Scheduling and Lot S treaming

in Two Machine No-Wait Openshops

Esaignani Selvaraj ah

A Thesis Submitted in Conformity with the Requirements for the Degree of Master of Applied Science

Graduate Department of Mechanical and Industrial Engineering University of Toronto

@ Copyright by Esaignani Selvarajah (1999)

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Dedicated to my beloved parents

Abstract

Lot streaming is the process of splitting a production lot into a number of sublots,

in order to allow overlapping of successive operations, in multi-machine manufacturing

systems. The well documented benefits of lot streaming include reductions in lead times

and work-in-process, and increasea in machine utilization rates. We study the problem of

minimizing the makespm in two machine no-wait openshops producing multiple products

. We assume that there is a fixed maximum number of sublots for each product, and that

lot streaming is possible. This intractable problem requires finding sublot &es, a product

sequence for each machine and a machine sequence for each product. We prove t hat optimal

solutions can always be obtained for single product continuous case in polynomiai tirne. A

heuristic aîgorithm, Global is developed for the multi-product integer-sized sublot problem.

For the multi-product case, we simultaneously find the optimal number of sublots from

a given upper bound on the number of sublots, sublots sizes and the product sequence.

This problem is shown to be equivalent to a Traveling Salesman Problem (TSP) with

a pseudepolynomial number of cities, for which computationally efficient heuristics are

tested. Finally we provide some possible extensions of our mode1 and the results of the

computational experiments on randomly generated problem instances.

Key Words and Phrases: machine scheduling, newait openshops, lot streaming, trav-

eling salesman problem.

Acknowledgment s

First and foremost, I express my sincere gratitude to Professor Chelliah Sriskandarajah,

who helped me profusely at various stages of this thesis work. His guidance an profound in-

volvement in rny research needs apprecWation. His support in arranging financial assistance

for me warrants special mention.

1 gratefully acknowledge the guidance and suggestions provided by Prof. Nicholas G.

Hall of Ohio State University. I would like to thank Professor A.K.S Jardine and Professor

V.Makis for their valuable comments and suggestions which irnproved the content and the

presentation of my theais. 1 am also thankful to Prof. Tapan Bagchi of Indian Institute

of Technology Kanpur, Prof. Rogers, Prof. Carter, Prof. M J M Posner and whole MIE

family who helped me at various stages.

This research was supported in part by Natural Sciences and Engineering Research

Council of Canada (Grant no. OGP0104900). 1 appreciate NSERC for its fund support to

pursue my thesis work. 1 wish to thank Professor Gilbert Laporte (University of Montreal)

for supplying a copy of the GENIUS program.

The administrative assistance provided by Prof. Beno Benhabib as Graduate Coor-

dinator, Nonna Dotto, Louisa and Brenda as Departmental staff and Oscar del Rio as

corn puter sys tems Adminis trator is deeply remernbered.

1 greatly acknowledge my friends Mr. Ganesharajah, Mr. Kohulan, Mr.Subodha Ku-

mar, Mr. Thampu Joseph, Mr. Devinder and Mr.Bimal for their induable help and

support, who made my stay at UofT an enjoyable period. My special thanks are for Mr.

Subodha Kumar for his timely assistance. Mr. and Mrs. Raman Patel deserve thanks for

the moral support they provided.

Laat but not the least 1 would like to express my deep gratitude to my parents, my

brothers, my sisters and my in-laws, who are always a source of motivation and inspiration

to me.

Contents

1 Introduction

2 Literature Review

3 Notations and Assumptions

4 Lot Streaming: Single product and continuous shed sublots 18

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Introduction 18

. . . . . . . . . . . . . . . . . . . . 4.2 Scheduling and Algorithm Development 19

. . . . . . . . . . . . . . . . . . . . . . . . 4.3 DifFerent Structures of Schedule 38

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Conclusion 39

5 Multiple Products 40

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction 40

. . . . . . . . . . . . . . . . . . . . 5.2 Scheduling and Algorithm Development 41

. . . . . . . . . . . . . . 5.2.1 Scheduling of Products wit h Given Profiles 41

. . . . . . . . . . . . . 5.2.2 Simultaneous Lot Streaming and Scheduling 42

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Conclusion 50

6 Computational Teilting 52

7 Some Related Results and Discussion 54

. . . . . . . . . . . . . . . . . . . . . . 7.1 Single Product Integer-sized Sublots 54

. . . . . . . . . . . . . . . . . . . . . 7.2 Problems with Attached Setup Times 54

. . . . . . . . . . . . . . . . . . . . . . . 7.3 Sublots with Restricted Capacity 55

. . . . . . . . . . . . . . . . . . . . . 7.4 Problems with Sublot Transfer Time 55

iii

7.5 Two Machine NeWait Flowshop . . . . . . . . . . . . . . . . . . . . . . . 55

8 Conclusion8 59

List of Figures

. . . . . . . . . A basic flowshop mode1 with 4 machines processing one lot 7

. . . . . . . A basic flowshop mode1 with 4 machines processing two sublots 7

. . . . Network representation of a single job m machine flow shop problem 10

Time lag mode1 for a given job on two machine problem . . . . . . . . . . 12

A busy schedule for Case 1 . a = b. n = 2k. Cm, = s + Wu . . . . . . . . . . 20

A busyschedulefor Case 2.1. a > b. s Lt .n=2k.CmaS = s + Wa . . . . . . 21

A typical schedule of sublots X2.- and XZr . . . . . . . . . . . . . . . . . . 23 A busy schedule for Case 2.2.1. a > b. s < t . Wa + s > Wb + t . n =

2k.C,,,., = s + Wu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

The busy schedule for Case 2.2.2. a > b. s < t . Wu + s = Wb + t . Cm., =

Wu+$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 A busy schedule for Case 2.2.3. a > 6. s < t . Wa + s < Wb + t . n =

2k. Cm., = s + Wu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

. . . . . . . . . . . . . . . . . . . . . . . . . . . . Alternating Pair Structure 38

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Flowshop Structure 38

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Hybrid Structure 39

Profiles of Openshop Newait Lot Streaming Problems . . . . . . . . . . . . 41

. . . . . . . . . . . . . . . . Example Schedule of N Products in Sequence o 43

. . . . . . . . . . . . . . . . . Example With Two Countries and Five Cities 44

. . . . . . . . . . . . . . . . . . . . . . . . . . . . Schedule from Solution 1 47

Scheduie €rom Solution 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

. . . . . . . . . . . . . . . . . . . . . . . . . . . . Schedule from Solution 3 48

Schedule from Solution 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Schedule from Solution 5 49

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9 Scheduie from Solution 6 49

. . . . . . . . . . . . . . . . . . . . . . . 5.10 Optimal Schedule from Solution 7 49

. . . . . . . . . . . . . . . . . 7.1 Forward flowshop structure with transfer time 55

. . . . . . . . . . . . . . . . 7.2 Backward flowshop structure with transfer time 56

7.3 Alternating pair structure with transfer time . . . . . . . . . . . . . . . . . 56

Chapter 1

Introduction

Scheduling, a decision making process, involves allocation of limited resources such as

equipment, labor and space to jobs, activities, tasks, or custorners through time. In a

manufacturing environment it includes worker and machine assignment, job sequencing,

and the coordination of material handling and maintenance support. Effectiveness of a

scheduling process is measured by performance measures on the production system. There

have been many measures used depending on the organization's objective. One of these

performance measures is makespan which is the total time required to process al1 the jobs

on ali machines.

The scheduling problem differs considerably based on the type of operation such as con-

tinuous production, m a s production and project works. Further, it depends on machine

environment and technological constraints. These different scheduling problems naturally

lead to different models.

The basic machine environments of a shop are flowahop, openshop and jobshop. Flow-

shop is a special case, where all the jobs follow the same processing order, where as in

jobshop there is no limitations on processing order and a machine can process more than

one tasks for a job. Openshop is a special case of jobshop where a machine can process

only one tmk for a job.

Some of the technological constraints in manufacturing systems are precedent require-

ments, blocking production and newai t production. Precedent requirements are p hysical

restrictions in the order in which operations are performed. In blocking production system

no b d e r exists between machines. Therefore, jobs have to wait on the machine till the

succeeding machine is free. Nwwait production is basicdy a blocking production system

except a job cannot wait on any machine in nwwait production system.

In m a s production case, items are produced in lots usually called economic lot. Lot

streaming problems consider splitting of these lots (jobs) into smaller sublots so that

machines idle time and the completion time of the lot is minimized. As in many cases,

there are different versions of lot streaming problems that exist according to the sublots'

nature. In discrete version sublots can have only integer number of items, whereas in

real version a sublot can have fraction of items. Further more, if the sublot sizes vary

between machines, it is called the variable version, and if the sublot sizes are maintained

across the machines, it is called the consistent sublots. There is one more possible version

regarding the lot process. When sublots of a given lot are processed continuously, it is

called continuous processing.

In addition to the advantage of minimum machine idle time and minimum average

inventory, lot streaming makes many organizations cope with the modern manufacturing

technique of JIT, wbere each sublot can be t reated as a kanban card. Furt her , lot streaming

can be accommodated in traditional production planning and control models without any

difficulties. These inherent merits of lot streaming have attracted many recent researchers.

This thesis studies a production environment with two machines, where products are

produced in lots and a lot must be processed from start to finish without any interruption

on or between machines. Further, each product requires two operations which can be done

in any order. Thus, this system can be referred to as a two machine n-wait openshop

maas productions system. We try to apply lot steaming technique on this production

system so that the makespan can be minimized. For the single product case, we try to

use 41 the number of sublots available, if possible. In the multi-product case, for each

product with its lot size, we find the optimal sublot sizes, the number of sublots from a

given upper bound on the nurnber of sublots and the job sequence that would minimize

the makespan. The newait constraint restricts the sublots to be consistent amoss the

machines. This problem is motivated from a real world problem encountered in integrated

chip manuficturing process in semi-conductor industries, where the chips under go a final

testing at high temperature in lots loaded on boards by automated testing equipments.

The boards are specific for most of the cases. There are different testings have to be done.

The order of testing is not specific. Further, the tirne required to test a lot depends on the

number of chips and the type of chips on it . The temperature requirement, the nonspecific

order of testing and the board specification for the chips demonstrate this problem as a

newait openshop lot streaming problem.

The problem of minimizing makespan in a two machine newait openshop is known

to be unary NP-hard (Sahni and Cho 1979). This result also applies to our problem if

only one sublot is allowed for each product. Therefore, lot streaming in openshop newait

problems with one sublot is unary hard. Thus, it is unlikely that our problem can be

solved by an optimal efficient algorithm.

This thesis is orgaaized as follows. We give a brief survey of literature on lot streaming

in Chapter 2. The notations and assurnptions of our mode1 appear in Chapter 3. Chapter

4 studies, the single product continuous case with given maximum number of sublots.

The analysis on continuous-sized sublots may give some insights for integer-sized sublots.

Further, the analysis can be appiied to integer problems when the number of items in a

lot is so large and the number of sublots are few. This chapter contains scheduiing and

algori t hm developrnents, and defini tions of several possible su blot structures for scheduling

a single product. These definitions are also useful in considering multiple products. We

consider the single product continuous case into different cases and provide algorithrns

which give optimal solutions for each case. We show that equal-sized sublots will not

always be optimal, but the equd-sized pain will be optimal in most of the cases. Even

though it is the maximum number of sublots, we assign maximum number of sublots, if

possible. Integer case is difficult to solve. However the dynarnic programming algorit hm,

which is provided for the multiple product integer case in Chapter 5, can also be utilized

with some modifications to solve the single product problem requiring integer sublot sizes.

Chapter 5 shows that the multiple product problem can be formulated as a large

traveling salesman problem (TSP) for which the number of sublots for each product is a

decision variable. The optimal schedule for individual products may not be optimal for

the multiple product case. Therefore for each product, we consider di possible scheddes

using dynamic progratnmjng algorithm. Then each product is consideted as a country and

the profiles of a product are considered as cit ies of the country. Now it can be treated as

a speciai case of traveling salesman problem where salesman has to visit only one city in

a country. We develop a cost matrix to restrict the travel only to one city in a country.

In Chapter 6 we provide detailed computational tests of our algorithm's ability to

solve the scheduling and lot streaming problem and in Chapter 7, we discuss some possible

extensions that can be derived fmm our algorithm Global. Chapter 8 contains a conclusion

and some suggestions for future reaearch. The "Cn coded dynamic programming algorithm,

a simple output of the dynamic programming and the TSP algorithm, and the schedule

of the output are given in the Appendix.

Chapter 2

Literature Review

In monufacturing orgznizations, job (product, lot) scheduling is one of the most important

tasks. Two major faetors to be considered in scheduling are demand for the product and

the resources (machines) availability. Usually schedulers assume, no machine can process

more than one job at a tirne. Each job requires different machines to procesa in different

or in the same order depending on the shop environment. Hence the scheduling problem

is to schedule jobs auch that each machine works on at most one job at a tirne.

Two major objectives of any manufacturing organizations are customer satisfaction

and profit rnauimization. They both are generally interrelated, Le., increase in customer

satisfaction results increase in turnover. The high turnover leads to lower unit cost due

to economies of scale, and gives high profit margin. Thus, the lower unit cost and high

sales volume booms the profit. On realizing this fact most of the organizations try to

satisfy customer requirements as much as possible to survive in the cornpetitive market

environment. High qudity products at lower prices, and delivery within due dates are the

critical requirements to be satisfied. Therefore, in order to rneet customer requirements,

products must be available when they axe required.

However, prducing a product as and when required will not be desitable if some

machines are commonly used by many jobs each of which ha9 a high volume of demand, and

the setup time for eacla product on machines are hi&. Then reseuchers proposed economic

lot size for eaeh product so that the total production and inventory cost is rninimized. In

traditional lot production, a lot is indivisible and once an operation is started on a lot,

the fidl lot is completed before it is tramformeci to its succesaor operation.

Even though the economic lot size yield minimum total production and inventory cort,

it has ita own demerite. The economic lot production needs relatively longer time to finish

the lot. Thezefore, prduct lead times are high which in turn requires high safety stock.

As o result average inventory is increased. If the lot size is s m d , the lead time will be

reduced, but the setup cmt will be increased.

In a multi-machine production environment, if an economic lot which consists of a

number of identical items (units), can be split into smdler sublots and each sublot is

transferred to its successor operation immediately upon complet ion, the lead time will be

reduced while maintainhg the minimum setup cost. This technique of splitting a lot into

smaller sublots to allow overlapping of operations so that the completion time of a job is

minimized is called lot streaming. Even if the overall criterion is something other than

the makespan, malespan minimization of individual jobs improves performance. We see,

how lot streaming minimizes the mdtespan, through an example on single job flowshop

model. If a job consists of 100 items and there are four machines wit h unit processing time

requirements of 8,3,6 and 4 time units, the job completion time will be 2100 time units if

no lot streaming is dlowed (Figure 2.1). However, if the job is split into two equal sublots,

the cornpletion time will be 1450 time units which is shown in Figure 2.2. It is a 30%

reduction in the time requirements. If the number of sublots are increaseâ, the percentage

reduction will be increased. However, the improvement is diminishing with the number of

sublots. Baker and Jia (1993) test the makespan irnprovement as a function of the number

of sublots for different versions of the lot streaming problem for three machine problems.

Lot streaming makes many organizations cope with the modern manufacturing tech-

nique JIT where each sublot can be treated as a kanban card. Further, lot streaming

can be accommodated in tradi tional production planning and control models wi t hout any

difficulty. For example, in the MRP system, order releaae date of sublots can be estimated

as usual. The inherent dvantages of lot streaming has attracted many recent researchers.

Unlike preemption of job, lot streaming deals with a particular job. There have b e n many

solution methods pro@ by difTerent researchers with their own asswnptioas. Before a

100 units I

800 1100 1700

Figure 2.1: A basic flowshop model with 4 machina processing one lot.

M2

M3

M4

50 units 50 units

100 units

1 1 I

400 550 850 1050 1250 1450

Figure 2.2: A basic flowshop model with 4 machines processing two sublots.

100 units a

I I

I I

5û units units

100 mite

. . . . . . . . O . O O O . . . . . . . . . . . . O . O

I 1

50 units I I I I 1

50 units . . . O . ....

. O O . . . . .... o . .

50 units I I I I I l l I

50 units . Y . . . . . . O

discussing the solution techniques of different researchers, we give different versions of lot

streaming problems.

In m a t of the practicd cases, sublots are assigned for integer number of items which

is calleci the discrete version of the lot streiuning problem. in the continuous version,

sublots are assigned for na1 number of items. Generdly, problems of diwete version cm

be formulated as an Integer Linear Piogram(1LP). However, solving ILP is difficult with

large n m & r of variables. Therefore, reseders solve for the continuous version and

round off the solutions uring heuristics to solve for the discrete version. When the lot size

is very large with small number of sublots, rounding the solution obtained in continuous

version will be a reaponable solution. In addition, the optimal makespan in continuous

version aerves as a lower bound on the optima makespan in the discrete version, and the

rnakespan found by rounding the continuous solutions serves as an upper bound. Thue,

the analysis on continuous version gives some insights on discrete version.

In generai, sublot sizes vary between machines and which is cailed variable sublots.

However, there are many technological constraints that may aftect the formation and

movement of sublots. For example, in a newait production system, any item in a sublot

canaot wait on a machine or for a machine. Hence the sublots are restricted to maintain the

same items across the machines which is called the consistent sublots. Further, consistent

sublots are easy for analysis.

In most of the cases, sublots of a given job are processed continuously and it is called

continuous processing. In this case, only one setup for each batch is required. When the

setup is separable from the job, it is called detacheci setup. Detached setup can be done

once the machine becornes available. However, if the setup is not separable from the job,

Le., attached setup, setup can be done only when the job is available. There have been

many lot streaming problems discussed in the li terature wit h different combinat ions of the

above versions.

The goal of the basic lot streaming problem is to find the optimal size of each of a prede

terminecl number of sublots that wiil minimize the objective funetion. The lot streamiq

problems, however, are difficult to hmdle mathematically because of the existence of a

large number of tramfer lots and the complicated interaction among lots, tranefer lots and

machines. Lot streaming wiil not improve the makespan or my other r@ar performance

measures when thue i i only a single machine production proass. Therefore, the simplat

lot streaming problem uises with two machine production processes uid m a t of the rc

searchers study two machine cases to get the insight of lot streaming and extend the resulta

obtained, if possible, to more than two machine cases. It is not always possible to reduce

the makaspan by lot streaming even for more than one machine. Sen and Benli (1998)

study two machine multiple job openshop problem and prove there will be improvement

on makeepan due to lot streaming oniy if the total processing time of any job on machine

1 and machine 2 is greater than the total processing time of al1 the jobs either on machine

1 or on machine 2.

Lot streaming approach reduces the average inventory in the system. Hence, the t r t

ditional Economic Production quantity (EPQ) will not yield minimum cost. Most of the

works on lot streaming assume a given lot size (EPQ) and fixed number of sublots. Szen-

drovits (1975) develops a model to estimate EPQ and the cycle time, which is the time

required to manufacture the complete lot, for a single job continuous processing with equal

sublot sizes in a flowshop production system. If the assumption of equal sublot sizes is

removed then the makespan may be reduced. Goyal (1976) Studies this model and find

the optimal sublot sizes.

Often in practice, the sublot size is kept the same across al1 the machines to preserve

the integrity of sublots throughout the process. Consistent sublots do not always yield the

minimum makespan except in some special cases. Potts and Baker (1989) prove consistent

sublots of single job in two machine Bowshop and three machine flowshop always yield

the minimum malespan. They prove that for the two machine Bowshop problems dl the

sublots of optimal sizes are geornetric, i.e., Xj+< = pXj, where Xi is the size of the ith

sublot and p is the ratio between time of processing a unit on machine 2 and machine 1.

Generdy the number of sublots are determined by the technologieal constraints such

as the number of pallets available. Thus most of the problems are studied on fixeù number

of sublots. Lot streaming problems are further restticted by transporter capacity. There

Figure 2.3: Network representation of a single job m machine flow shop probkm

are few papers which analyze the lot streaming problem with iimited transporter capacity.

Trietsch and Baker (1993) obtain an expression for optimal sublot sizes for the two machine

flowshop problem of continuous veniion with limited transporter capacity and provide an

algorithm for the discrete version.

Different mathematical tools and/or existing scheduling theories are used by different

researchers. Linear pmgramming mode1 and network mode1 are widely used mathematical

tools. Lot streaming problems in flowshops can be explained by a network. For example

a single job flowshop lot streaming problem can be represented by a network shown in

Figure 2.3, which contains a vertex (i, j) with weight piXi, for each sublot X j on every

machine Mi, where pi is the unit processing time on machine Mi. The directed arc from

vertex (i, j) to vertex (i + 1, j) restricts processing of sublot j on machine Mi+l only after

completion on Mi. The directed arc from vertex ( i , j) to vertex (i, j + 1) restricts to s t u t

sublot (j + 1) on machine Mi only after the completion of sublot j on it. Makespan is

given by the length of the longest path (critical path) in the network, where the length of

any path is the sum of the weights of the vertices on it. Any sublot corresponding to a

vertex on the critical path is called the critical sublot.

In a flowahop all jobs follow the same machine sequence, and each job has exactly

one operation on eoch machine. An openshop is similar to a flowshop, except that the

operations of a job may be petformecl in any order. Therefore in openshop problems,

we have to find the machine sequence in addition to the job sequence. A brief survey of

openshop scheduling problems can be found in Kubiak et al. (1991), Graham et al. (1979)

and Lawler et al. (1993). Useful general references on machine scheduling problems include

the books by Baker (1974) and Pinedo (1995).

Glsss, Gupta and Potts (1994) use network representation to solve single product lot

streaming problems in three machine flowshop and openshop problems and two machine

jobshop problems. They daim that their schedule possesses the newait and/or blocking

property tw. Initidy, three machine flowshop problem is solved and the result is useà

to get expressions for two machine jobshop problems. In two machine jobshop problem

with three operations, firat and third operations of a job are done on the same machine.

Therefore, an addi t ional const raint is insested t O prevent simult aneous first and t hird

operations of the same sublot. In the openshop case, routing of items through machines

must be specified in ddition to the sublot sizes. They prove that equal sublots of size

1 m for the first m aublots (when the number of sublots s is not less than the number of

machines m) and assigna zero items to the remaining sublots and the routing of sublot i

as ( i , ..., m, 1, ..., i - 1) will give optimal solution. Hence when the number of sublots is not

less than three, their model can be used to solve ( t w e and) three machine lot streaming

problem. They further, explain that the lot streaming of equal-sized two sublots, m

machine problem can be treated as m sublots and two machine problern which can be

solved by the algorithm of Gonzalez and Sahni (1976) to get optima sequencing. Thus,

their models solve three machine lot streaming problems.

Potts and Baker (1989) prove that for the two machine flowshop problems all the

sublots of optimal solution are criticd in the network model. Chen and Steiner (1998,

1995 and 1996) use network modela to solve lot streaming problems with discrete version

of the single job in two machine flowshops, three machine flowsho$s with detached setup

and three machine flowshops with attached setup respectively.

Baker (1993) utilizeil the existing theory for time l g and aetup times to analyze lot

streaming of equal sublots in two machine flowshop with setup times. In a two machine

Figure 2.4: Time lag model for a given job on two machine problem

problem with time lag, the processing of a job ( j ) can begin at machine 2 while the

later portion is king carried out on machine i as shown in Figure 2.4. di is deôned as

the transfer hg. Rom Figure 2.4, we found that there is an overlapping of operations.

Therefore, it c m be treated for lot streaming problems. Initially, for a two machine

flowshop problem, the aut hor finds the relationship between the Johnson alprithm (1954)

and the time lag model. Then he uses the result of this problem for the lot streaming

problems. He further discusses the extensions to more than two machines using the non-

bottleneck property used by Johnson for three machine flowsbop problem.

Usually, we assume there is no time to transfer a sublot from one machine to the

other. However, if the transfer times are large, then ignoring the transfer time will leid to

some emneous decision. Vickson (1995) provides a Linear Programming (LP) formulation

for the lot streaming problem of multiple products in a two machine flowshop with setup

t imes and sublot transfer time. He provides mathematical models for optimal lot streaming

and sequencing problems for continuous version with attached setup timeldetached setup

times andior lot transfer tirnes when regular performance meaaures are used, where regular

measure is a performance measure which is non-decreasing in the completion tirne. Finally,

the author preaents an algorithm for the integer version and illustrate it with an example.

Stephane and L~sliem (1997) provide a mathematical model to solve lot streamjng

problem with consistent sublots in general jobshop environment for a given number of

sublots. They daim this formulation can solve only s m d instances. Therefore, a t w c ~

level iterative proadun is proposed in order to determine the limits of a global a p p t d

precisely. The uppet level solves for finding the sublot sima for a given sequencing using

a Linear Programming (LP) model, and the lower level solvai for sequencing the sublots

with si= found in upper level with another LP model. Then an iterative procedure ici

used for mlving the above model starting with an initial sublot sizes. Since the iterative

procedure gives reai values for sublot Mzes, the authors presemt a rounding procedure to

get iinteger-sized sublotr. In rounding, they round the sublot si= to the cloeest integet for

al1 sublots except the lsst one. Then the last sublot size is found by deducting the total

size of d the sublots from the number of items available. They give a mixed integer LP

model to solve lot streaming problem when setup costs are involved. A heuristic is used to

solve the lot streaming (integer-sized sublots) and scheduling problems. In the heuristic,

the setup time L included in the processing time of each sublot, thereby the makespan is

overestimated. Using some experimental results on two sets of problems with and without

setup times, the authors conclude that small number of sublots is sufficient for jobshop

problems without setup.

In batch manuf'turing systerns, makespan is the cycle time. There are few researchers

who study the lot streaming problem to minimize some performance measures other than

makespan. Steiner and Tniscott(l993) consider the discrete version of lot streaming prob

lem of multi-machine batch processing system with equal-sized sublots and continuous

processing (one setup for each batch) with the objective of minimizing cycle time, flow

time ( the average length of the time periods during which units are in the system) and/or

the total processing cmt ( sum of work-in-process inventory carrying cost and the sublot

transportation cost). They obtain lower bounds for the objective d u e . If there is no

technobgical constraint , they argue, that unit size sublots will give optimal solution.

We study lot streaming and scheduling problem in openshop environment. In out

openshop model, ptocesaing of sublots is no-wait in the sense that the processing of a

sublot on a machine is commenceà as mon as the processing of the subIot on the preading

machine is finished. This may lead to machine idle times between successive sublots. In

a na-wait shop, esch sublot must be processed continuously fmm its start in the first

machine, to its completion in the last machine, without any intemption on machines and

without any waiting in between the machines. Therefore, n-wait lot streaming problems

require consistent sublota.

The characteristics of the procaising technology, or the absence of storage capacity

between operations of a job may couse! no-wait production environments. The scheduling

problems in newait shop arise in the chemicai processing ~ n d petrochemical production

environment. Anot her example of newai t situation arises in hot met al rolling industria,

where the metah have to be processed continuously at high temperatwe.

A considerable amount of interest has arisen in newait scheduling problems in the

recent years. Refer to the survey papers on no-wait scheduling by Hall and Sriskandarajah

(1995) and Goyai and Sriskandarajah (1988). This interest appears to be motivated by

applications as well as by questiona of research interest.

Let us consider the scheduling problems without lot strearning. Note that rninimizing

makespan in two machine newait openshop shown to be NP-Hard (Sahni and Cho 1979).

Thus, it is unlikely that the problem can be d v e d by optimal efficient algorithm (Garey

and Johnson 1978). More closely related to this paper are studies of scheduling problems

in newait flowshops. Gilmore and Gomory (1964) provide an efficient algorithm for the

two machine newait flowshop. Rock(1984) shows that the similar problem with three

machines is intractable.

Hsu and Stein, (1991) study a real world newait problem in chemicd manufacturing

system called 'onodizing line" which is made up of a series of chemical processing tanks.

The line produces many types of products for the commercial and automotive industries

auch as pipes, trim and truck grilles. The objective here is to produce the daily production

requirements of various products in the shortest possible time. In front of the line is a

racking iuea, where items of a product are loaded onto racks prior to chemical processing.

Since the shapes of prducts are different, each product ha9 its own racks. The number

of racks available for a product is limiteci. The products need to be processexi as newait

for two reasom : (i) there is no bufiat between tanks (ii) once a rack exits a tank it must

immediately enter the next one or the products will be ruined due to deterioration of the

items while expoaed to the atmosphete. Since it is a chemical process, the processing time

of a ta& (or a sublot) in a tank depends on the total surface area. Hence the procesahg

tirne in proportional to the number of items in a rack. There is a setup time to change

the chernical compoaitions of tanks when changing from the processing of one product to

another product .

Sriskandarajah and Wagneur (1999) study lot streaming in two machine newait flow-

shops with multiple products. An LP mode1 is formulated io find the optimal sublot sizes

for the single product esse. They use Gilmore and Gomory algorithm to get optima se-

quencing of multiple products with given sublot sizes and prove tbat the optimal sublot

size of a single product will be optimal for the simultaneous problem of lot streaming

and seguencing of multiple products. They provide heuristics for integer version of sin-

gle product and multiple product cases, and carryout computational experiments for the

performance evaluation of their heuristic algorithms. Findy, a tabu search technique is

given for the simultaneous problem of finding the number of sublots, integer sublot size

and sequencing of multiple products.

No atudies on lot streaming in newait openshop is found in the literature. This is the

topic for my thesis project which will be described in the next chapters.

Chapter 3

Notations and Assumptions

Data

Mi, M2: the two machines in the openshop.

Pj, j = 1 , . . . , N: the products to be produced.

aij bj > 0: the processing times of unit product Pj on machines Mi and Ma, respectively . nj: the maximum ailowable number of sublots for product Pj.

Wj: the number of units demanded for product Pj.

sj: the setup time of p t~duct Pj on Mi.

t j : the setup time of product Pj on Ma.

Variables

Xij: the totd number O€ unita of product Pj in ~ublot i , where Xij 2 1 and integer.

pj: the optimal number of sublots for product î'j (a decision variable 2 5 pj 5 nj).

A~sumptionr

1. Al1 producta are ;rMilable at time zero.

2. Each product has a specified maximum number of sublots.

3. The processing of sublots is newait.

4. A product's sublots are consistent.

5. Setup times are independent O€ the product sequence, and setup times are attached (in

attached setup times, setup can be done only when the product is available).

6. Setup times and processing times on machines are detenninistic. There is no breakdown

of machines.

7. There is no limitation on the p d e t capacity, i.e., tbere is no limitation on the Xkj

value8.

Asaumption 2 apeeifies thzt the maximum number of sublots for each product is fixed.

This aasumption is vllid for a production environment in which the number of palleta

is limiteci, and each product is allocated a specified number of pallets to transport the

sublot S.

Assumption 4 is a necessary condition in no-wait scheduling since in no-wait scheduling

items are not allowed to wait on machines or in between machines.

Chapter 4

Lot Streaming: Single product and continuous sized sublots

4.1 Introduction

The analysis on continuous-sized sublot may give some insights for integer-sized sublots.

Hence we analyze lot streaming of continuous-sized sublots and develop heuristics for

integet-eized lot s t reaming. Fur t ber, the anal ysis of cont inuous-sized sublot can be applied

to real problem, when the number of items in a product lot is so large (and the number of

sublots are few) that the product lot c m be treated as infinitely divisible or rounding off of

the continuous size will not affect the objective value considerably. Note that each product

has a specified maximum number of sublots according to assumption 2. This means that

the maximum number of sublots is fixed for each product. This assumption is d i d for

the production environment in which the nurnber of pallets is limited and each product

is provided a specified number of pallets to carry the sublots. It is convenient to handle

small lots in production environments. Thus, items need to be distributed in the &en

number of sublots. In order to distribute the items evenly, it is desirable to have positive

number of items in each sublot and maintain equal sublots whenever possible. We consider

the production case where setup times are attached with the processing and setup is done

once at the start of pmessing.

For convenience, we c d the first two sublots the first pair, the next two the second pair,

and so on. We can use any even number of sublots for the optimal solution. However in out

derivation, we use dl the sublots if the maximum number of sublots is even. Otherwiae,

we use one less number of sublota. As can be seen later that it is convenient to deal

with the problem with even n. In the openshop problems, we c m have multiple optimal

solutions. As can ba aeen later that the equal-sized sublots will not always be optimal, but

in mort cases, equai-sized pairs of sublots will give optimai solutions. ln'equ&sized pain

of sublots, eaeh pair of sublots are assigneci the sune number of items (5). We mdyze

the problern into t h main casai. Case 1 analyses the problems where a = 6. Case 2

analyses the problems where o > 6, and Case 3 analyses the problems where a < b.

4.2 Scheduling and Algorithm Development

In our discussion, we shall denote by aXi (bX;) both the time for sublot i on machine

Mi (machine M2) and the name of the operations itself. The use will be clear from the

content.

Definition : a 6 w y schedule is an optima schedule for an even number n = 2k of sublots,

and hss the following properties :

pi: Cma, = max{Li, La), where Li = s + a C:=, X; = s + a W , and 4 = t + b& Xi =

t + bW. f i : There are k time points T I , . . . , ~k such that for 1 5 i 5 k, operations and bX2i

end at time Ti.

pl is clearly a suflicient condition for optimality. It is easy to see that it is also necessary

for optimality, because for the optimal makespan, either machine 1 or machine 2 must not

have any idle time. The schedule structure satisfying the property p2 is called p2 structure

(or alternating pair schedule). There aie two ways to schedule sublots, single routing and

multiple routing. In single routing all sublots of a given job follow the same machine

sequence. Thus, it represents a flowshop structure. Then the optimal sublot sizes are

geometric as given by Potts and Baker (1989) and we cannot get equd-sized or equal-si4

pair sublots in any case unleas a = 6. In multiple routing sublots of a job can follow any

machine aequence. f i stmcture lies in multiple routing. Because of this multiple routing

Ml 01 . ml Ma m . - 1

Tl T1 ri,

Figure 4.1: A buey schedule for Case 1. a = b, n = 2k, Cm, = s + Wu.

it has the flexibility to get equal aublots &es or equd pair si= in many cases. Reader

c m see this flexibiüty of the f i structure through out this thesir. Therefore we introduce

the a, structure which tries to satisfy our quality requinment of equd-sized sublots as

much ae possible. We present algorithms to find the optimal sublot sizes for even number

of sublots.

Theorem 1 For a single product and an even number of sublots, there ezists an optimal

schedule that satisfies both properties pl and pl .

Proof: The relative d u e s of a, b, s and t define severai cases. Lemma 1 to Lemma 5

prove that the theorem is true for aii c e s . This is done by deciding the lot sizes and

constructing a schedule with property pl so that the property pi is satisfied.

Lemma 1 An optimal solution X' = (XI, Xi, . . . , X;) in Case 1 is given &y a busy

schedule, tuhen

Proof: Since a = b,using the sublots X , = n,i ' = 1,2, ... n, a busy schedule c m be

constructeci as in Figure 4.1. Note that there is no idle time on machine 1 or machine 2

during proceseing. However, there may be some idle time either on machine 1 or machine

2 in the beginning of the schedule which is unavoidable.

Since the properties pi and pl are satisfied, it is a busy schedule and is optimai. O

,ml 9

Mi, bXn-1

Ti '0 7,

Figure 4.2: A busy schedule for Case 2.1. a > b,s 2 t ,n = 2k,Cm, = s + Wa.

Case 3: a > b.

When s 2 t, no idle time on machine 1 will give an optimal schedule, whereas when

s < t, we may neeà to conaider no idle tirne on machine 1 and/or machine 2 depending on

the situation. Hence, we divide Case 2 into furt her two subcases. Case 2.1 analyses when

s 2 t and Case 2.2 anaiyses when s < t .

Case 2.1 a > b , s 2 t .

Lemma 2 An optimal solution X' = (Xi, Xi , . . . , X:) in Case 2.1 is giuen by a bwy

schedule, when

Proof: Using the sublots Xi' = 5, i = 1,2, ... n, a busy schedule as shown in Figure 4.2

can be constructed for Case 2.1.

Let sublots r and (t + 1) make a pair. Then,

b Also X,+i 2 -X,p = 1,3 ,..., n- 1

a

Therefore, !x, 5 Xr+i 5 fXr, where f < 1, and % > 1.

If we have Xr+i = y&, where 5 y 5 t, we can have optimal lot sizes. Since 7 = 1

f d s in the range (!, f), equd sublot size will be optima. Since the schedule satisîles the

properties pi and pl (Figure 4.2), it is a busy schedule and optimal. O

If Wa + s > Wb + t, there should not be any idle time on machine 1 whereas if

Wu + s < Wb + t there should not be any idle time on machine 2 in the optimal schedule.

When Wa + s = Wb + t, both machines must be fuily u t i l i d . Therefore, we anaiyze

this case into another three subcases. Case 2.2.1 analyses when Wu + s > Wb + t , Case

2.2.2 analyses when Wu + s = Wb + t and Case 2.2.3 analyeee when W u + s < Wb + t.

In order to deal with remaining Casea, we define A, as the difference between time when

machine 2 is ready for processing the rth sublor pair and the time when machine 1 is ready

for the rth sublot pair in the schedule satisfying pz. For example, initially, for the first

pair, Al = t - s, when there is no sublots scheduled but the settings have been pedormed

starting at time zero. Then for the second pair, A2 = bXl - a&, when the first pairs of

sublots have been processed conforming the schedule structure pl.

Case 2 . 2 . 1 a > b , s < t , W u + s > W b + t .

In Case 2.2.1, the lot sizes can be determined by the algorithm Lot1 given below which

provides a busy schedule. The algorithm Lot 1 assigns items to sublots such t hat machine

1 is kept busy. This is done in three steps corresponding to different conditions. Thus For

the firat pair, the setup time of machine 1 and the processing time of the first sublot on

machine 1 must be greater than the setup time of machine 2 and the processing time of the

second sublot on machine 2. Step 1 therefore, assigns equal-sized sublots if this condition

is aatiefied. If the first pair satisfies this condition, the remaining pairs will satisfy the

condition. The reader can see it in the proof given next to the algorithm. Step 2 checks

whether the setup tirne of machine 1 and the processing time of the first pair on machine

1 is greater than the setup time of machine 2 and the processing time of the first pair on

machine 2, if equal-sized pair is use& If this condition is satisfied for the first pair, it will

be satisfied for the remaining pairs too which is shown in the proof. Step 3 assips the

first pair so that there is no idle tirne due to this pair and the remaining sublots satisfy

condition for Lemmr 2. Since the condition for Lemma 2 is satisfied, it qually assigne the

remaining items to the remaining sublots.

Algorithm Lot l

Step 1 If 2 t - J, then set

x; = - 1 . . ; Stop. n

Step2 I f m z t - , , t h e n s e t n

x;=*+$ , X;= n(a+b) 2wa !S. a+b'

w-x:-xi - w x; = - - n-a , r = 3,4, ..., n; Stop. Step 3 Set X i = X i = -w~ (a -b )

W-x:-xi = W - x: = !-a n-2 n-2 (a-b)(n-2) 9 r = 3,4, ..., ta; Stop

Lemma 3 An optimal solution X' = (X;, Xi, . . . , X:), obtained b y the algorithm Lotl,

in Cuse 2.2.1 is giuen by a b w y schedule.

Proof: Let the idle tirne on machine 1 before uX2t-l is started on it be Il and the idle tirne

on machine 2 before bX2, is started on it be 12. Figure 4.3 describes a typical schedule

of sublota X2,-* and X2, . We have to show that the algorithm maintains Il = O for d l

sublot pairs.

The optimal schedule in generd for Case 2.2.1 is shown in Figure 4.4.

Step 1: When r = 1,

Ii+aX; = 12+bX;+At .

f2 = Il +uX; - bX; - ( t -s).

Since equd sized sublots, I2 W(a - 6) ri + - ( t - s).

n W(a - 6)

Note that t - s n

Therefon, I2 2 Il

A2 = bX; -aX;

Thuii, machine 1 can be fuily utiüzed by setting Il = O for the first pair. Since A2 < 0,

from Lemma 2, equal aublot size is possible for the remaining (n - 2) sublots.

Step 2: When r = 1,

Ia 2 W b t - s 2Wa t - s

--1-(t-8) = l i+ ' [n(a+b) + a l - b [ n ( a + b ) a + b

Since Il = O is possible, machine 1 can be fully utilized. Further, AI 5 O . Therefore,

from Lemrna 2, we eau have equal size for the remaining sublots.

Note that W(a - 6) > t - s (condition for Case 2.2.1). Therefore sublot sizes for the

case n = 2 is decided at least in Step 2. Hence the condition in Step 3 may occur for the

case with n > 2.

Step 3: #en r = 1,

f2 = I l + a X ; - b X ; - ( t - s ) (t - s)a ( t - ~ ) b

= I l + a - 6 (a' - bl) - (t - 3)

(a2 - P)

"WI 9 9 m

& bXn bXm-1

Tl "2 a F i p 4.4: A busy schedule for Case 2.2.1. a > b,s < t, Wu +s > Wb+ t, n = 2k, Cm, = s + wu.

Therefore machine 1 c m be fully utilized by setting Il = O for the first pair. Since

A2 = bX; - aX; = 0, froxn Lemma 2, equal sublot size c m be optimal for the remaining

sublots. Further, we need to show that the total items allocated for the first two sublots

are strictly les8 than W and al1 the items are assigned.

W(a - b) >

w >

w >

w >

Therefore, W >

Number of items assigned =

- -

- -

t - s

t - s - a - b

" w t - s X; + + C(n-2 - - b)(n - 2)

r=3 )

(1 - s)(a + b) t - s +W--

a2-b3 a - b

Case 1 . 2 . 2 a > b , s < t , W a + s = Wb+t

The optimal solution of Case 2.2.2 should maintain both machines busy. Figure 4.5

shows an optima schedule for the Case 2.2.2. Algorithm Lot2 provides a busy schedule

for this case. Since both machines have to be fully utilized, equal-sized sublots is not

possible. Therefore, the algorithm tries to assign equal pair size, if possible. We dehe

another term î, as the numbet of items left to be assigned once the rth pair ie scheduled.

When the rth pair is to be scheduled, there will be items remaining to be assigneci to

the (2k - 2r + 2) sublots. The algorithm Lot2 decides lot aizea for the construction of

type schedule such that both machine 1 and machine 2 are busy thtough out. This is done

in two stepe comesponding to different conditions. When the (r - 1)" pair is scheduled,

Step 2 tries to assign equd-sized sublot pain. This step is similar to Step 3 in algorithm

Lotl. It can be recalleù that 1 = k. If the equal pair is not possible for the rth pait, Step

3 assigis the rth pair ruch tbat both machines are busy and the total number of items

is lm than the total number of item8 available when the rth pair is not last. If it is the

last pair it select the aublots such that no idle tirne is ineerted on uly machine and dl the

items are assigned.

Algorithm Lot2

Step 1

Step 2

Step 3

Step 4

Lemma 4 An optimal solution X' = (X;, Xi,. . . , X:) in Case 2.2.2 is given by the

algorithm Lot2

Proof: We have to show that in each step Il = I2 = O is poesible for each pair, = O

and the condition for C e 2.2.2 is hold for al1 pairs.

Step 2: Since W(a - b) = t - s, the case n = 2 wiii fa11 in this Step.

Therefore, both machines can be maintained busy by setting Il = I2 = 0.

Therefore, Ar+, = rr(a - b) will be tme, if A, = i',& - b). We are given Ai =

r0(a - b). Hence by mathematical induction, al1 remaining sublots and the remaining

items will satisfy the condition for Case 2.2.2.

When r = k,

Step3: Case r < k

Therefore, both machines can be kept busy by set ting 4 = h = 0.

Ar(2a2 - b2 + ab) XG-, +Xir =

2a(a2 - V)

- -

But =

Therefore, A =

A,+l = rr(a - b) is tme, if Ar = rr-i(a - b). Further ro(a - 6) = Al. Therefore, by

mathematical induction, the condition for Case 2.2.2 is satisfieà for al1 remaining items

and sublot pain.

Figure 4.5: The busy schedule for Case 2.2.2. a > 6, s < t , Wa+s = Wb+t, Cm, = Wa+s.

Step 3: Case r = k

Therefore, both machines can be kept busy by set ting Il = I2 = O.

but Ak = r k - L ( ~ - b )

Thus al1 the items are allocated. O

Case 2.2.3 a > b,s < t , W a + a < W b + t

The sublot sizes for Case 2.2.3 cm be determined by the algorithm Lot%.

Algorithm Lot3

Step 1

Step 2

Step 3

Step 4

Step 5

Step 6

Else set

- Ar(a+2b) + I'.-I (a+ x à r - ~ 2a(a+b) 2(a+b) ' Go to Step 6.

If r = k, then set r a C ' b

X;iC-1= 3, Xik = I=L; a t b

Else set - rr-1 (2a2-b2j + rr-l (2a-b)

Xàr-1 O 4+ t b ) 4a 3

Set,

Stop.

Lemma 6 An optimd solution X' = (X;, Xi, . . . , X:), obtained b y the algorithm Lot3,

in Corc 2.2.3 U giucn by a bus y schedule.

Figure 4.3 describer a typical schedule of sublots X2r-1 and Xa by dgonthm Lot%

Prwt: We have to show that the algorithm maintains I2 = O for all pairs, and if a pair

fds in Step 2 or Step 3 then the remaining pairs fall in the aame step.

Step 2:

Thus Il = O when I2 = O. Hence, machine 2 can be fully utilized.

LI BU^, r, = - k - r + 1

Therefore > rr(a - b)

Thetefore by mathemat ical induction, remaining items will satisfy the condit ion for

Case 2.2.3.

But, z', = k - r + i k - r -

L',a Tberefore A,+i < -

k - r

r O Therefore, if A, < &, then Ar+i < z. Further, Al < y. Therefore, by mathe

matical induction, the remaining items and sublots will satisfy the condition for Step 2.

When r = k,

Hence dl the items are assigneci.

Ah+l = A& - r k - l ( a - b)

> O

Step 3:

11 > 1 2

Hence, machine 2 can be fully utilized.

M h e r , Al > ro(a - 6)

Therefore, by mat hematical induction, condition for Case 2.2.3 is sat isfied.

rra Therefore. -

Therefore, by mat hematical induction, the condition for Step 3 is satisfied.

Therefore, ail the items axe assigned.

Therefore, machine 2 can be kept busy.

Therefore, dl the items are allocated.

Therefore, I2 = O is possible.

Ar But - 2

>

Therefore Ar+i >

Therefore, if Ar > l'r,l(a - b), then Ar+* > rr(a - b). Rirther, AI > îo(a - b). Hence,

by mathematicid induction, the condition for Case 2.2.3 will be satisfied by all remaining

sublots and items.

Step 5 Case r = k

Therefore, machine 2 can be fuîiy u t i l id .

Therefore, ail items are assigned.

Step 5 Caae r < k

- ï ,-l (361 - 4a2 - ab) - 1 2 + 4(0 + b) + Ar

Therefore, Il > I2

'I 'a T&

Figure4.6: A busy sehedulefor Case 2.2.3. a > b,s < t , Wa+s < Wb+t,n = 2k,Cm, = s + Wu.

> r , ( a œ b ) .a

Figure 4.6 depicts an optimal schedule.

Figure 4.7: Alternating Pair Structure.

Figure 4.8: Flowshop Structure.

C a s e S a d .

Problems of Case 3 are the mirror image of the problems of Case 2. Therefore, optimal

achedules are similar to those of Case 2.

4.3 Different Structures of Schedule

Different structures can be generateà for a given product. These different structures play

an important role in multi-product scheduling. We can generalize that there are three

possible lot size structures:

1. Altemating pairs (p2 schedule). An alternating pair schedule is one which has the

following properties: there are k time points T I , . . . , sk such that for 1 5 i 5 k, operations

u X ~ ~ - ~ and bXai complete at time ri, as shown in Figure 4.7.

2. Forward Bowshop. Here ali sublots of a given product follow the sequence Mi + Ma, ap shown in Figure 4.8.

3. Backward Bowshop. Here al1 sublots of a given product follow the sequence M2 -t Ml,

as shown by considering the reverse schedule in Figure 4.8.

The above three stmctures c m dao be combined into a hybrid structure. Figure 4.9

shows a hybrid structure where sublots 1 and 7 are processed as forward flowshops, sublots

Figure 4.9: Hybrid Structure.

2,3,5,md 6 are procemed as altemating pairs, and sublot 4 is processed as a backward

flowshop.

4.4 Conclusion

We have shown that there exists a busy schedule for each case. Hence there is aiways

an optimal schedule for the single product continuous-sized sublots. We have used the

maximum number of sublots dlowed or one less than that value if the maximum number

of sublots is odd. However, optimal solutions can be obtained with any even nurnber of

sublots in the given number of maximum sublots. This is obvious by substituting n by

any even number which is less than n. Thus, single product continuous case c m have

multiple optimal solutions. Equal sublot size will not always give optimal schedule, but

the equal-sized pairs give optimal schedules in most of the cases. The algorithms we

provided for different cases c m be solved in polynomial time. Further, these algorithms

can be successfully used for integer case, when the number of items are very large and the

nurnber of sublots are feu so that the effect of rounding off is negligible.

Chapter 5

Multiple Products

5.1 Introduction

Sriskandatajah and Wsgneur (1999) explain how the schedule of a given lot can be divided

into a head (H), a body (B) and a tail (T) for a newait flowsbop problem. The head is

the beginning portion of the schedule, in which machine M2 (respectively, machine Mi)

is ide throughout; the tail is the laat portion of the schedule, where machine Ml (resp.,

machine Ma) is idle throughout; the body is the middle portion of the schedule between

the head and tail. In openshop problems, we can have a head or tail on either machine

(see Profiles 1,2,3 and 4 in Figure 5.1).

In this chapter, we develop an algorithm to find al1 possible profiles. Then a cost matrix

is generated and a TSP algorithm is used to get optimal product profile and product

sequence. We consider the case without setup time. However, the algorithm can be

successfully applied for the case wi t h set up t ime wi t h small modification.

Problern F&ao - waitlC,. can be formulated as a solvable case of the TSP with

the cost matrix: cj j+l = max{O, pj - aj+i}, where job j is followed by job j + 1 in the

production sequence, and aj and Pi are the processing times of job j on mathine 1 and

machine 2, respectively. cj,j+i is the idle time on machine 1 after completion of job j on

machine 1 until the start of job j + 1 on machine 1. cj,j+i can also be interpreted as the

cost of traveling ftom city j to city j + 1 in TSP terminology. The algorithm of Gilmore

and Gornory (1964) minimizes the makespan in TSP instances with the above cost matrix.

We now show that, for givcn p d u c t pmfiles, an optimal product sequence for our problem

ean be obtained using the algorithm of Gilmore and Gornory (1964).

Profle 1: Hi 2 0, Ti 2 O. Profile 2: Hi < O, Ti 5 0.

Profile 3: Hi 2 0, Ti < O. Profüe 4: Hi < O, > O.

Figure 5.1: Profiles of Openshop Nwwait Lot Streaming Problems.

5.2 Scheduling and Algorithm Development

In this section, we andyze how product sequence can be obtained for given profile of each

product, Le., when the number and the sizes of sublots are known for each product. If

the sublots sizes are given for products, Gilmore and Gomory algorithm gives the optimal

product sequence. For the simultaneous problem of sublot sizes and product sequence, we

develop a dynamic programming algorithm to get al1 the profiles and use a TSP a.lgorithm

b select the optimal profile of eacb product and product sequence.

5.2.1 Scheduling of Products with Given Profiles

When the profile of each product is known, i.e., the sublot sizes are already selected, we

need to find the Optimal product sequence. The algorithm of Gilmore and Gomory (9164)

provides optima product sequence in this case.

Theorem 2 For given product profiles, an optimal product sequence for the two machine

no-wait openshop ptoblem with lot stnaming can be obtained using the algorithm of Gilmon

and Gomoy (1964).

Prwf. In order to prove this result, we let heads (a,) on machine Mi and tails (Pi) on mûchine 4 as pi t ive , and heads on machine Ma and tails on machine Mi as negative.

Then, we show that product sequencing problem in newait openshop can also be for-

mulated p(l a TSP with the COS^ matrix cj,j+l = ma{O,oj - ~ t ~ + ~ ) . We prove that this

cost matriv is correct for dl possible combinations of profdee, where product j + 1 follows

product j .

If Ti = pj 2 O and Hj+i = aj+l 2 O, then produet j has either profile 1 or 4 and

product j + 1 hm either profile 1 or 3. The idle time OR Y is cj,j+l = max(0, Ti - Hj+i} = mw(0, - ai+*)* If Ti = Pi 2 O and Hj+l = -Qj+i > O, then produet j has either profile 1 or 4, and product j + 1 has either profile 2 or 4. The idle time on Ml ii,

cj,j+l = Tj + Hj+l = Pi - aj+i = mw(0, pj-aj+l)* If Ti = opj > O and = a,+l 2 O,

then product j has either profile 2 or 3 and product j + 1 has either profile 1 or 3. The

idle tirne Ml is cj,j+l = O = rnax{O,pj - crj+l). Finally, if T j = -Pi > O and

Hj+l = -aj+i > O, then product j has either profile 2 or 3 and product j + 1 bas either

profile 2 or 4. The idle time on Ml is cj,j+l = max(0, - Ti) = max{O, Pi - aj+l).

Thus, the cûet expression cj, j+l = mw{O, pj-aj+l) is volid for al1 profile combinations.

Hence the algorithm of Gilrnore and Gornory (1964) finds optimal product sequence in a

two machine no-wait openshop for any given product profiles. O

5.2.2 Simultaneous Lot Streaming and Scheduling

When the profile is not given, then we have to find the optimal number of sublots, sublot

sizes and the product sequence. We develop a dynamic programming algorithm and then

incorporate it with a TSP algorithm to solve this difficult problem.

We let Ah and At are the length of head and tail, respectively of a given product profile.

Note that we have to consider al1 possible profiles of each product for Our scheduling

pmblemi, with multiple products. First, the different profiles for each prduct must be

found, as explained below.

Given a product profile for each of N products, let us consider a product sequence a

which is the order of product indices, i.e., products are scheduled in the order: P,(o),

E ( 1 ) - , Pu(,-,) , , Pe(r+l) , - - - , K(N) 9 P'(N+I) as in Fi W e 5-29 where ~ ( i ) denotes the

index of the product that is scheduled in the ith position of the sequence a. P,(o),

Figure 5.2: Example Schedule of N Products in Sequence o.

are artificial products having zero length of head, body and tail. For convenient, we define

süghtly different cust matrix than the one described above, that is, distance from city

(product) @(i ) to City ( p r d ~ ~ t ) @(i + 1) is du(i)s(i+~) + (&(il + &(i+1))/2 where dU(i),(i+i)

is the time between the start of tail of product Pu(i) and finish of head of product Pg(i+ll

(see Figure 5.2). The reamn for the choice becomes clear shortly.

Our problem is now simultaneous selection of profiles of products and scheduling. In

order to get the optimal schedule, we introduce a TSP formulation. Each product is

repreaented by a country, and each profile for each product is represented by a city in

that country. Then o u problem is: Starting and finishing at a given city in a given

country, visit each country exactly once and visit exactly one city in each country. Let

Bi, denote the cwt to visit city i, of country i , and let di,, denote the travel cost from

city i, to city ji. Since a TSP tout visits each city in each country, we need to d a p t

the TSP matrix to ensure that, on entering country i at city i,, the tour next visits cities

i,+i, iq+2, . . . , iq+k.i(dki), where ki is the number of cities in country i , at zero cost.

Note that io = iki. This is accomplished by specifying zero costs for that sequence and

prohibitively large coetr (M2) for any other sequence within country i. Furthemore, since

the tour wiu leove country i from city ic+ki-i(di) when it enters that country at city i,,

the coet of travel from city ig+ki-l(,,,oditi) must be specified as Ci,?, . This is accomplished by

setting 4 , = di,,(a)tjl + + Bjl)/2], where the term in square brackets ia

hslf the total size of the bodies of i, and j,. Finally, reentering a given country is prevented

by the addition of a large eost (M) when traveling between any pair of countries. The

Countryi. Countryj

Figure 5.3: Example With Two Countries and Five Cities.

optimd cost, C, will satisfy NM < C < (N + 1)M.

Example

Consider two countries i and j with thme and two cities respectively, as in Figure 5.3. We

obtain the foliowing TSP cost matrix.

For example, if our solution follows the path il + j2, then the corresponding TSP

solution follows the path il + il + i3 + j2 + jl, with total cost dith + (Bi, + Bja)/2.

We now describe our overall algorithm, Global, for out problem. Let vector v =

(vi, va, %, 4, where integers V I and q (q and v4) represent the aizes of first two (last

two) sublots ïnd vh 2 O, k = 1,2,3,4. Since pj 2 2, we have Wj - 1 2 VI + ~2 > O and

Wj - 1 2 t~ + ~4 > O. Given a vector v for a ptoduct sudi that V I + v2 + US + v4 5 Wj,

the head and tail of the product ia dehed as Ah = av1- At = h3 --a,.

Algorithm Global

For product j, j = 1, ..., N: For id v SU& that + ~2 + ~3 + U, 5 Wj:

Run D P ( j ) to h d optimal integer sublot sizes conesponding to W .

Defme city (Le., prduct profile) k in country j by those optimal sublot sizes.

End For

End For

Set up the TSP cost matrix to enforce a visit to exactly one city in each country.

Run the TSP heuristic, output the solution and stop.

It remains to define D P v ( j ) , which is a dynamic progamming algorithm. For each

product j, this algorithm finds the optimal sublot sizes for a given vector u which defines

An and At. It chooses the sublot sizes such that idle time on machine Mi is minimized.

By doing so, it generatea ali the possible profiles (Le., cities) for al1 the products (Le.,

countries), as required in our TSP formulation.

In order to describe our dynamic programming algorithm, we defuie a pair of integers

fik-l,&k > O, for k = l , . . .,nj. We rquire that at l e a t two and at m a t nj d u e s

are positive integers. These positive Y, values represent the optimal sublot sizes. Let

N ( S ) denote the number of poaitive valued elements in the set S. In order to simplify the

description and example that follow, we suppress v and j.

Algonthm DPv(j)

Optimal Value Function

Let fik(&k-l, Kr, w, r) denote the minimum idle tirne on Mi for the partial schedule

formecl by sublots hk+1,. . . , &, , given that the (2k - I ) ' ~ sublot has size 15k-1 and

the 2kth aublot has size Kr, and that there are m unite and r sublots left to allocate.

[Comment: &k-i and hk are both nonzero if the schedule has alternat ing pair structure;

Kk-1 is nonzero and & is zero if the scheduie ha9 forward flowshop structure; hk-l i s

zero and is nonzero if the scheduie hm backward flowshop structure.]

It is important to note that three possible situations may occur at each stage:

i. > O and = O, where the schedule follows the forward flowshop structure.

ii. l$k-l = O and Kk > O, where the schedule follows the backward flowshop structure.

iii. Kr-, > O and Kk > O, where the schedule follows the alternating pair structure.

The amount of computation time required by Algorithm Global can be calculateci as

follows. The number of possible dulues of j is O ( N ) . The number of possible vectors v

is O(Wf). The number of possible d u e s of k is O(nj). The number of possible values

of w is O(Wj). Finally, the number of possible values of Kk-, and hk in the recurrence

relation is O ( q ) . Thus, the overalî computation time of Global is O(N ziN,, njw;).

Example: Consider a single product j, where Wj = 4, nj = 4, a = 5, and b = 4. We

now iilustrate DPV( j ) using v = (1,0,0, l), i.e. vl = 1, v2 = 0,v3 = O, VI = 1, therefore

Ah =ou1 - bu2 = 5 and At = bV3-uu4 = -5.

The boundary conditions give:

fZ(3 0,1,2) = rnw{O, 4(2) + 4(1) - 5(0) - 5(0)) = 12.

f2(0, 2,1,2) = mw{O, 4(0) + 4(1) - 5(2) - 5(0)) = 0.

f2(i, 1, 1 , l ) = mw{O, 4(1) + 4(1) - 5(1) - 5(0)) = 3.

f4(l, O, 1, l ) = mu<{O, 4(1) + 4(1) - 5(0) - 5(0)) = 8.

f4(0, 1, 1, l ) = miuc{O, 4(O) + 4(1) - 5(1) - 5(0)) = 0.

Figure 5.4: Schedule from Solution 1.

Then, using the recurrence relation at stage k = 1, we have:

f2(1, O, 2,2) = min{mw{0,4(1) + 4(0) - 5(0) - 5(1)) + f4(l, 0,1, l) , max{0,4(1) + 4(1) - 5(0) - 5(0)) + !,(O, 1,1,1)) = min{& 8}, where the choice is between Xj = 1, X4 = O and

XÎ = O, X4 = 1> respectively.

f2(0,1,2,2) = min{max{0,4(0) + 4(0) - 5(1) - 5(1)) + fr(1, O, 1, l), max{0,4(0) + 4(1) - 5(1) - 5(0)) + f4(0, 1,1,1)) = min{& O), where the choice is the sarne as above.

Similady, using the recurrence relation at stage k = O, we have:

f0(1,0,3,3) = min{max(0,4(1) + 4(0) - 5(0) - 5(1)) + f2(1,0,2,2), max{O,4(l) + 4(0) - 5(0) -5(2)) + 12@, O, m, m={O, 4(1) +4(1) - 50)) - 5(0)} +fi@, 1,2,2), max{0,4(1) + 4(2) - 5(0) - 5(0)) + h(0,2,1,2), m={0,4(1) + 4(1) - 5(0) - 5(1)) + /2(1,1,1?1)) =

min{8,12,8,12,6), where the choice is between Xi = 1, X2 = O; Xi = 2, X2 = O; Xi =

O, X2 = 1; Xi = 0,X2 = 2; and Xl = 1,X2 = 1, respectively.

There are seven possible solutions for the example, as we now discuss. In Figures 5.4

through 5.10, the makespan (Cm,) equals the processing time on machine Mi of a Wj = 20

units plus the idle tirne found by each schedule using D PV( j ) .

Solution 1: ( 1 O 3 , + ( 1 O, 2 , + ( 1 O, 1 1). This gives a solution with Yi =

1, f i = O, = 1, f i = 0, as shown in Figure 5.4.

Solution 2: fo(l, 0,3,3) -t f2(0, 1,2,2) + f4(l, O, 1,l). This gives a solution with =

O,b = l,& = 1,G = 0, as shown in Figure 5.5.

Solution 3: fO(l, 0,3,3) + fi(2, 0,1,2). This gives a solution with Yi = 2, = = YI =

0, as shown in Figure 5.6.

Figure 5.5: Schedule from Solution 2.

-- -- -

Figure 5.6: Schedule from Solution 3.

Solution 4: fo(1,0,3,3) + f2(1, 0,2,2) + f,(O, 1,1,1). This gives a solution with Yi =

1, = YJ = O, YI = 1, as shown in Figure 5.7.

Solution 5: fo(l, 0,3,3) + f2(0, 1,2,2) + f4(0, 1,1,1). This giva a solution with =

O& = 1,YJ = O,h = 1, as shown in Figure 5.8.

Solution 6: fo(l, 0,3,3) + f2(0, 2,1,2). This gives a solution with Yi = O, f i = 2, Y3 =

= O, as shown in Figure 5.9.

Figure 5.7: Schedule from Solution 4.

-

Figure 5.8: Schedule from Solution 5.

Figure 5.9: Schedule from Solution 6.

Solution 7: f ( l , 0 , 3 , ) + ( 1 l , l l ) This gives an optima sohtion with Yi = & =

1, YJ = YI = O, as shown in Figure 5.10.

We now prove that Algorithm D P v ( j ) provides optimal sublot sizes for a given vector

Lemma 8 If X' = (Xi,. . . , X:) is the uector of sublot sires /or a product given by D P U ( j )

Figure 5.10: Optimal Schedule from Solution 7.

for a given v (and thw given Ah and At values) and j , and X = (Xi,. . . , X,) is any other

vector of sublot sizes for the same An and At, then we have B ( X m ) 5 B(X) , when B(X*)

is the length of M y the ptoduct for sublot sizes X'.

Proof. Since DPV(j ) minimizes the coet over dl possible state transitions, X e minimizea

the value of idle time (Il) on Mi. Now, since Ah + B ( X 8 ) = Il + Wu, where Ah and WU

are constanta, it foUows that B ( X 8 ) 5 B ( X ) , O

Theorem S The lot sircs X'(j) = (X;,j,. . . , XZjj) obtained by D P v ( j ) minimire the

makespan for given v (and thw given Ah and At values) and j .

where is the body length of the product scheduled in the position a(i).

Suppose there exists a product k scheduled in position a(rj for which the sublot si-

are X(k) # X*(k) . We replace the lot sizes of X ( k ) with X'(k) for this product. The A

m b p w F,., cm be Witten F ' = Fe + G(,- ,),(,) + <P,(,),(,- ,) + B,&) a

Since Ah and At are given, Hz(,) = &(r) and T:(r) = TU(,(* Thus, e(r-I),(,) = dr(,-i),(r)

cP,(r),(r+I) = de(r)e(r+i)* Algo, from Lemma 6, B:(,) I Be(,)* Thus, F,' I Fe- Similady, we replace the lot sizes of X(j) with X a ( j ) for each product j in the sequence

o. Each such replacement will not increase the makespan. Thus, the sublot sizes X 8 ( j ) =

(Xi, j, . . . , Xi, ,j), j = 1,2, . . . , N given by D Pu ( j ) minimise the makespan for the given u

(i.e., An and At dues) . O

5.3 Conclusion

In this chapter, we have studied the problem of scheduling products with given sublot

sizes as well as simultaneous lot streaming and scheduling when sublot sizes are not given.

For the former case, we proved that the algorithm due to Gilmore and Gomory provides

an optimal schedule. Latter case b formulated as TSP for whi& there are many efficient

solution techniques available. We have considered the problem without setup times. How-

ever, the problems with setup times cm be solved by inserting the setup times with the

kst seublot pairs.

Chapter 6

Computational Testing

In orcler to evaluate the performance of Algorithm Global, we test a large number of tan-

domly generated two machine multi-product problem instances running ou Sun computer.

The dgorithm GENIUS for TSP (Genreau, Hertz and Laporte, 1992) is received in Pascd

and then translated into C language using the p2c translater. The algorithm Global is

written in C lanyage and the translated GENIUS is adopted for our problern.

For dl problems tested, the processing times aj and bj axe generated randornly from a

unifom distribution over the interval 1 5 a,, bj 5 10. When the number of products N is

in the range 2 < N 5 8, the number of sublots for each product nj is generated randomly

from a uniform distribution over the interval 2 5 n, 5 6 and the total number of unit8

for each product, Wj, is generated randomly from a uniform distribution over the intenial

2 5 Wj 5 6. When the number of products is in the range 9 5 N < 13, the number of

sublots for each product nj is generated randomly from a uniform distribution over the

interval 2 5 nj 5 and the total number of unjts for each product, Wj, is generated

randomly from a uniform distribution over the interval 2 < Wj 5 5. When the number

of pmducts is in the range 14 5 N 5 20, the number of sublots for each product nj is

generated randomly fiom a uniform distribution over the intervai 2 5 nj 5 4 and the total

nurnber of units for each product, Wj, is generated randomly from a unifom distribution

over the interval 2 5 Wj 5 4.

Each line in Table 6.1 represents results over ten raadomly generated problems. The

fourth column in Table 6.1 shows the mean CPU time required by Global.

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

ma{Wj) Of Cities

Table 6.1: Performance of Global.

Chapter 7

Sorne Related Results and Discussion

In this Chapter, we discuss some possible extensions of Our dgorithm Global. The al-

gorithm Globai can be utilized for single product integer-sized sublots, problems with

at tached setup times, sublots wit h pallet capacity, problems wit h transfer tirne between

machines and two machine newait flowshops. We briefly discuss on these possibilities.

7.1 Single Product Integer-sized Sublots

In the single product case, first we can find al1 possible profiles using the dynamic prw

pamming algorithm. Then the cycle time can be calculated for each profile and the profile

which gives the minimum cycle time is the optimal schedule.

7.2 Problems with Attached Setup Times

The algorithm Global can solve problems with attached setup time with a srnall modifica-

tion. First dynamic programming dgorithm can find the profiles without considering the

setup times. The setup times only affect the head values. Let us cal1 the head of profile k

of product i after adjusting for the setup times as (adjusted head):. Then

(adjusted head): = (head); + s i - t i , where (head); is the head of profile k of product

i befon adjustment for setup times. Now these profiles adjusted for setup can be used in

TSP to obtain the product sequence. The same adjustment can also be done for single

product case.

Figure 7.1: Forward flowshop structure with transfer time.

7.3 Sublots with Restricted Capacity

We have assumeci that sublots (pallets) have unlimited capacity. However, it is not true

in al1 cases. The dynamic pmgramming algorithm c m be restricted to assign items to

sublots within its capacity. This restriction reduces the number of possible profilea for a

product. As a result the problem can be solved more easily than our original problem.

Here we may assume that the pailets of a given product are identical.

7.4 Problems with Sublot Xkansfer Time

Generaily, tramferring sublots between machines consumes time. If this time is very

large, we may need to reduce the number of sublots. Our mode1 can consider such sublot

transfer time with a s m d modification. In our explanation let Tlp be the transfer time

frorn machine 1 to machine 2 and TZI be the transfer time from machine 2 to machine 1.

The Figures 7.1, 7.2 and 7.3 show how transfer time can be accommodated in a sublot

pair. Since we use dynamic propamrning, transfer times can be simply included for each

pair. Then the algorithm Global will solve for product sequence.

7.5 Two Machine No-Wait Flowshop

We have formdy modeIed the problem of simultaneous lot strearning and scheduling of

multiple products in two machine newait openshops to minimize malcespan. It has to be

noted that the flowshop version of the problem is a speciai case of o u openshop problem

Figure 7.2: Backward flowshop structure with transfer time.

Figure 7.3: Alternating pair structure with transfer time.

since we need to considet only the foruard flowshop during the construction of a achedule.

Thus, the dgorithm Global can eaeily be modified to solve two machine newait flowshops.

We now desaibe our overd dgorithm, Global Flow, for flowshop problem. Let vector

v = (vi, v4), where integer vl (v4) repreaents the sizes of first (last) sublot and uk 1 0,

k = 1 and k = 4. Since pj 1 2, we have Wj - 1 2 V I > O md Wj - 1 2 v4 > O. Given o

vector v for a pmduct such that + v4 W,, the head and tail of the product is defined

as XI = au!, AI = h 4 .

Algorithm Global Flow

For ptoduct j, j = 1, ..., N: For dl v S U C ~ that UI + ~4 5 Wj:

Run DPv( j ) to Çnd optimal integer sublot sizes corresponding to v.

Define city (Le., product profile) k in country j by those optimal sublot sizes.

End For

End For

Set up the TSP cost matrix to enforce a visit to exactly one city in eadi country.

Run the TSP heuristic, output the solution and stop.

It remPins to define DPv(j) , which is a dynamic programming algorithm. For each

product j, t his algorithm h d s the optimal sublot sizes for a given vector v which dehes

An and At. It chooses the sublot sizes such that idle time on machine Mi is minimized.

By doing so, it generates al1 the possible profiles (i.e., cities) for al1 the products (i.e.,

countries), as required in our TSP formulation.

In order to describe Our dynarnic programming algorithm, we defme an integer & 1 0,

for k = 1,. . . , nj. The positive & values represent the optimal sublot sizes.

Algorithm DPv( j )

Optimal Value knct ion

Let ji(Yk, w, r ) denote the minimum idle tirne on Ml for the partial schedule formed by

sublots K+l,. . . , Y,,. , given that the (k - 1)th sublot has size and that there are w

unita and r sublots left to docate.

Boundaty Conditions

f&,ur,r) = rnax{O,bK - avd) fot 1 5 r 5 nj - 1, t E (1,. . . ,nj - 1); S.(*,*, * , O ) = +OO

otherwise.

The amount of computation time required by Algorithm Global Flow can be calculated

as follows. The nurnber of possible values of j is O(N) . The number of possible vectors

v is O(W!). The number of possible values of k is O(nj). The number of possible values

of w is O(Wj)e Findly, the number of possible values of in the recurrence relation is

O(Wj). Thus, the overail computation tirne of Global Flow is O (N xgl n w!) .

For the problems with pallet capacity, the dynamic programming will keep track of

the items allocated to each sublot not to exceed the capacity. Thus, this will reduce the

number of possible cities and hence the CPU time will be reduced.

Chapter 8

Conclusions

In this thesis we develop algorithma for minimizing the makespan in two machine newait

openehop8 with lot streaming. For the single product continuous case we provide solution

techniques which give optimal solution. We show that multi-product integer-sized sublot

problem can be formdated as a traveling salesman problem with a pseudopolynomiai

number of cities. Many solution methods are available for solving large instances of the

TSP efkiently (Lawler et al., 1985). Our computatiooai tests of one such method indicate

that it can routinely deliver close to optimal solutions to scheduling and lot streaming

problem for instances with up to 20 products. The algorithm Global can aiso be used with

some modification to d v e for single product integer-sized sublot problem, problems with

limited pallet capacity, problems with attached setup times and problems with transfer

times . Further we have shown that the idea can be efficiently used for two machine nct

wait flowshop problem. The problem studied here bas the assumption of deterministic

production environment. Possible future reseaxch could study the more general problems

obtained by relaxing t hie assumpt ion.

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Lawler, E.L., J.K. Lenetta, A.H.G. Rinnooy Kan and D.B. Shmoys. 1993. Sequencing and Scheduling: AIgotithm~ and Complexity, in Handbooka in Operations Research and Management Science, 4, 445-552.

Pinedo, M. 1995. Scheduling: Thcoq, Algorithms, and Systeme, Prentice Hall, Englewood Clifh, N.J.

Potts, C.N. and K.R. Baker. 1989. Flowshop Scheduling with Lot Streaming, ope ratio^ Researrh Lettera, 8, 4, 297-303.

R k k , H. 1984. The Three Machine NeWait Flowshop Problem is NP-Complete, Journal of the Association of Cornputer Macliinery, 31,336345.

Sahni, S. and Y. Cho. 1979. Complexity of Scheduling Shops with NeWait in Process, Mathematics of Operations Researeli 4, 448-457.

Sen Alper and Orner S.Benli. 1998. Lot Steaming in Opeoshops. Working paper, Inform* tion and Operatioas management Department, Marshail School of Business, University of Southern California.

Sriskandarajah, C. and E. Wagneur. 1999. Lot Streaming and Scheduling Multiple Proci- ucts in Tw*machine Newait Flowshops. IIE Tronsacttow, To appear.

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Stephane Dauzere-Peres and jean-Bernard Lasserre. 1997. Lot Streaming in Jobshop Scheduling, Operatione Research, 45, 584-595.

Szendrovits, A.Z. 1975. Manufacturing Cycle Time Determination for a Multi-S tage b nornic Production Quantity Model, Management Science, 22, 298-308.

Trietsch, Dg and K.R. Baker. 1993. Basic Techniques for Lot Streaming, Opemtiow Reseamh, 41,6, 106!&1076.

Vickson, R.G. 1995. Optimal Lot Streaming for Multiple Products in a Two-Machine Flowshop, European Journal of Operational Research, 85, 556-575.

Appendices

Appendix 1 : C Coded Prognun for the Dynamic Programming Algorithm

Next next;

1 ht; FILE *output$

Global pmd:

if (items - 0)

I lenigd = b*i 1 +b*evens-a*odds-a*i2;

rntempdf i ) *mpb*

temps = y+temp6;

if (temps < kst-value)

I kt-value = temps;

Fina@, ml, oddB , evem ,lots,items ,besttvaiue);

1 1 retura s;

1

l8 aitemüng stnictipe is possible if w>l rrrd nuO */

I* previau ooes fdlows famrd flowshop slnrture */

/* peviais one fdlows bakwnrd flowshop structure *1

I* peviou8 one follows f m d flowshap structure V

I* previaus one fdlows bachard flowshop structure */

CREATING DISTANCE MATRIX

Appendix 2: Sample Output of a problem

We provide sample output of a problem with two products. For product 1, W = 4, n = 5,a = l , b = 2 and for product 2, W = 4, n = 6, a = 2, b = 3. The Output "Profile" in Appendix 2.1, explains there are 85 pionles for product 1 and 155 profila for product 2. Thus in our cost rnutix, fist 85, i.e.,l-85, cities(profles) axe of country 1 (product l), and the reéunining 155,i.e.,8&240, cities (profiles) are of country 2 (product 2).

The output 'Path" in Appendix 2.2, explains that solesman has to leave product 1 at profile 24 and enter product 2 at profile 1. Therefore teh selected profile for product 1 is 25, and for product 2 is 1.

Sublot sizes of profiles are given in the output UProNes". It shows tbat for product 1, sublot si- are (0,3), (0,l) and (0,l); Thus the schedule will be aa follows:

for product 2, sublot sizes are (OJ), (1,3) and (0,l).

cycle time = 32 C m

Schedule for the sample output

Appendix 2.1 : Output "Rofiie"

The number of subbts = 4

The number of items = 5

Pmcessing T b on Mschine 1 = 1

Rocessing Time on Machine 2 = 2

The number of sublots = 4

The number of items = 6

PnrwisingTimeonMIichine 1 = 2

R.ocessing T h e on Machine 2 = 3

Idie t i d

Ide t1-5

Ide t1-6

Idle t1-7

ldle t1-2

Ide t1-3

Ide time=3

ldle t i d

Ide time=û

Idle t1-1

Idle time= 1

Ide time=l

Ide ti@

Idle t i d

Idle t1-2

Ide tians3

ldle ticne4

ldle time=O

Idle t i m d

Idle ti-t

Idie tiiiiP--n

Idle ti&

IdletimP--n

Idie ti-1

Hetime=I

Me ti&

Mie tlme=û

Idle t i d

Idle ti&

Ide ti&

Idle time=7

Ide ti-8

Idie ti&

Ide time=lO

Ide t i d

Ide time=S

ldle ti-6

Ide ti-7

Ide a-3

Ide time=3

Idle tinm4

Idle time=4

Idle tirne= 1

Idle time=û

ldle ti-5

Idle ti-6

Ide ti-7

Idle ti-3

Ide tirnec3

Idie t i d

Idie tirtiP--"

Idle time=l

Ide ti&

Me ti-3

bdle t i d

Ide -1

Ide time=l

ldletime=O

kûe timetl

Idle ti&

Me tJ-8 Idlettme=9

Idle tiinel0

Zdle tiras5

Idie ti-6

Idle ti-7

Me time=6

Idle t i b

Idle timezl

Idle tirne4

Idle ti-7

Idle ti-3

Idle t i m d

Idie ti-1

Idie time=4

Idle ti-1

Idie time=9

Idle ti-10

Ide ti-8

ldle ti-7

Idle time=4

Ide ti-7

Idle timecll

Idle tirne= IO

Idle ti-7

Roeie: m(0 , l),(O , S),(O , O),(O , O) Idle tirne13

Rofiïe: 91:(0 , 1),(2 .2),(1 , O),(O ,O) Ide timsû

Roûk 92:(0 , l),(O .3),(1 , 1),(0 ,O) Ide time=7

praeie: 93:(0 , 1),(0 .2),(1 ,2),(0 ,O) ïdie t l m d

Ro6îe: W:(O , 1),(0 , 1),(1 ,3),(0,0) Me time=6

ndEle: 9S:(O , 1),(1 ,4),(0 , O),(O , O) Ide ti-8

notae. %:(O , 1),(2 , 1),(2 , O),(O , O) IQe tim

ProHle: W:(O , l),(O ,2),(2 ,1),(0 , O) Ide tim-4

R d k 98:(0 , 1),(0 , 1),(2 ,2),(0,0) Me ti-1

ProBie: 99:(0 , 1),(2 ,3),(0 , O),(O ,O) Idle ti1ne=3

noiBle: lW(0 , 1),(1 ,1),(3 , O),(O , O) Ide ri&

IWk 101:(0, 1),(1 ,0),(3, 1),(0,0) Idetirne0

ProtUe: 102:(0 , lX(3 ,2),(0 ,0),(0 , O) Ide tiW

Roflle: 103:(0 , 1),(0 , 1),(4 ,0),(0 ,O) ldle tirne= 1

Roflle: lM:(O , 1),(4 , l),(O ,0),(0 , O) Idle tiW

Profîle: 105:(0 , I),(S ,0),(0 ,0),(0 , O) Ide ti-

RoBie: 106:(0 ,2),(1 ,2),(0 , 1),(0 , O) Ide time=2

RolBle: 107:(0 ,2),(0 ,2),(0 ,2),(0 ,O) We time=4

R d e : 108:(0 ,2),(0 , 1),(0,3),(0 ,O) Idle time't

Rosle: 109:(0 ,2),(O ,4),(0,0),(0 , O) Idle tirne8

Rotlle: 1 10:(0 ,2),(1 ,2),(1,0),(0 , O) Mie time=O

Profile: 111:(0,2),(0,2),(1, 1),(0, O) Ide timts2

Praflle: 112:(0,2),(0, 1),(1,2),(0,0) fdletime=2

hofile: 113:(0,2),(1 ,3),(0,0),(0 ,O) Idletime=3

M e : 114:(0,2),(1 , 1),(2,0),(0,0) Idietime=û

Roük: 11S:(O, 2),(0,1),(2, 1),(0,0) ldletimd

R d e : 1 16:(0 ,2),(2 ,2),(0 , O),(O , O) Idle tirne0

Profile: 1 17:(0 ,2),(0 , 1),(3 ,0),(0 , O) Ide timeO

R d e : 118:(0,2),(3,1),(0,0),(0,0) IdletimeO

Roflle: 1 19:(0 ,2),(4 ,0),(0 ,0),(0 , O) Idle time=O

Proate: lU):(O ,3),(O ,2),(0, 1),(0 , O) Ide tlmeO

RoBfe: 121:(0,3),(0,1),(0,2),(0 ,O) Idletirne4

Roâle: 122:(0 ,3),(0 ,3),(0,0),(0 , O) Ide tirne3

RoBie: 123:(0 ,3),(0 ,2),(1,0),(0 ,O) Idle tirne0

Rotlle: 124:(0 ,3),(0,1),(1, 1),(0, O) Idle timgO

m e : a:(o .3),(1. 2),(0 , O)XO , O) t i d

RoLilc: 126:(0 ,3),(0 , l),(2 , O),(O , O) Ide t i m d

~olilc: in:(o .3),(2 , t),(o, o),(o , O) 1ae tiw

t>tofile.. 128:(0,3),(3 , O),(O , O),(O ,O) Idle ti&

ProBle: 129:(0 ,4),(0 , 1),(0 , l),(O , O) me ti-1 RotW: l a ( 0 ,4),(0 ,2),(0 , O),(O , O) Me time=û

RoBJe: 131:(0,4),(0,1),(1 ,0),(0,0) M e t i d

notiie: 132:(0 ,4),(1 , 1),(0 , O),(O , O) fdle timeO

Protüe: 133:(0,4),(2,0),(0,0),(0, O) M e t i W

Rdae: 1 M:(O , S),(O , 1),(0 , O),(O , O) Idle t i d

Rdk 135:(0, SX(1 , O),(O , O),(O , O) Idle t i m d

nofilc: 1W.(l ,O).(l ,3),(0,1),(0 ,O) fQetime=lO

Profile: 137:(1,0),(0 ,3),(0 ,2),(0 ,O) Ide ti-12

Rdk: 138:(1 , O),(O, 2),(0,3),(0 ,O) IUieti-14

Profile: 1 N:(l ,0),(0 , 1),(0 ,4),(0 , O) Idle time 16

Raflle: 140:(1 , O),(O ,5),(0 , O),(O , O) Idie ti- 18

FMUe: 141 :(1 .0),(2 ,2),(1 , O),(O , O) Idle ti-5

Rdile: 142:(1,0),(3,0),(1 , 1),(0 ,O) Iâîetime=lO

RorUe: i43:(1 ,0),(0,2),(1 ,2),(0, O) Idletime=9

noBle: 144:(1 ,0),(1 ,0),(1 ,3),(0 , O) Idle tirne= 1 1

Rofile: 145:(1 ,0),(1 ,4),(0 ,0),(0 ,O) Ide time=l3

M e : 146:(1 ,0),(2 , 1),(2 ,0),(0 , O) Ide ti-2

M e : 147:(1,0),(2,0),(2,1),(0, O) Idletinus5

Rofile: 148:(1,0),(1 ,0),(2,2),(0 ,O) ldletims6

notrie: 149:(1 ,0),(2 ,3),(0 ,0),(0 , O) Idîe ti-8

Rofiie: 150:(1 ,0),(1 , 1),(3 ,0),(0 ,O) Idle tirne4

Profile: 151:(1 ,0),(1 ,0),(3, 1),(0,0) Idle ti-1

Rotîle: 152:(1 ,0),(3 ,2),(0 ,0),(0 , O) Idle ti-3

Roflle: 153:(1 ,0),(0 , 1),(4,0),(0 ,O) Idle ti-6

Profile: 154:(1 ,0),(4, 1),(0,0),(0, O) I d l e t i m

Rofile: 155:(1 ,0),(5 ,0),(0,0),(0 ,O) Idie ti&

M e : 156:(1 , 1),(0 ,3),(0 , 1),(0 , O) Idle tirne10

M e : 157:(1 , 1),(0 ,2),(0 .2),(0 , O) Idle ti&

M e : 158:(1 , 1),(O , 1),(0 ,3),(0 , O) Idle tirne=ll

noSue: 159:(1 , 1),(0 ,4),(0 , O),(O , O) Idle -1 3

Ronle: 160:(1 , 1),(3 ,0),(1 ,0),(0 ,O) Idle tims7

H e : 161:(1 , 1),(0 ,2),(1 , 1),(0 , O) Idle ti-7

noBie: 162:(1 , 1),(0 , 1),(1 ,2),(0 ,O) Idle tirne0

nafTle: 163:(1, 1),(1 ,3),(0 , O),(O ,O) Idle ti-8

F M ï k : 164:(1 , 1142 ,0),(2 , O),(O ,O) Idle t imed

PM&: 165:(1,1),(0 , 1),(2 , 1),(0 , O) Me t i m d

M e : 166:(1 , 1),(2 ,2),(0 ,0).(0 , O) Mie tirne3

noaie: 167:(1 , 1),(1 ,0),(3 , O),(O , O) Idle time=O

Prof& 168:(1 , lX(3 , 1),(0 , O),(O , O) Ide timeO

nanie: 169:(1 , lX(4 , O),(O , O),(O , O) Ide t i m d

Rotîk: l7O:(l ,2),(0 ,2),(0 , 1),(0 , O) Ide tirne5

Plonle: 171:(1 ,2X(O .1),(0,2),(0, O) ide time=6

FWîk: 172:(1 ,2),(0 ,3),(0,0),(0 , O) Mie ti-8

RdW 173:(1 ,2),(2 ,0),(1,0).(0 , O) Ide t i d

Rotiie: 174:(1 ,2),(0 , 1),(1 , 1),(0 , O) Ide tirne2

FWik 175:(1 ,2X(1 ,2),(0,0),(0 , O) Me tirne3

nofile: 176:(1 ,2),(1 ,0),(2 , O),(O , O) Ide tirne0

noaie: l77:(1 ,2),(2 , 1),(0 , O),(O , O) Idle time=û

nafile: 178:(1 ,2),(3 ,0),(0 ,0),(0 ,O) Idle ti&

~ o a i e : 179:(1 ,3),(0 , 1).(0 , 1),(0 , O) Ide time 1

Rotlle: 180:(1 ,3),(0 ,2),(0 ,0),(0 , O) Ide tirne3

Rofile: 181:(1 ,3),(0, 1),(1,0),(0 ,O) IdietimeO

Rofile: 182:(1 ,3),(1 , 1),(0,0),(0, O) Idle time=û

Profile: l83:(l ,3),(2 ,0),(0 ,0),(0 , O) Idle time=O

R e 184:(1 ,4),(0,1),(0,0),(0, O) Idle time=O

R&e: 185:(1 ,4),(1 ,@,(O, 0).(0,0) Idle ti&

Rotlie: 186:(2 ,0),(1 ,2),(0 , 1),(0 , O) Iâie tirne= 12

Rotlle: 187:(2 , O),@ ,2),(0 ,2),(0 , O) Idle ti-14

Rot&: l88:(2 , O),@ , 1),(0 ,3),(0 , O) Idle tirne16

M e : l89:(2 ,0),(0 ,4),(0 ,0),(0 , O) Ide time i 8

Rome: 190:(2 ,0),(2 , 1),(1,0),(0 , O) Ide tirne=7

RoBIe; 191 :(2 ,0),(2 ,0),(1 , 1),(0 , O) Ide tirne9

M e : 192:(2 , O),(l ,0),(1,2),(0 , O) Ide tirne 1 1

RoSlle: 193:(2 ,0),(1 ,3),(0 , O),(O , O) fdle tirne= 1 3

Rofile: 1 W:(2 , O),(t ,1),(2 ,0),(0 , O) Ide ti-7

M e : 195:(2 ,0),(1 ,0),(2 , 1),(0 , O) Iâie time=6

Profile: 1%:(2 ,0),(2 ,2),(0 , O),(O , O) Idie time=8

RatEle: 197:(2 , O),(O , 1),(3 , O),(O , O) Idle tiW

Rot&: 198:(2 , O),(3 , 1),(0,0),(0, 0) Ide ti-3

M e : 199:(2 ,0),(4 , O),(O , O),(O ,O) Idle tirne0

Raffle: AWl:(2 , 1142 , (&(O, 1),(0,0) Idle tirne9

no&ie* 201:(2 , lL(1 , O),(O ,2).(0 , O) Ide tirne 1 1

Aoa)e: A)2:(2 , l),(O ,3),(0 , O),(O , O) Idîe tirne13

Prof&: 2û3:(2 , 1142 , 0),(1 , O),(O , O) Idle tirne4

P M k 204:(2 , l),(O .1),(1,1).(0 . 0) Me -7

PWW: 20!!:(2 . 1),(1,2),(0,0),(0 , 0) Me -8

mtk 206:(2, lMl .O),(2 .O).(O. 0) IQtlm3=2

PmtW 207:(2 . 1M2 .1),(0,0),(0 , 0) We timet3

RoBle: 208:(2 . lX(3 . O).(O . O).(O . 0) Idc time0

Rofile: m ( 2 ,2),(1 , O),(O , l),(O . 0) Idc time6

RatPc: 210;(2 .2),(0 ,Z),(O, 0),(0 ,0) Ide t i m 8

mak 211:(2.2),(1.0),(1 .O),(O .O) Idctimgl

Roiile: 212:(2 .2),(1 , 1),(0 ,0),(0 ,0) kne ti-3

RolPe: 213:(2 , ZX(2 . O),(O , O).(O , 0) Idle ti&

Rolile: 214:(2 .3),(0 , 1),(0 , O),(O , 0) Me time=3

R&: 2U:(2 .3),(1 , O),(O , O),(O , 0) Idle time0

M e : 216:(3 , O).(l . 1),(0 , 1),(0 , 0) Idle ti-14

Rofile: 217:(3.0),(0,1),(0.2),(0 ,O) Idle time16

M e : 218:(3.0),(0.3).(0.0),(0 ,O) Idletim4~18

Rotne: 219:(3 ,0),(1 , 1)41 , O),(O , 0) Idle t i m l 0

Rotlle: 220.(3 , O),(l ,0),(1 , 1),(0 , 0) Idle t ime t 1

RoBle: 221 :(3 , O),(l ,2),(0 , O).(O , 0) M e time 13

Profile: 222:(3,0),(0, 1),(2,0),(0 ,O) Idleti-12

Rotlle.. 223:(3 , OM2 , l),(O , O),(O , 0) Idle t i w 8

Rot&: 224:(3 ,0),(3 , O),(O , O),(O , 0) Idle time3

R a e : 225:(3 , 1),(1 , O),(O , 1),(0 , 0) Idle t ime1 1

RoBle: 226:(3 , 1),(0 .2),(0 , O),(O , 0) Me time= 13

RoBle: 227:(3 , 1),(1 ,0).(1 , O),(O , 0) Idle time6

R a e : 228:(3 , 1),(1 , l),(O , O),(O , 0) Idle time8

Roflle: 229:(3 , 1),(2 , O),(O , O),(O , 0) We t ime3

Profile: 230:(3 ,2),(0 . 1),(0 , O),(O , 0) Idle t icw8

Ro&: 231:(3 ,2),(1 , OM0 , O),(O , 0) Idle timts3

RoBle: 232:(4 , OM0 , 1),(0 , 1),(0 . 0) Idle 1111~~16

Rotile: 233:(4 . O),(O ,2),(0 , O),(O , 0) Idle time1 8

R a e : 234:(4 ,0),(0 , I),(1 ,0),(0 ,0) Idle time15

RoBle: 235:(4 . 0)Xl , 1),(0,0).(0 , 0) Idle ti-13

Rofilc: 2=(4 ,0),(2 , O),(O , O),(O , 0) Idle t i n e 8

PlnQ: 237:(4 , 1MO , 1),(0,0),(0 , 0) Me 11-13

RoBk: 238:(4 , 1),(1 , O),(O , O),(O . 0) Idle tixne=B

Rofilc: 239:(5 , OM0 ,1),(0,0),(0 , 0) Ide ti-18

RdPe: a ( 5 . 0),(1 , O),(O , O),(O , 0) Idle t i m l 3

Appendix 2.2: Output Tath"