.00 3378.22 3058.00 172.11 3230.11 24.78 123.33 0.24 3370.078, 40, 3, 15 72 3600.00 3704.87 2836.00 395.80 3231.80 35.67 437.40 5.08 3525.829, 45, 2, 9 57 3600.00 4520.22 3956.00 224.00 4180.00 30.00 310.22 7.13 4219.459, 45, 5, 9 59 3600.00 4373.55 3956.00 221.11 4177.11 27.00 169.44 1.23 4320.489, 45, 2, 15 51 3600.00 5029.00 3782.00 390.33 4172.33 41.93 814.73 10.23 4562.24


In this chapter, we proposed a new production methodology, ATO-CST, to solve SD-AJSSP. It is designed to exercise a successful implementation of a mass customizationsystem in production by collectively aiming to (1) keep the production costs low byleveraging upon commonality of sub-assemblies in products’ BOM and by producingsub-assemblies on a mass scale during one of the two stages of production, specifically


Stage 1, (2) minimize the loss due to excess inventory built up in anticipation ofstochastic demand of products by postponement of production of certain apex sub-assemblies in products’ BOM till actual demand realization, and (3) reduce the timeof delivery for products to the customers, in order for the companies that operate inthe arena of Mass Customization to benefit by adopting this methodology.

We accomplished the development of ATO-CST in stages. We first proposed ATO-CS that does not consider the timing aspect of production. We demonstrated its usethrough an example. We also obtained a solution for the example using some otherpopular production methodolgies, namely, MTS, MTO, and ATO. We obtained sat-isfactory preliminary results for the proposed methodology, thereby proceeding toinclude the timing aspect of the problem statement as well (i.e., scheduling of sub-assemblies over machines). The ATO-CST was developed to solve a productionenvironment of a full-fledged assembly job shop with two stages of production. Weformulated ATO-CST as a mathematical model. Then, we designed an algorithmthat can effectively solve this mathematical model for large-sized data-set as well.The algorithm is based on the solution of a Lagrangian dual problem using subgradi-ent method, wherein the inner Lagrangian relaxation problem is further solved usingBenders decomposition technique.

We have also shown analytically that the ATO-CST would always outperform theother three production methodologies in terms of the total cost. Based on theseresults, we have also demonstrated the effectiveness of the proposed algorithm inobtaining a lower cost solution for the test cases pertaining to a large-sized data-setover the direct solution of the mathematical model using state-of-the-art commercialsolver CPLEX R©. For all the test cases considered, the proposed algorithm not onlyout-performed direct solution method in terms of total cost, but also attained a loweroptimality gap.

Chapter 5

Concluding Remarks and Ideas forFuture Work

In this dissertation, we address three problems in the areas of logistics and schedulingthat are encountered in real-life.

The first of these problems pertains to setup and operation of biomass logistics thatis encountered for the production of ethanol from biomass switchgrass. The BiomassLogistics Problem (BLP) addressed in Chapter 2 seeks to determine an optimal mixof biomass harvested from multiple locations, each of which is covered by a Bio-energyPlant (BeP), the selection of production fields for each location and their assignmentto Satellite Storage Locations (SSLs), and the routing of equipment sets among theSSLs, for the objective of minimizing the total annual cost due to the ownership ofequipment sets, fixed setups and land rental cost, and the cost of transportation ofbiomass from the fields to the BePs and bio-crude oil from the BePs to the refinery.We first formulated a mixed-integer model of the problem. To solve it effectively,we devised a decomposition method based on Nested Benders decomposition. Con-sequently, we identified three stages of decomposition. The first stage deals withselection of BePs and the amount of biomass harvested at each of the BeP, selectionof the fields and the SSLs, and assignment of the fields to the SSLs. The remainingmodel constitutes multiple Capacitated Vehicle Routing Problems (CVRPs), and isseparable over individual BePs. For each BeP, the CVRP is further decomposed intostage two and stage three sub-problem, and is solved following the use of standardBenders decomposition method, except for the difference that the sub-problem isan integer program rather than a linear program (note that, a standard Benders

106

Sanchit Singh Chapter 5. Concluding Remarks 107

decomposition approach is limited in application because of the assumption that thesub-pproblem is a linear program with continuous variables). Therefore, we proposeda multi-cut version of optimality cuts, that can capture solution value at an integersolution for the sub-problems. We first investigated operation of a single Bio-energyPlant (BeP) without the consideration of rail transportation cost of bio-crude oilfrom BeP to the nearest refinery. We considered a single bi-annual harvest scenarioand drew comparison among three different types of equipment systems that are usedfor loading and unloading of biomass on to the trailers/trucks at the Satellite StorageLocations (SSLs). The ‘side-loading’ rack system gave the least total cost of 10.23M(and a unit cost of $21.37/Mg), and we chose to conduct the rest of the tests withthis type of equipment system. Then, we made comparison between the performanceof our decomposition scheme with that of a heuristic approach (based on the work ofResop et al. [2011]) to locate SSLs in this region, for a total of six harvest scenarios.Our approach resulted in the least total cost, ranging from $10.23M to $16.38M,compared to $10.77M to $17.96M for the heuristic method. We then presented re-sults for the complete problem by incorporating the rail transportation cost incurredfrom BePs to refinery and considered logistics operations at two locations, one atGretna, VA and the other at Bedford, VA. Each location is dedicated to a singleBeP operation at its center. We replicated a detailed land-use pattern from the Geo-graphic Information System (GIS) data for Gretna, VA to the other location as well.We set up five test cases with increasing amount of total biomass intake required atthe refinery. The maximum amount of ethanol that can be produced by using twoBePs at both the locations is calculated to be 78.1 million gallons (MG), incurringa total cost of $50.13M, which amounts to a unit production cost for ethanol to be$0.64/gallon.

For future work, we propose to: (1) incorporation of more locations such as Keysvillein Virginia and Reidsville and Liberty in North Carolina as part of the Piedmontregion, (2) study of a cost-based model that incorporates tactical and operationaldecisions and deals with the time-related constraints w.r.t. scheduling of pickup anddelivery of biomass by trucks, and (3) investigation of the logistics system proposedto study the Biomass Logistics Problem (BLP) in our work and the Nested Bendersdecomposition scheme developed for its solution to other biomass types, such ascorn-stover, wood chips, and sorghum.

The second problem discussed in Chapter 3 is the 1 + m Two-stage Hybrid FlowShop Lot-streaming Problem (1 +m TSHFS-LSP), which pertains to the schedulingof a lot over a two-stage HFS with one machine at Stage 1 and m parallel andidentical machines at Stage 2, using the lot-streaming concept, for the objective

Sanchit Singh Chapter 5. Concluding Remarks 108

of minimizing the makespan, under a deterministic setting. Such a manufacturingconfiguration is encountered in continuous processing industries such as textile, foodprocessing, chemicals and pharmaceutical, as well as, processing of discrete productssuch as electronics, furniture, and steel industries. We incorporated the conceptof lot-streaming, thereby splitting the production lot into multiple sublots, each ofwhich is processed on machines or transferred over from one machine to the next inan overlapping manner. The novelty of our work is in obtaining an optimal schedulein polynomial time, O(m3 +mn), where n is the specified number of sublots the lotis to be split into for scheduling over m machines at Stage 2, for the case when thesublot sizes are relaxed to be continuous. We also presented a pseudo-polynomialtime algorithm of time complexity, O(m3 + mn) (where n is the maximum numberof sublots given as a problem’s parameter), to determine optimal number of sublotsand optimal schedule for processing sublots on the Stage 2 machines, for the case ofcontinuous sublot sizes. We then developed a branch-and-bound-based method tosolve the problem for the case when the sublot sizes are discrete. The method is aheuristic approach that relies on the use of a tight lower bound on makespan owingto the use of results from the continuous case. Its performance is evaluated againstthat of the direct solution of a mixed-integer model formulation of 1+m TSHFS-LSPby CPLEX R©. Our method obtained solution within 0.10% of the optimality gap in83 out of 90 instances, and required an average of 315.00 secs before termination,whereas the direct solution method obtained solution within 0.10% of the optimalitygap in only 40 out of 90 instances, and required as much as 1095.00 seconds on anaverage before termination.

Some potential areas for future work include: (1) consideration of Problem 1 + mTSHFS-LSP (N) for a single-lot with m non-identical machines at Stage 2, i.e., theyhave different unit processing times, p1 ≤ p2 ≤, . . . pm, (2) consideration of Problem1+m TSHFS-LSP (A) for a single-lot with m identical machines at Stage 2 that havedifferent earliest availability times, i.e., 0 ≤ a1 ≤ a2 ≤ · · · am, and (3) consideration ofProblem 1+m TSHFS-LSP (M) for multiple lots. Other configurations of HFSs thatdeserve attention in the context of scheduling with an application of lot-streaming arem1 +m2 two-stage HFS and m1 +m2 +m3 three-stage HFS, where m1,m2,m3 ≥ 1,in general.

The third problem discussed in Chapter 4 is the Stochastic Demand AssemblyJob Shop Scheduling Problem (SD-AJSSP) in the presence of commonality of sub-assemblies across the products. We proposed a new production methodology, named,Assemble-To-Order with Commonality of Sub-assemblies and Timing Aspect (ATO-CST) to not only solve SD-AJSSP, but that which is designed to achieve a successful

Sanchit Singh 109

implementation of a Mass Customization system with a focus on production aspectby collectively aiming to (1) keep the production costs low by leveraging upon com-monality of sub-assemblies in products’ BOM and by producing sub-assemblies on amass scale during one of the two stages of production, specifically Stage 1, (2) min-imize the loss due to excess inventory built-up in anticipation of stochastic demandof products by postponement of production of certain apex sub-assemblies in prod-ucts’ BOM until the actual demand is realized, and (3) reduce the time of products’delivery to the customers. ATO-CST determines the optimum levels of productiontogether with scheduling assembly operations/jobs over the machines at each stageof production, where the second stage is a full-fledged assembly job shop. We showedthat ATO-CST outperforms the other three production methodologies, namely Make-To-Stock (MTS), Make-To-Order (MTO), and Assemble-To-Order (ATO) to solvethe SD-AJSSP, from the viewpoint of total cost. We formulated ATO-CST as amathematical model and further designed an algorithm that can effectively solvethis mathematical model for large-sized data-set. The proposed algorithm is basedon the solution of a Lagrangian dual problem using subgradient method. We alsodemonstrated the effectiveness of the proposed algorithm in obtaining a lower costsolution for the test cases pertaining to a large-sized data-set over the direct solutionof the mathematical model using the state-of-the-art commercial solver CPLEX R©.For all the test cases considered, the proposed algorithm not only out-performeddirect solution method from the viewpoint of total cost, but also attained a loweroptimality gap.

Some potential areas for future work pertain to extending the scope of SD-AJSSP to:(1) consider a representation of the assembly structure of products that is based onidentifying the type of an operation (which is either dedicated to one sub-assemblyonly or related to assembly of different sub-assemblies) together with identifyingthe type of a sub-assembly, (2) incorporate different assembly time and cost modelswhich are not necessarily based on setup and/or are linear with the number of itemsconsidered, and (3) a multi-period or multi-stage stochastic model.

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Appendices

118

Appendix A

Proof of Theorem 3.1

Proof.Case A1. n ≤ m. We prove the result by contradiction. Let sublot sizes, si,∀i =1, . . . , n be obtained using Theorem 3.1 (in particular (3.14) and (3.15)). Note thatonly n machines are required at Stage 2 to schedule n sublots, where each of then machines processes only a single sublot immediately upon its transfer from Stage1. We denote the makespan value corresponding to this schedule by MC . Thecompletion times of the ith sublot (from the end) on Stage 1 and Stage 2 machinesare denoted by C1,i and C2,i,i respectively, assuming w.l.o.g. that it gets assigned tomachine i on Stage 2. For i = 2, . . . , n, we have,

C2,i,i = C1,i + psi

= (C1,i−1 − (si−1 − t)) + psi

= (C1,i−1 − (si−1 − t)) + p(ρsi−1 + t/p)

= C1,i−1 + psi−1

= C2,i−1,i−1

Thus, the completion times at Stage 2 machines are the same for all the sublots, andwe have MC = C2,i,i,∀i = 1, . . . n.Now, suppose that this schedule is not optimal. Therefore, there exists anotherschedule that gives a lower makespan value than MC . The size of the ith sublot(from the end) of this schedule, ˆsi, is as follows: ˆsi = si + xi, ∀, i = 1, . . . , n with∑

1≤i≤n xi = 0 =⇒∑

1≤i≤n ˆsi =∑

1≤i≤n si = U , with not all xi’s being zero. Onceagain, we can continue to use only n machines at Stage 2 instead of m to obtain lowest

119

Sanchit Singh Appendix A. Proof of Theorem 3.1 120

possible makespan value for this schedule, where each of the n machines processesonly a single sublot immediately upon its transfer from Stage 1. The makespan valuecorresponding to this schedule is denoted by MC . The completion times of the ith

sublot (from the end) on the Stage 1 and Stage 2 machines are denoted by ˆC1,i andˆC2,i,i respectively, assuming w.l.o.g. that it gets assigned to machine i at Stage 2.Then, ∀i = 1, . . . , n, we have,

ˆC2,i,i = ˆC1,i + pˆsi

= (ˆsn + ˆsn−1 + . . .+ ˆsi + (n− i+ 1)t) + pˆsi

= C1,i + (xn + xn−1 + . . .+ xi) + psi + pxi

= MC + (xn + xn−1 + . . .+ xi) + pxi

= MC + ∆i

where, ∆i = pxi + (xi + . . .+ xn)

=

{px1 if i = 1

pxi − (x1 + . . .+ xi−1) if 2 ≤ i ≤ n

and, MC = max1≤i≤n

( ˆC2,i,i)

= max1≤i≤n

(MC + ∆i)

According to the claim made earlier, we have MC < MC . Therefore, ∆i ≤ 0,∀i =1, . . . , n, which implies, ∆1 ≤ 0, and therefore x1 ≤ 0. Similarly, ∆2 ≤ 0 =⇒px2 − (x1) ≤ 0 =⇒ x2 ≤ 0, and so on. Thus, xi ≤ 0,∀i = 1, . . . , n. Together withthe condition that

∑1≤i≤n xi = 0, we obtain xi = 0,∀i = 1, . . . , n, thus violating the

condition that not all of the xi’s are zero.

Case A2. n ≥ m. We prove this case by induction. Let r (≥ m) be an indexfor number of sublots. We show that if the schedule obtained by using Theorem3.1 for n = r or lower is optimal (note that we have already proved the base casewith n = r = m), then the schedule specified by Theorem 3.1 provides an optimalschedule for n = r + 1 is optimal as well, thus completing the proof for this case.Let si,∀i = 1, . . . , r + 1 be the sublot sizes obtained by using Theorem 3.1 forn = r + 1. Suppose there exists another schedule for r + 1 number of sublots thatis optimal. Let the sublot sizes of this schedule be denoted by ˆsi), and they aresuch that not all ˆsi = si,∀i = 1, . . . , r + 1. It is noteworthy that the conditionˆsr+1 6= sr+1, i.e., the first sublots (recall the sublots are numbered backwards) are ofdifferent sizes is not only a sufficient condition for the two schedules to be different,


but also a necessary condition, because if they are equal, we could continue to splitthe remaining lot of size U − ˆsr+1 = U − sr+1 into r sublots in the claimed optimalschedule, considering it started its processing on Stage 1 at time ˆsr+1 + t = sr+1 + twith the goal of minimizing the makespan value. In accordance with the inductionstep 1, we can use Theorem 3.1 to obtain an optimal schedule for a lot to be split intor sublots. Thus, we will obtain sizes of the remaining r sublots using Theorem 3.1for the claimed optimal schedule, which would be exactly the same as in the givenschedule. Therefore, let the size of the first sublot of the claimed optimal schedule,ˆsr+1 = sr+1 + x with x 6= 0. So far, our claim has been that the claimed optimalschedule can provide a lower makespan value than that for the schedule obtainedusing Theorem 3.1. We will show this to be untrue, thus completing the overallproof. Upon fixing the size of the (r + 1)th sublot from the end, the leftover lot sizeis U− ˆsr+1 which we denote by ur. Depending upon the sign of x, we have two cases.

Case A2.1 x < 0.(see Figure A.1) Once we have fixed size of the first sublot, wesplit the leftover lot of size ur (=U− ˆsr+1) into r sublots so as to minimize makespanfor the overall schedule. The optimal makespan value for a lot of size U available attime T = 0, and split into r + 1 sublots is the same as the optimal makespan valuefor a lot of size ur available at time T = ˆsr+1 + t = sr+1 + x + t and split into rsublots. The solution for ˆsi,∀i = 1, . . . , r can be obtained based on the assumptionmade in induction step 1, i.e. using Theorem 3.1 provided there is no overlap (δ1)at Stage 2 machines, i.e.,

δ1 = ˆsr + . . .+ ˆsr+1−m +mt− pˆsr+1

= ˆsr + . . .+ ˆsr+1−m +mt− p(sr+1 + x)

= (ˆsr − sr) + . . .+ (ˆsr+1−m − sr+1−m)− px [by (3.17)]

> 0

Before we proceed further, it is necessary to make the following remark, which willbe used in the relevant context in the proof.

Remark A.1. Note that, here, we use the result of Theorem 3.2 according to whichthe sublot sizes increase (decrease) with increment (decrement) in lot size. However,the proof of Theorem 3.2 relies on the validity of Theorem 3.1 to hold true for n ≤ rin our induction step, therefore, we use the result of Theorem 3.2 whenever n ≤ r.

Since x < 0, it follows ur > ur, and we have, ˆsi > si,∀i = 1, . . . , r by Theorem3.2 (Remark A.1). Hence, overlap δ1 > 0. Next, we will show that the changein makespan value for this schedule, δM (shown in Figure A.1(ii)), is positive thus


1rs rs 1s

1rps

rps

1ps

x

1r mps

1

ru rt

( 1)U r t

M

1 1ˆ ˆ(ii) , 0; 1, ,

ˆspecified by Theorem 1 with replaced by

r r i

r

s s x x s i ran

u

d

u

(i)

1rs x

1( )rp s x

0T

Sublots obtained using Theorem 1

Sublots not necessarily obtained using Theorem 1

2r ms

2r mps

1r ms

ˆrs 1s2

ˆr ms 1

ˆr ms

ˆru rt

ˆrps

2ˆk mps

1ˆr mps 1

ˆps

(i) Lot of size , split into 1sublots

speci

,

1, 1, fied by Theorem 1i

U r

s i r

(ii)

Figure A.1: Proof of Theorem 3.1, Case A2.1: A schedule where ˆsr+1 = sr+1 +x, x <0.


1rs rs

1rps

rps

x

ru rt

( 1)U r t

1 1ˆ(ii) , 0,

ˆ

x-(y ) ; r z r+2-

,

m

2 ,

0

r r

i i i

r z

s s x x

s s y i r

y

r m


specified by Theorem

,

1, 1, 1i

U r

s i r

1rs x

1( )rp s x

0T

2r ms

2r mps

ˆrs 2

ˆr ms

ˆru rt

ˆrps

2ˆr mps

x ry

2x )( r r my y

1 (ˆ 1 )r m r m tu

1I

2I

mI

1I

1I

ˆmI



(i)

(ii)

Figure A.2: Proof of Theorem 3.1, Case A2.2.1: A schedule where ˆsr+1 = sr+1+x, x >0, x− (vr + . . .+ vz) ≥ 0,∀r ≤ z ≤ r + 2−m.

contradicting the claim that it can provide a lower makespan value compared withthat for the schedule obtained using Theorem 3.1.The completion time of all the sublots at Stage 1 is the same being equal to U+(r+1)tfor both the cases (see Figure A.1(i) and Figure A.1(ii)); therefore, a change in themakespan value is given by the difference in the processing time of the first sublot(from the end), i.e. δM = p(ˆs1 − s1) > 0 by Theorem 3.2 (Remark A.1).

Case A2.2 x > 0. Apart from fixing size of the first sublot, we also fix sizesof the next m − 1 sublots, i.e. let ˆsi = si − vi,∀i = r, . . . , r + 2 − m, such that(1) ˆsi > 0 =⇒ vi < si, ∀i = r, . . . , r + 2 − m and (2) ˆsr + . . . + ˆsr+2−m <ur =⇒ vr + . . .+ vr+2−m > (sr + . . .+ sr+2−m)− ur. Depending upon the values ofvi,∀i = r, . . . , r + 2−m, we have the following two cases.


Case A2.2.1 x > 0, x − (vr + . . . + vz) ≥ 0,∀r ≤ z ≤ r + 2 −m (see Figure A.2).Note that the makespan is comprised of the idle time and total processing times onthe machines at Stage 2. However, the latter is constant for all the schedules onthe machines at Stage 2. Hence, we only need to show that, for the claimed optimalschedule, the total idle time on the machines at Stage 2 (Figure A.2(ii)) has increasedover that for the schedule with r+1 sublots obtained by Theorem 3.1 (Figure A.2(i)).

Let Îj,∀j = 1, . . . ,m be the total idle time encountered on machine j at Stage 2 forthe schedule in Figure A.2(ii), and similarly, Ij,∀j = 1, . . . ,m be the idle times forthe schedule in Figure A.2(i). We have,

Î1 = I1 + x > I1, and Îj ≥ Ij + x− (vr + . . .+ vr+2−j) ≥ Ij, ∀j = 2, . . . ,m


1rs rs

1rps

rps

x

( 1)U r t

1 1

1 1

ˆ ˆ(ii) , 0, , 2 ; :

ˆ0 1, 1 solved to provide

an optimal mak

, r z r+2-

espan,

m,

x-(y )

besides preventing an overlap at Stage 2 with other

;

sublots ot(n

r r i i i

r z z z i

s s x x s s y i r r m z

u u s r my i

necessarily provided by Theorem 1)


specified by The

,

1, 1 r , o em 1i

r

s r

U

i

1rs x

1( )rp s x

2r ms

zps

ˆrs 2

ˆr ms

ˆrps

1 (ˆ 1)z zu t

zs

2r mps

ˆzs

( ) xr zy y x ry

1 ( 1)zu z t

ˆzps

2ˆr mps

1rs x

1( )rp s x

0T

ˆrs

ˆrps

1 (ˆ 1)z zu t

ˆzs

ˆzps

2ˆr mps

1ˆzs 2

ˆzs

M

1ˆzps

2ˆzps

1

1

ˆ(iii) Optimal makespan resulting from a lot of size available at

ˆ ˆ split into z-1 sublots

ˆ(size , 1, 1), obtained using Theorem 1; it provides a lower

bound

,

( 2 ) ,

on t

z

r z

i

u

T s s

s i

r

z

z t

he makespan claimed optimal in (ii)

1s

1ps

1ˆs

1ˆ

ps



(i)

(ii)

(iii)

Figure A.3: Proof of Theorem 3.1, Case A2.2.2: A schedule where ˆsr+1 = sr+1+x, x >0, x− (vr + . . .+ vz) < 0, for atleast one value of z, r ≤ z ≤ r + 2−m.


Case A2.2.2 x > 0, x − (vr + . . . + vz) < 0, for atleast one value of z, r ≤ z ≤r + 2 −m. (see Figure A.3). Let uz−1 = ˆsz−1 + . . . + ˆs1 > sz−1 + . . . + s1 = uz−1.The partial lot of size uz−1 (> uz−1) starts its processing on Stage 1 at time T =ˆsr+1 + . . . + ˆsz + (r + 2 − z)t, and finishes at time T = U + (r + 1)t (the samefinish time as that for the schedule obtained using Theorem 3.1). It needs to be splitinto z − 1 sublots so as to minimize makespan for the overall schedule. The optimalmakespan value for a lot of size U available at time T = 0, and split into r + 1sublots (with sublots z to r + 1, from the end, fixed in size and scheduled over bothstages) is the same as the optimal makespan value for a lot of size uz−1 available attime T = ˆsr+1 + . . . + ˆsz + (r + 2 − z)t and split into z − 1 sublots. The solutionfor ˆsi,∀i = 1, . . . , z − 1 can be obtained based on the assumption made in inductionstep 1, i.e. using Theorem 3.1 (since n = z − 1 ≤ r). But using this solution couldresult into an overlap among sublots at Stage 2 considering that r + 2 − z sublotshave already been fixed and scheduled over both stages prior to implementing it.Nonetheless, the makespan value resulting from this overall schedule (possibly aninfeasible one due to overlap, see Figure A.3(iii)) will provide a lower bound onthe true lowest makespan value for the claimed optimal schedule in Figure A.3(ii).Hence, it is sufficient to show that the makespan for the schedule in Figure A.3(iii) is

greater than that in Figure A.3(i). And, indeed, the difference, δM = p(ˆs1− s1) > 0,

since uz−1 > uz−1, which implies, ˆs1 > s1 (Remark A.1).

Appendix B

Determination of s1

B.1 Determination of s1, case: p 6= m

Let yi,k be the coefficient of sk and γi be a constant term, when si,∀i = 1, . . . , n, iswritten as a linear combination of sk,∀k = 1, . . . ,m− 1, i.e.,

si = yi,1s1 + . . .+ yi,m−1sm−1 + γi (B.1)

We can determine the values of yi,k and γi, ∀i = 1, . . . ,m − 1,∀k = 1, . . . ,m − 1 asfollows. By (3.16), we have, sm = (1+1/p)s1 +(1/p)s2 + . . .+(1/p)sm−1 +(m−1)t/p,which gives us the values of ym,k,∀k = 1, . . . ,m − 1, and γm. We write the initialvalues of yi,k and γi,∀i = 1, . . . ,m, ∀k = 1, . . . ,m− 1,

y1,1 y1,2 . . . y1,m−1

y2,1 y2,2 . . . y2,m−1...

.... . .

...ym−1,1 ym−1,2 . . . ym−1,m−1

ym,1 ym,2 . . . ym,m−1

=

1

1. . .

1ρ 1/p . . . 1/p

(B.2)

[γ1 γ2 . . . γm−1 γm

]=[0 0 . . . 0 (m− 1)t/p

](B.3)

By (3.17), we have, ∀i = m+ 1, . . . , n,

si = (si−1 + . . .+ si−m)/p+mt/p

127

Sanchit Singh Appendix B. Determination of s1 128

= ((yi−1,1s1 + . . .+ yi−1,m−1sm−1 + γi−1) + . . .+ (yi−m,1s1 + . . .+ yi−m,m−1sm−1 + γi−m))/p+mt/p

= ((yi−1,1 + . . .+ yi−m,1)/p)s1 + . . .+ ((yi−1,m−1 + . . .+ yi−m,m−1)/p)sm−1

+ (γi−1 + . . .+ γi−m)/p+mt/p

= yi,1s1 + . . .+ yi,m−1sm−1 + γi,

where,

yi,k = (yi−1,k + . . .+ yi−m,k)/p, ∀k = 1, . . . ,m− 1, and, (B.4)

γi = (γi−1 + . . .+ γi−m)/p+mt/p (B.5)

Let,

σi,k =i∑

j=1

yj,k, ∀i = 1, . . . , n, ∀k = 1, . . . ,m− 1 (B.6)

We have, ∀i = m+ 1, . . . , n, ∀k = 1, . . . ,m− 1,

σi,k − (y1,k + y2,k + . . .+ ym,k)

= ym+1,k + ym+2,k + . . .+ yi,k

= (y1,k + y2,k + . . .+ ym,k)/p+ (y2,k + y3,k + . . .+ ym+1,k)/p+ . . .

+ (yi−m,k + yi−m+1,k + . . .+ yi−1,k)/p [from (B.4)]

= (σi−m,k + (σi−m+1,k − y1,k) + . . .+ (σi−1,k − (y1,k + y2,k + . . .+ ym−1,k)))/p

= (σi−1,k + σi−2,k + . . .+ σi−m,k)/p− ((m− 1)y1,k + (m− 2)y2,k + . . .+ (1)ym−1,k)/p

=⇒ σi,k = (σi−1,k + . . .+ σi−m,k)/p+ ak, (B.7)

where, ak =

{(2p+ 2−m)/p for k = 1

(p+ 1−m+ k)/p for k ≥ 2(B.8)

Let,

βi,k = σi,k + bk, ∀i = 1, . . . , n, ∀k = 1, . . . ,m− 1 (B.9)

By (B.7) and (B.9), we have,


βi,k = (βi−1,k + . . .+ βi−m,k)/p, ∀i = m+ 1, . . . , n, ∀k = 1, . . . ,m− 1, and,(B.10)

bk = akp/(m− p) (B.11)

By (B.2), (B.6) and (B.9), we can write the initial values of βi,k, ∀i = 1, . . . ,m, ∀k =1, . . . ,m− 1,

β1,1 β1,2 . . . β1,m−1

β2,1 β2,2 . . . β2,m−1...

.... . .

...βm−1,1 βm−1,2 . . . βm−1,m−1

βm,1 βm,2 . . . βm,m−1

=

1 + b1

1 + b1 1 + b2...

.... . .

1 + b1 1 + b2 . . . 1 + bm−1

ρ+ 1 + b1 ρ+ b2 . . . ρ+ bm−1

(B.12)

We designate the above matrix on the left as Bm,m−1. Consider the constant term in(B.1). Let,

τi = γi + c, ∀i = 1, . . . , n (B.13)


τi = (τi−1 + . . .+ τi−m)/p, ∀i = m+ 1, . . . , n, and, (B.14)

c = mt/(m− p) (B.15)

Let,

πi =i∑

j=1

τj, ∀i = 1, . . . , n (B.16)

We have, ∀i = m+ 1, . . . , n,

πi − (τ1 + τ2 + . . .+ τm)

= τm+1 + τm+2 + . . .+ τi

= (τ1 + τ2 + . . .+ τm)/p+ (τ2 + τ3 + . . .+ τm+1)/p+ . . .

+ (τi−m + τi−m+1 + . . .+ τi−1)/p [from (B.14)]


= (πi−m + (πi−m+1 − τ1) + . . .+ (πi−1 − (τ1 + τ2 + . . .+ τm−1)))/p

= (πi−1 + πi−2 + . . .+ πi−m)/p− ((m− 1)τ1 + (m− 2)τ2 + . . .+ (1)τm−1)/p

=⇒ πi = (πi−1 + . . .+ πi−m)/p+ d, (B.17)

where, d = (m−m(m− 1)/(2p))c+ (m− 1)t/p (B.18)

Let,

ωi = πi + e, ∀i = 1, . . . , n (B.19)


ωi = (ωi−1 + . . .+ ωi−m)/p, ∀i = m+ 1, . . . , n, and, (B.20)

e = pd/(m− p) (B.21)

By (B.3), (B.13), (B.16) and (B.19), we can write the initial values of ωi,∀i =1, . . . ,m,[ω1 ω2 . . . ωm−1 ωm

]=[c+ e 2c+ e . . . (m− 1)c+ e (m− 1)t/p+mc+ e

](B.22)

We designate the vector on the left side as Wm. By (3.18) and (B.1), we have,

n∑i=1

yi,1s1 +n∑i=1

yi,2s2 + . . .+n∑i=1

yi,m−1sm−1 +n∑i=1

γi = U, (B.23)

and, using (3.27), we have,

s1 =U −

(∑ni=1 γi + t

(∑m−1k=2 (

∑ni=1 yi,k)(ρ

k−1 − 1)))∑m−1

k=1 (∑n

i=1 yi,k)ρk−1

(B.24)

By (B.6), (B.9), (B.13), (B.16) and (B.19), we have,

s1 =U −

((ωn − e− nc) + t

(∑m−1k=2 (βn,k − bk)(ρk−1 − 1)

))∑m−1k=1 (βn,k − bk)ρk−1

(B.25)


The values of βn,k and ωn, ∀k = 1, . . . ,m − 1, are determined using a recurrencesequence that is presented next.

B.1.1 Linear homogenous recurrence sequence with constantcoefficients

An order d linear homogeneous recurrence sequence with constant coefficients is arelation of the form,

ai = c1ai−1 + c2ai−2 + . . .+ cdai−d, (B.26)

where coefficients cj, ∀j = 1, . . . , d are constants, and i ≥ d+ 1. There are d degreesof freedom for the above sequence, i.e., the initial values a1, . . . , ad can be assumedto take any values. Then, the linear recurrence determines the sequence uniquely.The coefficients cj yield the characteristic polynomial (or ”auxiliary polynomial”),

p(t) = td − c1td−1 − c2t

d−2 − . . .− cd (B.27)

The recurrence is stable (in the sense that the iterates converge asymptotically toa fixed value) if and only if the roots of the characteristic equation, whether realor complex, are all less than unity in absolute value. Then, we can write ai as acombination of all the distinct roots rk (k = 1, . . . , λ , with qk being the multiplicity ofroot rk, such that q1 + q2 + . . .+ qλ = d) as shown below with d unknown coefficients,kl, ∀l = 1, . . . , d,

ai = (k1 + k2i+ . . .+ kq1iq1−1)ri1 + (kq1+1 + kq1+2i+ . . .+ kq1+q2i

q2−1)ri2

+ . . .+ (kd−qλ+1 + kd−qλ+2i+ . . .+ kdiqλ−1)riλ (B.28)

The coefficients kl are determined in order to fit the initial values of the sequence.Say, we have the initial values ai,∀i = 1, . . . , d. Then, we can write the following,


[a1 a2 . . . ad

]=

[k1 k2 . . . kd

]

r1 r21 . . . rd1

......

. . ....

r1 2q1−1r21 . . . dq1−1rd1

......

. . ....

rλ r2λ . . . rdλ

......

. . ....

rλ 2qλ−1r2λ . . . dqλ−1rdλ

,

or, when the roots are all distinct,[a1 a2 . . . ad

]=

[k1 k2 . . . kd

] r1 r2

1 . . . rd1r2 r2

2 . . . rd2...

.... . .

...rd r2

d . . . rdd

We designate the vector on the left side as Ad, the one in middle as Kd, and thematrix on the right side as Rd,d (regardless of whether the roots are all distinct ornot). Then, we have,

Kd = Ad(Rd,d)−, (B.29)

where (.)− represents inverse of a matrix inside parenthesis. A general ith (i ≥ 1)term of the sequence can now be written as,

ai = KdRid = Ad(Rd,d)

−Rid, (B.30)

where Rid is either

[ri1 . . . iq1−1ri1 . . . riλ . . . iqλ−1riλ

]Tor[ri1 ri2 . . . rid

]T, de-

pending upon whether the roots are all distinct or not.

B.1.2 Determination of βn,k and ωn, ∀k = 1, . . . ,m− 1

For each k = 1, . . . ,m−1, Equation (B.10) takes the same form as (B.26), where theorder of recurrence d = m, and coefficients cj = 1/p, ∀j = 1, . . . ,m. We construct amatrix Rm,m (analogous to Rd,d in §B.1.1) from the roots of (B.27), with the values of


d and cj as set above. Note, that, Rm,m is a function of p and m alone, and remainsunchanged with different k values, although the initial values in the βik sequence dochange with different k values. We designate matrix Bkm (analogous to Ad in §B.1.1)as a row vector of size m with initial set of values βki ,∀i = 1, . . . ,m, for each kvalue. As done in §B.1.1, we designate matrix Kk

m (analogous to Kd in §B.1.1) as arow vector of size m with unknown coefficients, kkl ,∀l = 1, . . . ,m, for each k value.Then, we can write,

Kkm = Bkm(Rm,m)−, ∀k = 1, . . . ,m− 1, (B.31)

thus obtaining nth term βn,k in the kth recurrence sequence as,

βn,k = KkmRn

m = Bkm((Rm,m)−Rnm), ∀k = 1, . . . ,m− 1, (B.32)

where matrix Rnm (analogous to Ri

d, for i = n, in §B.1.1) is common for all k values.Note that, the expression above in the outer parenthesis is independent of k value,so we can also write the above in a more concise manner as,[

βn,1 βn,2 . . . βn,m−1

]T= (Bm,m−1)T ((Rm,m)−Rn

m), (B.33)

where Bm,m−1 is shown in (B.12), and (.)T represents the transpose of a matrix insideparenthesis. The procedure to obtain ωn is similar to that for βn,k, since ωi followsthe same recurrence relationship (B.20) as in case for βi,k (B.10). We have,

ωn = Wm((Rm,m)−Rnm), (B.34)

where Wm is as shown in (B.22).

B.2 Determination of s1, case: p = m

Note that, (B.9) in conjugation with (B.7) leads to (B.10) and (B.11). However, thedenominator of (B.11) is (m − p), and hence, we cannot use the relationship whenp = m. Therefore, let,

β′

i,k = σi,k + ib′

k, ∀i = 1, . . . , n, ∀k = 1, . . . ,m− 1 (B.35)


Now, by (B.7) and (B.35), we have,

β′

i,k = (β′

i−1,k + . . .+ β′

i−m,k)/p, ∀i = m+ 1, . . . , n, ∀k = 1, . . . ,m− 1, and,

(B.36)

b′

k = −2ak/(m+ 1) (B.37)

Proceeding in a similar manner as before, by (B.2), (B.6) and (B.35), we can writethe initial values of β

′

i,k,∀i = 1, . . . ,m, ∀k = 1, . . . ,m− 1,β′1,1 β

′1,2 . . . β

′1,m−1

β′2,1 β

′2,2 . . . β

′2,m−1

......

. . ....

β′m−1,1 β

′m−1,2 . . . β

′m−1,m−1

β′m,1 β

′m,2 . . . β

′m,m−1

=

1 + b

′1

1 + 2b′1 1 + 2b

′2

......

. . .

1 + (m− 1)b′1 1 + (m− 1)b

′2 . . . 1 + (m− 1)b

′m−1

ρ+ 1 +mb′1 ρ+mb

′2 . . . ρ+mb

′m−1

(B.38)

We designate the above matrix on the left side as B′m,m−1. To determine the constantterm in (B.1), let,

τ′

i = γi + ic′, ∀i = 1, . . . , n (B.39)


τ′

i = (τ′

i−1 + . . .+ τ′

i−m)/p, ∀i = m+ 1, . . . , n, and, (B.40)

c′= −2t/(m+ 1) (B.41)

Let,

π′

i =i∑

j=1

τ′

j , ∀i = 1, . . . , n (B.42)

We have, ∀i = m+ 1, . . . , n,

π′

i − (τ′

1 + τ′

2 + . . .+ τ′

m)

= τ′

m+1 + τ′

m+2 + . . .+ τ′

i


= (τ′

1 + τ′

2 + . . .+ τ′

m)/p+ (τ′

2 + τ′

3 + . . .+ τ′

m+1)/p+ . . .

+ (τ′

i−m + τ′

i−m+1 + . . .+ τ′

i−1)/p [from (B.40)]

= (π′

i−m + (π′

i−m+1 − τ′

1) + . . .+ (π′

i−1 − (τ′

1 + τ′

2 + . . .+ τ′

m−1)))/p

= (π′

i−1 + π′

i−2 + . . .+ π′

i−m)/p− ((m− 1)τ′

1 + (m− 2)τ′

2 + . . .+ (1)τ′

m−1)/p

=⇒ π′

i = (π′

i−1 + . . .+ π′

i−m)/p+ d′, (B.43)

where, d′= ((m− 1)/m+m(m− 4)/3)t (B.44)

Let,

ω′

i = π′

i + ie′, ∀i = 1, . . . , n (B.45)

By (B.43) and (B.45), we have,ω′

i = (ω′

i−1 + . . .+ ω′

i−m)/p, ∀i = m+ 1, . . . , n, and, (B.46)

e′= −2d

′/(m+ 1) (B.47)

By (B.3), (B.39), (B.42) and (B.45), the initial values of ω′i,∀i = 1, . . . ,m, are as

follows.[ω′1 ω

′2 . . . ω

′m−1 ω

′m

]=[c′+ e

′3c′+ 2e

′. . . m(m− 1)c

′/2 + (m− 1)e

′(m− 1)t/p+m(m+ 1)c

′/2 +me

′](B.48)

We designate the above vector as W′m. In view of (B.24), and using (B.6), (B.35),

(B.39), (B.42) and (B.45), we have,

s1 =U −

((ω′n − ne

′ − n(n+ 1)c′/2) + t

(∑m−1k=2 (β

′

n,k − nb′

k)(ρk−1 − 1)

))∑m−1k=1 (β

′n,k − nb

′k)ρ

k−1(B.49)

Following (B.33) and (B.34), we have,[β′n,1 β

′n,2 . . . β

′n,m−1

]T= (B′m,m−1)

T((Rm,m)−Rn

m), and, (B.50)

ω′

n = W′

m((Rm,m)−Rnm), (B.51)

where B′m,m−1 and W′m are given by (B.38) and (B.48) respectively.

Appendix C

Proof of Corollary 3.1

Proof. We present the proof of this corollary only for the second part of Theorem3.1 since the proof for the first part is similar. Since (ρi−1 − 1) > 0,∀i > 2, itfollows by (3.27) that if s1 ≥ 0, we will also have si > 0,∀i = 2, . . . ,m. Sublot sizessm+1, . . . , sn can be obtained from repeated application of (3.17), and by inductionproperty, we have, that they are strictly positive, since the right hand side of (3.17)contains terms that are all strictly positive.

136

Appendix D


Proof. We first prove that all the sublots are critical for either of the two cases, i.e.whether n ≤ m or n > m. When n ≤ m, we require only n machines at Stage 2to obtain an optimal schedule, and each of these machines processes only a singlesublot; therefore, all the sublots can start to be processed at Stage 2 immediatelyupon their transfer from Stage 1. For the case when n > m, we follow the ‘alternateassignment’ rule to assign sublots to machines at Stage 2. Let the machine at Stage2 on which sublot i (from the end) gets assigned to be designated by m(i). Ifi · mod · m 6= 0, then m(i) = i · mod · m, else m(i) = m. For i ≥ m + 1, subloti − m (from the end) is processed immediately after sublot i (from the end) onmachine m(i). In order to show that all sublots are critical, we now just need toshow that C2,i,m(i) = C1,i−m,∀i = m + 1, . . . n, i.e. sublot i −m (from the end) canstart its processing at Stage 2 immediately upon its completion at Stage 1. We have,C2,i,m(i) = C1,i + psi = C1,i + (si−1 + . . . si−m) +mt = C1,i−m,∀i = m+ 1, . . . n, using(3.17).

Next, we prove that the completion times of the last sublots on all machines at Stage2 are the same. Note that whether we have the case that n ≤ m (in which case wejust utilize m = n machines to obtain an optimal schedule) or n ≥ m, the completiontimes of the last sublots on all the m machines are given as, C2,i,i,∀i = 1, . . .m. Sinceall the sublots are also critical, we can write,

C2,i,i = C1,i + psi

= (C1,i−1 − (si−1 − t)) + psi

= (C1,i−1 − (si−1 − t)) + p(ρsi−1 + t/p) (using either (3.14) or (3.16))

137

Sanchit Singh Appendix D. Proof of Corollary 3.2 138

= C1,i−1 + psi−1

= C2,i−1,i−1,∀i = 2, . . . , n

Since, the completion times of the last sublots on all machines at Stage 2 are thesame, we have the makespan value given as M(U, n,m) = C2,i,i,∀i = 1, . . .m =⇒M(U, n,m) = C2,1,1 = U + nt+ ps1(U, n,m).

Appendix E


Proof. For a given number of machines, m (> 2), and number of sublots, n , firstconsider the case when n ≤ m. In that case, s1(U, n,m) is given by (3.19). Thus,s1(U + δ, n,m)− s1(U, n,m) = δ(ρ−1)/(ρn−1), has the same sign as that of δ, sincethe denominator is strictly positive.

Then, by (3.20), we have, si(U + δ, n,m) − si(U, n,m) = ρi−1(s1(U + δ, n,m) −s1(U, n,m)) = ρi−1δ(ρ − 1)/(ρn − 1), ∀i = 2, . . . n, and it follows the same sign asthat of δ, which proves the desired result for sublots numbered 1 to n (from the end).

Next, consider the case when n ≥ m. We further have two sub-cases, when eitherp 6= m or p = m. For both of these sub-cases, we have, s1 given by (B.24) as follows.

s1(U, n,m) =U −

(∑ni=1 γi + t

(∑m−1k=2 (

∑ni=1 yi,k)(ρ

k−1 − 1)))∑m−1

k=1 (∑n

i=1 yi,k)ρk−1

Note that, the expressions(∑n

i=1 γi + t(∑m−1

k=2 (∑n

i=1 yi,k)(ρk−1 − 1)

))in the numer-

ator, and∑m−1

k=1 (∑n

i=1 yi,k)ρk−1 in the denominator do not vary with U . Hence,

s1(U + δ, n,m)− s1(U, n,m) = δ/(∑m−1

k=1 (∑n

i=1 yi,k)ρk−1), has the same sign as that

of δ, since the denominator is strictly positive.

Then, by (3.27), we have, si(U + δ, n,m) − si(U, n,m) = ρi−1(s1(U + δ, n,m) −s1(U, n,m)) = δ/((

∑m−1k=1 (

∑ni=1 yi,k)ρ

k−1)/ρi−1), ∀i = 2, . . .m, and it follows thesame sign as that of δ, which proves the desired result for sublots numbered 1 to m(from the end). We use induction to extend this result to other sublots. Suppose thetheorem holds true for sublots 1 to r(≥ m) (from the end). We want to show that

139

Sanchit Singh Appendix E. Proof of Theorem 3.2 140

it holds true for sublot r + 1 (from the end) as well. By (3.17), sr+1(U + δ, n,m)−sr+1(U, n,m) = (

∑ri=r+1−m(si(U + δ, n,m)− si(U, n,m)))/p, and by our assumption,

we know that, si(U + δ, n,m) − si(U, n,m), ∀i = k + 1 −m, . . . k, follows the samesign as that of δ, which implies that, sr+1(U + δ, n,m) − sr+1(U, n,m) also followsthe same sign as that of δ.

Appendix F


Proof. For a given lot of size, U , and a number of machines, m (≥ 2), first considerthe case when n ≤ m. In this case, s1(U, n,m) is given by (3.19). Thus, ∀ 2 ≤ n ≤ m,

s1(U, n,m)− s1(U, n− 1,m)

=U(ρ− 1)− t(ρn − nρ+ n− 1)

ρn − 1− U(ρ− 1)− t(ρn−1 − (n− 1)ρ+ (n− 1)− 1)

ρn−1 − 1

<U(ρ− 1)− t(ρn − nρ+ n− 1)

ρn − 1− U(ρ− 1)− t(ρn−1 − (n− 1)ρ+ (n− 1)− 1)

ρn − 1

=−t(ρn − ρn−1 − ρ+ 1)

ρn − 1

=−t(ρn−1 − 1)(ρ− 1)

ρn − 1< 0

We now consider the case when n ≥ m. We further have two sub-cases, when eitherp 6= m or p = m. For both of these sub-cases, we have, s1, given by (B.24) as follows.

s1(U, n,m) =U −

(∑ni=1 γi + t

(∑m−1k=2 (

∑ni=1 yi,k)(ρ

k−1 − 1)))∑m−1

k=1 (∑n

i=1 yi,k)ρk−1

We will show that the expressions(∑n

i=1 γi + t(∑m−1

k=2 (∑n

i=1 yi,k)(ρk−1 − 1)

))in the

numerator and∑m−1

k=1 (∑n

i=1 yi,k)ρk−1 in the denominator increase with increment in

n, where the former increases strictly in n. This will be sufficient to prove the desired

141

Sanchit Singh Appendix F. Proof of Theorem 3.3 142

result. Note that, for a given number of machines, m, and a fixed value of p and t,these expressions are only a function of n, and are independent of U .

Let fγ(j) =∑j

i=1 γi, and fy(j, k) =∑j

i=1 yi,k, ∀j ≥ 1,∀1 ≤ k ≤ m− 1.

Consider, fγ(n) first. From (B.3), we have the initial conditions for γi recurrencesequence, γi = 0,∀i = 1, . . .m − 1,, and γm = (m − 1)t/p > 0. For i ≥ m + 1, γiis given by (B.5), so if γj ≥ 0,∀j < i, we have that γi > 0. Therfore, by induction,γi > 0, ∀i ≥ m+ 1 as well. Hence,

fγ(i+ 1)− fγ(i) = γi+1 > 0, ∀i ≥ m ≥ 2

=⇒ fγ(n+ 1)− fγ(n) > 0, ∀n ≥ m ≥ 2

Thus, fγ(n) is a strictly increasing function in n.Next, we analyze the nature of fy(n, k). By (B.2), we have the initial conditionsfor the yi,k recursion sequence, yi,k ≥ 0,∀i = 1, . . .m,∀k = 1, . . .m − 1. For i ≥m+ 1,∀k = 1, . . .m− 1, yi,k is given by (B.4), so if yj,k ≥ 0,∀j < i,∀k = 1, . . .m− 1,we have that yi,k ≥ 0,∀k = 1, . . .m − 1, and by induction property, yi,k ≥ 0,∀i ≥m+ 1,∀k = 1, . . .m− 1 as well. Hence,

fy(i+ 1, k)− fy(i, k) = yi+1,k ≥ 0, ∀i ≥ m ≥ 2; ∀k = 1, . . . ,m− 1

=⇒ fy(n+ 1, k)− fy(n, k) ≥ 0, ∀n ≥ m ≥ 2;∀k = 1, . . . ,m− 1,

which implies fy(n, k) to be a monotonically increasing function in n, ∀k = 1, . . .m−1.

Therefore, the expression,(n∑i=1

γi + t

(m−1∑k=2

(n∑i=1

yi,k)(ρk−1 − 1)

))=

(fγ(n) + t

(m−1∑k=2

(fy(n, k))(ρk−1 − 1)

)),

is strictly increasing in n. Similarly, the expression,

m−1∑k=1

(n∑i=1

yi,k)ρk−1 =

m−1∑k=1

(fy(n, k))ρk−1,

is monotonically increasing in n. And, limn→∞(∑n

i=1 γi + t(∑m−1

k=2 (∑n

i=1 yi,k)(ρk−1 − 1)

))=

limn→∞(fγ(n) + t

(∑m−1k=2 (fy(n, k))(ρk−1 − 1)

))→∞. This implies that there exists

Sanchit Singh Appendix F. Proof of Theorem 3.3 143

an ni, s.t. ∀n > ni, s1(U, n,m) is negative, and the schedule given by Theorem 3.1,with number of sublots n > ni is infeasible.

Appendix G


Proof. We first prove the first part of the corollary. By Theorem 3.3, we have,s1(U, n,m) ≤ s1(U, no,m),∀n ≥ no. Thus, if s1(U, no,m) < 0, it follows that ∀n ≥no, s1(U, n,m) < 0 and as such, the schedule obtained using Theorem 3.1 is notfeasible.

The proof of the second part of the corollary follows naturally from the first. We doso by contradiction. Suppose that if s1(U, no,m) ≥ 0, then ∃n : n < no, such thatthe schedule obtained using Theorem 3.1 with n number of sublots is not feasible,i.e. s1(U, n,m) < 0 (following Corollary 3.1). Following the results from the firstpart of this proof, we must then have s1(U, no,m) < 0, which is a contradiction.

144

Appendix H


Proof. For a given U , n (≤ nf ), and m (≥ 2), we have, s1(U, n,m) ≥ 0; therefore,optimal makespan value,M(U, n,m) = U+nt+ps1(U, n,m) corresponds to a feasiblesolution. Consider another value for the number of sublots, ns. Let Mns be theoptimal makespan value corresponding to ns. A lower bound on Mns is Mlb

ns =U + nst. If we have, Mlb

ns ≥ M(U, n,m), it implies that Mns ≥ M(U, n,m). Thisfurther reduces to ns ≥ n+ps1(U, n,m)/t. Therefore, for any such ns, the makespanobtained is at-least as large as that obtained for n. Thus, we can fix ns at aninteger value, ns = n + bps1(U, n,m)/tc, s.t. the makespan for the lot does notimprove for number of sublots larger than ns, over that obtained with n sublots, i.e.M(U, n,m). Note that ns does not exist independently, and it is rather a functionof an appropriate value of n.

145

Appendix I

Data Files Used For Table 4.16and Table 4.17

Data-set referenced in the Table 4.16 and Table 4.17 by |K|, |J |, |M |, |S| is as follows.

1. <|K|, |J |, |M |, |S|> = <8, 40, 3, 9>.

|K|

8

|J|

40

|Mc|

3

|Sc|

9

----------------------

L_i

i= 1 : 14 => 1 4 7 10 13 16 19 22 25 28 31 34 37 40

i= 2 : 13 => 2 5 8 11 14 17 20 23 26 29 32 35 38

i= 3 : 13 => 3 6 9 12 15 18 21 24 27 30 33 36 39

----------------------

r_k

k= 1 => 40

k= 2 => 39

k= 3 => 38

k= 4 => 37

k= 5 => 36

k= 6 => 35

k= 7 => 34

k= 8 => 33

----------------------

R_j

j= 1 : 0 =>

j= 2 : 0 =>

j= 3 : 0 =>

j= 4 : 0 =>

j= 5 : 0 =>

j= 6 : 0 =>

j= 7 : 0 =>

j= 8 : 0 =>

j= 9 : 3 => 3 7 8

j= 10 : 2 => 1 8

j= 11 : 2 => 3 7

j= 12 : 2 => 1 7

146

Sanchit Singh Appendix I. Data Files - Table 4.16 and Table 4.17 147

j= 13 : 2 => 4 8

j= 14 : 3 => 2 4 6

j= 15 : 3 => 1 3 4

j= 16 : 3 => 3 5 8

j= 17 : 2 => 9 15

j= 18 : 3 => 6 7 10

j= 19 : 3 => 7 12 15

j= 20 : 3 => 2 8 15

j= 21 : 2 => 2 11

j= 22 : 2 => 10 15

j= 23 : 2 => 7 9

j= 24 : 2 => 11 13

j= 25 : 2 => 5 20

j= 26 : 3 => 15 20 23

j= 27 : 4 => 7 9 22 23

j= 28 : 3 => 3 7 24

j= 29 : 3 => 4 7 23

j= 30 : 3 => 15 17 19

j= 31 : 2 => 1 20

j= 32 : 4 => 8 12 18 21

j= 33 : 3 => 8 23 31

j= 34 : 3 => 12 19 26

j= 35 : 2 => 3 32

j= 36 : 3 => 11 26 30

j= 37 : 2 => 11 32

j= 38 : 3 => 3 4 27

j= 39 : 4 => 16 25 28 29

j= 40 : 3 => 14 28 32

----------------------

j a_j_fix a_j_unit b_j_fix b_j_unit

1 0 1 0 0

2 0 1 0 0

3 0 1 0 0

4 0 1 0 0

5 0 1 0 0

6 0 1 0 0

7 0 1 0 0

8 0 1 0 0

9 1 3 1 3

10 1 2 1 2

11 1 2 2 2

12 1 2 2 2

13 1 2 1 2

14 1 3 1 3

15 1 3 2 3

16 1 3 2 3

17 1 2 1 2

18 1 3 1 3

19 1 3 2 3

20 1 3 1 3

21 1 2 1 2

22 1 2 1 2

23 1 2 1 2

24 1 2 2 2

25 1 2 2 2

26 1 3 1 3

27 1 4 1 4

28 1 3 1 3

29 1 3 2 3

30 1 3 2 3

31 1 2 2 2

32 1 4 1 4

33 1 3 1 3

34 1 3 1 3

35 1 2 2 2

36 1 3 2 3

37 1 2 1 2

38 1 3 2 3

39 1 4 2 4

40 1 3 1 3

----------------------

p_s

s= 1 => 0.111111

s= 2 => 0.111111

s= 3 => 0.111111

s= 4 => 0.111111

s= 5 => 0.111111

s= 6 => 0.111111


s= 7 => 0.111111

s= 8 => 0.111111

s= 9 => 0.111111

----------------------

d_ks s=1 s=2 s=3 s=4 s=5 s=6 s=7 s=8 s=9

k= 1 8 9 8 12 8 12 9 10 12

k= 2 9 9 12 12 12 9 9 8 12

k= 3 12 12 12 11 8 12 12 8 12

k= 4 9 8 11 8 11 10 9 11 12

k= 5 9 12 11 8 8 9 12 12 8

k= 6 8 10 11 8 12 8 8 11 12

k= 7 11 8 8 12 11 9 8 11 12

k= 8 6 8 10 9 12 14 6 9 8

----------------------

e_k

k= 1 => 1

k= 2 => 1

k= 3 => 1

k= 4 => 1

k= 5 => 1

k= 6 => 1

k= 7 => 1

k= 8 => 1

----------------------

h_j

j= 1 => 1

j= 2 => 1

j= 3 => 1

j= 4 => 1

j= 5 => 1

j= 6 => 1

j= 7 => 1

j= 8 => 1

j= 9 => 1

j= 10 => 1

j= 11 => 1

j= 12 => 1

j= 13 => 1

j= 14 => 1

j= 15 => 1

j= 16 => 1

j= 17 => 1

j= 18 => 1

j= 19 => 1

j= 20 => 1

j= 21 => 1

j= 22 => 1

j= 23 => 1

j= 24 => 1

j= 25 => 1

j= 26 => 1

j= 27 => 1

j= 28 => 1

j= 29 => 1

j= 30 => 1

j= 31 => 1

j= 32 => 1

j= 33 => 1

j= 34 => 1

j= 35 => 1

j= 36 => 1

j= 37 => 1

j= 38 => 1

j= 39 => 1

j= 40 => 1

2. <|K|, |J|, |M|, |S|> = <8, 40, 5, 9>.

|K|

8

|J|

40

|Mc|

5

|Sc|

9

----------------------


L_i

i= 1 : 8 => 1 6 11 16 21 26 31 36

i= 2 : 8 => 2 7 12 17 22 27 32 37

i= 3 : 8 => 3 8 13 18 23 28 33 38

i= 4 : 8 => 4 9 14 19 24 29 34 39

i= 5 : 8 => 5 10 15 20 25 30 35 40

----------------------

r_k

k= 1 => 40

k= 2 => 39

k= 3 => 38

k= 4 => 37

k= 5 => 36

k= 6 => 35

k= 7 => 34

k= 8 => 33

----------------------

R_j

j= 1 : 0 =>

j= 2 : 0 =>

j= 3 : 0 =>

j= 4 : 0 =>

j= 5 : 0 =>

j= 6 : 0 =>

j= 7 : 0 =>

j= 8 : 0 =>

j= 9 : 3 => 3 7 8

j= 10 : 2 => 1 8

j= 11 : 2 => 3 7

j= 12 : 2 => 1 7

j= 13 : 2 => 4 8

j= 14 : 3 => 2 4 6

j= 15 : 3 => 1 3 4

j= 16 : 3 => 3 5 8

j= 17 : 2 => 9 15

j= 18 : 3 => 6 7 10

j= 19 : 3 => 7 12 15

j= 20 : 3 => 2 8 15

j= 21 : 2 => 2 11

j= 22 : 2 => 10 15

j= 23 : 2 => 7 9

j= 24 : 2 => 11 13

j= 25 : 2 => 5 20

j= 26 : 3 => 15 20 23

j= 27 : 4 => 7 9 22 23

j= 28 : 3 => 3 7 24

j= 29 : 3 => 4 7 23

j= 30 : 3 => 15 17 19

j= 31 : 2 => 1 20

j= 32 : 4 => 8 12 18 21

j= 33 : 3 => 8 23 31

j= 34 : 3 => 12 19 26

j= 35 : 2 => 3 32

j= 36 : 3 => 11 26 30

j= 37 : 2 => 11 32

j= 38 : 3 => 3 4 27

j= 39 : 4 => 16 25 28 29

j= 40 : 3 => 14 28 32

----------------------


1 0 1 0 0

2 0 1 0 0

3 0 1 0 0

4 0 1 0 0

5 0 1 0 0

6 0 1 0 0

7 0 1 0 0

8 0 1 0 0

9 1 3 1 3

10 1 2 1 2

11 1 2 2 2

12 1 2 2 2

13 1 2 1 2

14 1 3 1 3

15 1 3 2 3

16 1 3 2 3

17 1 2 1 2

18 1 3 1 3


19 1 3 2 3

20 1 3 1 3

21 1 2 1 2

22 1 2 1 2

23 1 2 1 2

24 1 2 2 2

25 1 2 2 2

26 1 3 1 3

27 1 4 1 4

28 1 3 1 3

29 1 3 2 3

30 1 3 2 3

31 1 2 2 2

32 1 4 1 4

33 1 3 1 3

34 1 3 1 3

35 1 2 2 2

36 1 3 2 3

37 1 2 1 2

38 1 3 2 3

39 1 4 2 4

40 1 3 1 3

----------------------

p_s

s= 1 => 0.111111

s= 2 => 0.111111

s= 3 => 0.111111

s= 4 => 0.111111

s= 5 => 0.111111

s= 6 => 0.111111

s= 7 => 0.111111

s= 8 => 0.111111

s= 9 => 0.111111

----------------------

d_ks s=1 s=2 s=3 s=4 s=5 s=6 s=7 s=8 s=9

k= 1 8 9 8 12 8 12 9 10 12

k= 2 9 9 12 12 12 9 9 8 12

k= 3 12 12 12 11 8 12 12 8 12

k= 4 9 8 11 8 11 10 9 11 12

k= 5 9 12 11 8 8 9 12 12 8

k= 6 8 10 11 8 12 8 8 11 12

k= 7 11 8 8 12 11 9 8 11 12

k= 8 6 8 10 9 12 14 6 9 8

----------------------

e_k

k= 1 => 1

k= 2 => 1

k= 3 => 1

k= 4 => 1

k= 5 => 1

k= 6 => 1

k= 7 => 1

k= 8 => 1

----------------------

h_j

j= 1 => 1

j= 2 => 1

j= 3 => 1

j= 4 => 1

j= 5 => 1

j= 6 => 1

j= 7 => 1

j= 8 => 1

j= 9 => 1

j= 10 => 1

j= 11 => 1

j= 12 => 1

j= 13 => 1

j= 14 => 1

j= 15 => 1

j= 16 => 1

j= 17 => 1

j= 18 => 1

j= 19 => 1

j= 20 => 1

j= 21 => 1

j= 22 => 1

j= 23 => 1


j= 24 => 1

j= 25 => 1

j= 26 => 1

j= 27 => 1

j= 28 => 1

j= 29 => 1

j= 30 => 1

j= 31 => 1

j= 32 => 1

j= 33 => 1

j= 34 => 1

j= 35 => 1

j= 36 => 1

j= 37 => 1

j= 38 => 1

j= 39 => 1

j= 40 => 1

3. <|K|, |J|, |M|, |S|> = <8, 40, 3, 15>.

|K|

8

|J|

40

|Mc|

3

|Sc|

15

----------------------

L_i

i= 1 : 14 => 1 4 7 10 13 16 19 22 25 28 31 34 37 40

i= 2 : 13 => 2 5 8 11 14 17 20 23 26 29 32 35 38

i= 3 : 13 => 3 6 9 12 15 18 21 24 27 30 33 36 39

----------------------

r_k

k= 1 => 40

k= 2 => 39

k= 3 => 38

k= 4 => 37

k= 5 => 36

k= 6 => 35

k= 7 => 34

k= 8 => 33

----------------------

R_j

j= 1 : 0 =>

j= 2 : 0 =>

j= 3 : 0 =>

j= 4 : 0 =>

j= 5 : 0 =>

j= 6 : 0 =>

j= 7 : 0 =>

j= 8 : 0 =>

j= 9 : 3 => 3 7 8

j= 10 : 2 => 1 8

j= 11 : 2 => 3 7

j= 12 : 2 => 1 7

j= 13 : 2 => 4 8

j= 14 : 3 => 2 4 6

j= 15 : 3 => 1 3 4

j= 16 : 3 => 3 5 8

j= 17 : 2 => 9 15

j= 18 : 3 => 6 7 10

j= 19 : 3 => 7 12 15

j= 20 : 3 => 2 8 15

j= 21 : 2 => 2 11

j= 22 : 2 => 10 15

j= 23 : 2 => 7 9

j= 24 : 2 => 11 13

j= 25 : 2 => 5 20

j= 26 : 3 => 15 20 23

j= 27 : 4 => 7 9 22 23

j= 28 : 3 => 3 7 24

j= 29 : 3 => 4 7 23

j= 30 : 3 => 15 17 19

j= 31 : 2 => 1 20

j= 32 : 4 => 8 12 18 21


j= 33 : 3 => 8 23 31

j= 34 : 3 => 12 19 26

j= 35 : 2 => 3 32

j= 36 : 3 => 11 26 30

j= 37 : 2 => 11 32

j= 38 : 3 => 3 4 27

j= 39 : 4 => 16 25 28 29

j= 40 : 3 => 14 28 32

----------------------


1 0 1 0 0

2 0 1 0 0

3 0 1 0 0

4 0 1 0 0

5 0 1 0 0

6 0 1 0 0

7 0 1 0 0

8 0 1 0 0

9 1 3 1 3

10 1 2 1 2

11 1 2 2 2

12 1 2 2 2

13 1 2 1 2

14 1 3 1 3

15 1 3 2 3

16 1 3 2 3

17 1 2 1 2

18 1 3 1 3

19 1 3 2 3

20 1 3 1 3

21 1 2 1 2

22 1 2 1 2

23 1 2 1 2

24 1 2 2 2

25 1 2 2 2

26 1 3 1 3

27 1 4 1 4

28 1 3 1 3

29 1 3 2 3

30 1 3 2 3

31 1 2 2 2

32 1 4 1 4

33 1 3 1 3

34 1 3 1 3

35 1 2 2 2

36 1 3 2 3

37 1 2 1 2

38 1 3 2 3

39 1 4 2 4

40 1 3 1 3

----------------------

p_s

s= 1 => 0.0666667

s= 2 => 0.0666667

s= 3 => 0.0666667

s= 4 => 0.0666667

s= 5 => 0.0666667

s= 6 => 0.0666667

s= 7 => 0.0666667

s= 8 => 0.0666667

s= 9 => 0.0666667

s= 10 => 0.0666667

s= 11 => 0.0666667

s= 12 => 0.0666667

s= 13 => 0.0666667

s= 14 => 0.0666667

s= 15 => 0.0666667

----------------------

d_ks s=1 s=2 s=3 s=4 s=5 s=6 s=7 s=8 s=9 s=10 s=11 s=12 s=13 s=14 s=15

k= 1 8 9 8 12 8 12 9 10 12 12 9 9 12 12 12

k= 2 13 7 6 11 10 10 10 9 14 10 13 6 10 10 11

k= 3 14 9 6 13 9 11 7 11 11 9 14 7 10 9 14

k= 4 9 12 12 8 10 8 10 11 8 12 8 8 11 12 11

k= 5 14 14 9 13 6 7 6 8 10 9 12 14 6 9 8

k= 6 8 12 9 8 12 10 11 11 11 11 12 10 9 10 8

k= 7 9 9 9 12 9 12 9 8 10 9 8 12 9 10 11

k= 8 10 12 6 14 11 7 12 7 14 14 10 13 9 7 9

----------------------


e_k

k= 1 => 1

k= 2 => 1

k= 3 => 1

k= 4 => 1

k= 5 => 1

k= 6 => 1

k= 7 => 1

k= 8 => 1

----------------------

h_j

j= 1 => 1

j= 2 => 1

j= 3 => 1

j= 4 => 1

j= 5 => 1

j= 6 => 1

j= 7 => 1

j= 8 => 1

j= 9 => 1

j= 10 => 1

j= 11 => 1

j= 12 => 1

j= 13 => 1

j= 14 => 1

j= 15 => 1

j= 16 => 1

j= 17 => 1

j= 18 => 1

j= 19 => 1

j= 20 => 1

j= 21 => 1

j= 22 => 1

j= 23 => 1

j= 24 => 1

j= 25 => 1

j= 26 => 1

j= 27 => 1

j= 28 => 1

j= 29 => 1

j= 30 => 1

j= 31 => 1

j= 32 => 1

j= 33 => 1

j= 34 => 1

j= 35 => 1

j= 36 => 1

j= 37 => 1

j= 38 => 1

j= 39 => 1

j= 40 => 1

4. <|K|, |J|, |M|, |S|> = <9, 45, 2, 9>.

|K|

9

|J|

45

|Mc|

2

|Sc|

9

----------------------

L_i

i= 1 : 23 => 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45

i= 2 : 22 => 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44

----------------------

r_k

k= 1 => 45

k= 2 => 44

k= 3 => 43

k= 4 => 42

k= 5 => 41

k= 6 => 40

k= 7 => 39

k= 8 => 38

k= 9 => 37


----------------------

R_j

j= 1 : 0 =>

j= 2 : 0 =>

j= 3 : 0 =>

j= 4 : 0 =>

j= 5 : 0 =>

j= 6 : 0 =>

j= 7 : 0 =>

j= 8 : 0 =>

j= 9 : 0 =>

j= 10 : 3 => 2 3 8

j= 11 : 2 => 3 7

j= 12 : 2 => 4 5

j= 13 : 3 => 4 5 9

j= 14 : 3 => 2 6 9

j= 15 : 2 => 1 5

j= 16 : 2 => 1 6

j= 17 : 3 => 3 6 9

j= 18 : 3 => 3 8 9

j= 19 : 3 => 2 11 12

j= 20 : 4 => 5 14 15 17

j= 21 : 3 => 1 13 14

j= 22 : 3 => 15 16 18

j= 23 : 2 => 3 10

j= 24 : 3 => 4 14 15

j= 25 : 2 => 7 11

j= 26 : 2 => 14 16

j= 27 : 3 => 1 10 14

j= 28 : 2 => 12 20

j= 29 : 3 => 3 8 26

j= 30 : 3 => 5 14 26

j= 31 : 2 => 15 26

j= 32 : 2 => 20 26

j= 33 : 2 => 19 25

j= 34 : 3 => 20 22 27

j= 35 : 2 => 21 23

j= 36 : 3 => 2 14 26

j= 37 : 2 => 22 29

j= 38 : 3 => 11 16 29

j= 39 : 3 => 15 30 34

j= 40 : 2 => 9 36

j= 41 : 3 => 13 24 36

j= 42 : 3 => 7 22 32

j= 43 : 4 => 19 28 31 35

j= 44 : 3 => 2 30 33

j= 45 : 2 => 35 36

----------------------


1 0 1 0 0

2 0 1 0 0

3 0 1 0 0

4 0 1 0 0

5 0 1 0 0

6 0 1 0 0

7 0 1 0 0

8 0 1 0 0

9 0 1 0 0

10 1 3 2 3

11 1 2 2 2

12 1 2 2 2

13 1 3 2 3

14 1 3 2 3

15 1 2 1 2

16 1 2 2 2

17 1 3 2 3

18 1 3 1 3

19 1 3 1 3

20 1 4 2 4

21 1 3 1 3

22 1 3 1 3

23 1 2 1 2

24 1 3 1 3

25 1 2 1 2

26 1 2 1 2

27 1 3 1 3

28 1 2 2 2

29 1 3 2 3


30 1 3 2 3

31 1 2 1 2

32 1 2 1 2

33 1 2 1 2

34 1 3 1 3

35 1 2 2 2

36 1 3 1 3

37 1 2 2 2

38 1 3 1 3

39 1 3 1 3

40 1 2 2 2

41 1 3 1 3

42 1 3 2 3

43 1 4 1 4

44 1 3 2 3

45 1 2 1 2

----------------------

p_s

s= 1 => 0.111111

s= 2 => 0.111111

s= 3 => 0.111111

s= 4 => 0.111111

s= 5 => 0.111111

s= 6 => 0.111111

s= 7 => 0.111111

s= 8 => 0.111111

s= 9 => 0.111111

----------------------

d_ks s=1 s=2 s=3 s=4 s=5 s=6 s=7 s=8 s=9

k= 1 10 12 12 9 8 8 9 8 8

k= 2 10 12 12 9 11 8 9 11 10

k= 3 12 9 8 10 10 10 12 12 10

k= 4 11 10 8 11 8 12 11 11 10

k= 5 11 9 8 12 9 12 12 8 12

k= 6 10 11 12 8 12 12 10 9 8

k= 7 12 10 11 10 11 11 9 10 9

k= 8 14 11 8 6 10 10 8 13 13

k= 9 10 12 14 9 7 13 7 9 14

----------------------

e_k

k= 1 => 1

k= 2 => 1

k= 3 => 1

k= 4 => 1

k= 5 => 1

k= 6 => 1

k= 7 => 1

k= 8 => 1

k= 9 => 1

----------------------

h_j

j= 1 => 1

j= 2 => 1

j= 3 => 1

j= 4 => 1

j= 5 => 1

j= 6 => 1

j= 7 => 1

j= 8 => 1

j= 9 => 1

j= 10 => 1

j= 11 => 1

j= 12 => 1

j= 13 => 1

j= 14 => 1

j= 15 => 1

j= 16 => 1

j= 17 => 1

j= 18 => 1

j= 19 => 1

j= 20 => 1

j= 21 => 1

j= 22 => 1

j= 23 => 1

j= 24 => 1

j= 25 => 1

j= 26 => 1

j= 27 => 1


j= 28 => 1

j= 29 => 1

j= 30 => 1

j= 31 => 1

j= 32 => 1

j= 33 => 1

j= 34 => 1

j= 35 => 1

j= 36 => 1

j= 37 => 1

j= 38 => 1

j= 39 => 1

j= 40 => 1

j= 41 => 1

j= 42 => 1

j= 43 => 1

j= 44 => 1

j= 45 => 1

5. <|K|, |J|, |M|, |S|> = <9, 45, 5, 9>.

|K|

9

|J|

45

|Mc|

5

|Sc|

9

----------------------

L_i

i= 1 : 9 => 1 6 11 16 21 26 31 36 41

i= 2 : 9 => 2 7 12 17 22 27 32 37 42

i= 3 : 9 => 3 8 13 18 23 28 33 38 43

i= 4 : 9 => 4 9 14 19 24 29 34 39 44

i= 5 : 9 => 5 10 15 20 25 30 35 40 45

----------------------

r_k

k= 1 => 45

k= 2 => 44

k= 3 => 43

k= 4 => 42

k= 5 => 41

k= 6 => 40

k= 7 => 39

k= 8 => 38

k= 9 => 37

----------------------

R_j

j= 1 : 0 =>

j= 2 : 0 =>

j= 3 : 0 =>

j= 4 : 0 =>

j= 5 : 0 =>

j= 6 : 0 =>

j= 7 : 0 =>

j= 8 : 0 =>

j= 9 : 0 =>

j= 10 : 3 => 2 3 8

j= 11 : 2 => 3 7

j= 12 : 2 => 4 5

j= 13 : 3 => 4 5 9

j= 14 : 3 => 2 6 9

j= 15 : 2 => 1 5

j= 16 : 2 => 1 6

j= 17 : 3 => 3 6 9

j= 18 : 3 => 3 8 9

j= 19 : 3 => 2 11 12

j= 20 : 4 => 5 14 15 17

j= 21 : 3 => 1 13 14

j= 22 : 3 => 15 16 18

j= 23 : 2 => 3 10

j= 24 : 3 => 4 14 15

j= 25 : 2 => 7 11

j= 26 : 2 => 14 16

j= 27 : 3 => 1 10 14

j= 28 : 2 => 12 20


j= 29 : 3 => 3 8 26

j= 30 : 3 => 5 14 26

j= 31 : 2 => 15 26

j= 32 : 2 => 20 26

j= 33 : 2 => 19 25

j= 34 : 3 => 20 22 27

j= 35 : 2 => 21 23

j= 36 : 3 => 2 14 26

j= 37 : 2 => 22 29

j= 38 : 3 => 11 16 29

j= 39 : 3 => 15 30 34

j= 40 : 2 => 9 36

j= 41 : 3 => 13 24 36

j= 42 : 3 => 7 22 32

j= 43 : 4 => 19 28 31 35

j= 44 : 3 => 2 30 33

j= 45 : 2 => 35 36

----------------------


1 0 1 0 0

2 0 1 0 0

3 0 1 0 0

4 0 1 0 0

5 0 1 0 0

6 0 1 0 0

7 0 1 0 0

8 0 1 0 0

9 0 1 0 0

10 1 3 2 3

11 1 2 2 2

12 1 2 2 2

13 1 3 2 3

14 1 3 2 3

15 1 2 1 2

16 1 2 2 2

17 1 3 2 3

18 1 3 1 3

19 1 3 1 3

20 1 4 2 4

21 1 3 1 3

22 1 3 1 3

23 1 2 1 2

24 1 3 1 3

25 1 2 1 2

26 1 2 1 2

27 1 3 1 3

28 1 2 2 2

29 1 3 2 3

30 1 3 2 3

31 1 2 1 2

32 1 2 1 2

33 1 2 1 2

34 1 3 1 3

35 1 2 2 2

36 1 3 1 3

37 1 2 2 2

38 1 3 1 3

39 1 3 1 3

40 1 2 2 2

41 1 3 1 3

42 1 3 2 3

43 1 4 1 4

44 1 3 2 3

45 1 2 1 2

----------------------

p_s

s= 1 => 0.111111

s= 2 => 0.111111

s= 3 => 0.111111

s= 4 => 0.111111

s= 5 => 0.111111

s= 6 => 0.111111

s= 7 => 0.111111

s= 8 => 0.111111

s= 9 => 0.111111

----------------------

d_ks s=1 s=2 s=3 s=4 s=5 s=6 s=7 s=8 s=9

k= 1 10 12 12 9 8 8 9 8 8


k= 2 10 12 12 9 11 8 9 11 10

k= 3 12 9 8 10 10 10 12 12 10

k= 4 11 10 8 11 8 12 11 11 10

k= 5 11 9 8 12 9 12 12 8 12

k= 6 10 11 12 8 12 12 10 9 8

k= 7 12 10 11 10 11 11 9 10 9

k= 8 14 11 8 6 10 10 8 13 13

k= 9 10 12 14 9 7 13 7 9 14

----------------------

e_k

k= 1 => 1

k= 2 => 1

k= 3 => 1

k= 4 => 1

k= 5 => 1

k= 6 => 1

k= 7 => 1

k= 8 => 1

k= 9 => 1

----------------------

h_j

j= 1 => 1

j= 2 => 1

j= 3 => 1

j= 4 => 1

j= 5 => 1

j= 6 => 1

j= 7 => 1

j= 8 => 1

j= 9 => 1

j= 10 => 1

j= 11 => 1

j= 12 => 1

j= 13 => 1

j= 14 => 1

j= 15 => 1

j= 16 => 1

j= 17 => 1

j= 18 => 1

j= 19 => 1

j= 20 => 1

j= 21 => 1

j= 22 => 1

j= 23 => 1

j= 24 => 1

j= 25 => 1

j= 26 => 1

j= 27 => 1

j= 28 => 1

j= 29 => 1

j= 30 => 1

j= 31 => 1

j= 32 => 1

j= 33 => 1

j= 34 => 1

j= 35 => 1

j= 36 => 1

j= 37 => 1

j= 38 => 1

j= 39 => 1

j= 40 => 1

j= 41 => 1

j= 42 => 1

j= 43 => 1

j= 44 => 1

j= 45 => 1

6. <|K|, |J|, |M|, |S|> = <9, 45, 2, 15>.

|K|

9

|J|

45

|Mc|

2

|Sc|

15


----------------------

L_i

i= 1 : 23 => 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45

i= 2 : 22 => 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44

----------------------

r_k

k= 1 => 45

k= 2 => 44

k= 3 => 43

k= 4 => 42

k= 5 => 41

k= 6 => 40

k= 7 => 39

k= 8 => 38

k= 9 => 37

----------------------

R_j

j= 1 : 0 =>

j= 2 : 0 =>

j= 3 : 0 =>

j= 4 : 0 =>

j= 5 : 0 =>

j= 6 : 0 =>

j= 7 : 0 =>

j= 8 : 0 =>

j= 9 : 0 =>

j= 10 : 3 => 2 3 8

j= 11 : 2 => 3 7

j= 12 : 2 => 4 5

j= 13 : 3 => 4 5 9

j= 14 : 3 => 2 6 9

j= 15 : 2 => 1 5

j= 16 : 2 => 1 6

j= 17 : 3 => 3 6 9

j= 18 : 3 => 3 8 9

j= 19 : 3 => 2 11 12

j= 20 : 4 => 5 14 15 17

j= 21 : 3 => 1 13 14

j= 22 : 3 => 15 16 18

j= 23 : 2 => 3 10

j= 24 : 3 => 4 14 15

j= 25 : 2 => 7 11

j= 26 : 2 => 14 16

j= 27 : 3 => 1 10 14

j= 28 : 2 => 12 20

j= 29 : 3 => 3 8 26

j= 30 : 3 => 5 14 26

j= 31 : 2 => 15 26

j= 32 : 2 => 20 26

j= 33 : 2 => 19 25

j= 34 : 3 => 20 22 27

j= 35 : 2 => 21 23

j= 36 : 3 => 2 14 26

j= 37 : 2 => 22 29

j= 38 : 3 => 11 16 29

j= 39 : 3 => 15 30 34

j= 40 : 2 => 9 36

j= 41 : 3 => 13 24 36

j= 42 : 3 => 7 22 32

j= 43 : 4 => 19 28 31 35

j= 44 : 3 => 2 30 33

j= 45 : 2 => 35 36

----------------------


1 0 1 0 0

2 0 1 0 0

3 0 1 0 0

4 0 1 0 0

5 0 1 0 0

6 0 1 0 0

7 0 1 0 0

8 0 1 0 0

9 0 1 0 0

10 1 3 2 3

11 1 2 2 2

12 1 2 2 2

13 1 3 2 3

14 1 3 2 3


15 1 2 1 2

16 1 2 2 2

17 1 3 2 3

18 1 3 1 3

19 1 3 1 3

20 1 4 2 4

21 1 3 1 3

22 1 3 1 3

23 1 2 1 2

24 1 3 1 3

25 1 2 1 2

26 1 2 1 2

27 1 3 1 3

28 1 2 2 2

29 1 3 2 3

30 1 3 2 3

31 1 2 1 2

32 1 2 1 2

33 1 2 1 2

34 1 3 1 3

35 1 2 2 2

36 1 3 1 3

37 1 2 2 2

38 1 3 1 3

39 1 3 1 3

40 1 2 2 2

41 1 3 1 3

42 1 3 2 3

43 1 4 1 4

44 1 3 2 3

45 1 2 1 2

----------------------

p_s

s= 1 => 0.0666667

s= 2 => 0.0666667

s= 3 => 0.0666667

s= 4 => 0.0666667

s= 5 => 0.0666667

s= 6 => 0.0666667

s= 7 => 0.0666667

s= 8 => 0.0666667

s= 9 => 0.0666667

s= 10 => 0.0666667

s= 11 => 0.0666667

s= 12 => 0.0666667

s= 13 => 0.0666667

s= 14 => 0.0666667

s= 15 => 0.0666667

----------------------

d_ks s=1 s=2 s=3 s=4 s=5 s=6 s=7 s=8 s=9 s=10 s=11 s=12 s=13 s=14 s=15

k= 1 10 12 12 9 8 8 9 8 8 8 10 12 12 9 11

k= 2 9 11 10 12 9 8 10 10 10 12 12 10 8 11 10

k= 3 11 8 12 11 11 10 8 11 9 8 12 9 12 12 8

k= 4 12 10 11 12 8 12 12 10 9 8 10 12 10 11 10

k= 5 13 11 9 9 12 8 13 13 7 11 14 11 8 6 10

k= 6 12 12 10 11 9 11 12 12 8 10 11 9 9 11 9

k= 7 14 12 8 14 7 8 9 11 12 12 9 14 11 11 10

k= 8 7 6 14 11 9 11 13 8 8 10 7 12 9 12 6

k= 9 11 9 11 11 8 9 12 9 8 9 11 8 11 8 12

----------------------

e_k

k= 1 => 1

k= 2 => 1

k= 3 => 1

k= 4 => 1

k= 5 => 1

k= 6 => 1

k= 7 => 1

k= 8 => 1

k= 9 => 1

----------------------

h_j

j= 1 => 1

j= 2 => 1

j= 3 => 1

j= 4 => 1

j= 5 => 1

j= 6 => 1


j= 7 => 1

j= 8 => 1

j= 9 => 1

j= 10 => 1

j= 11 => 1

j= 12 => 1

j= 13 => 1

j= 14 => 1

j= 15 => 1

j= 16 => 1

j= 17 => 1

j= 18 => 1

j= 19 => 1

j= 20 => 1

j= 21 => 1

j= 22 => 1

j= 23 => 1

j= 24 => 1

j= 25 => 1

j= 26 => 1

j= 27 => 1

j= 28 => 1

j= 29 => 1

j= 30 => 1

j= 31 => 1

j= 32 => 1

j= 33 => 1

j= 34 => 1

j= 35 => 1

j= 36 => 1

j= 37 => 1

j= 38 => 1

j= 39 => 1

j= 40 => 1

j= 41 => 1

j= 42 => 1

j= 43 => 1

j= 44 => 1

j= 45 => 1

modeling, analysis, and algorithmic development of some ... · the biomass logistics problem is a...

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