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This is to certify that the

dissertation entitled

PRODUCTION SCHEDULING 0F CUTTING STOCK
IN A MULTI-PERIOD, MULTI-PROCESS FACILITY
WITH SEQUENCE DEPENDENT SETUPS

presented by
Michael P. D'Itri
has been accepted towards fulfillment

ofthe requirements for

Ph.D. degree in Business Administration

flit/ . 14/ , ydét;

 

Major professor

Date June 29, 1994

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chMma-DJ

 

 

   

PRODUCTION SCHEDULING OF CUTTING STOCK
IN A MULTI-PERIOD, MULTI-PROCESS FACILITY
WITH SEQUENCE DEPENDENT SETUPS

BY

Michael Payne D’Itri

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Management

1994

ABSTRACT
PRODUCTION SCHEDULING OF CUTTING STOCK
IN A MULTI-PERIOD, MULTI-PROCESS FACILITY
WITH SEQUENCE DEPENDENT SETUPS
BY
Michael P. D’Itri

This paper introduces an explicit formulation for
sequence dependent production scheduling where the produced
stock will be cut into finished products and a: heuristic
solution procedure that obtains good, but not necessarily
optimal, solutions to a production scheduling problem common
in the paper industry. The heuristic begins by adding
additional constraints and a few binary variables that limit
the number of decision variables likely to take fractional
values when the formulation is solved with the integrality
restrictions relaxed. Next, the heuristic proceeds to
sequence production on each. machine one period at time.
Sequencing is accomplished with an iterative procedure that
successively increases the number of transition variables
required to be binary until a complete production sequence,
for that period, is achieved. Once a production sequence is
established for a period, the heuristic defines a range of
integers describing possible production quantities for that
period. Another mixed integer program is formulated that
establishes integer values for the production in the period

just sequenced and begins the sequencing for the next period.

 

Copyright by
MICHAEL P. D’ITRI
1994

 

This dissertation is dedicated to my grandparents
Payne and Catherine Ward

iv

ACKNOWLEDGMENTS

I would like to thank the many individuals who, through
their generous contributions of time and expertise, have made
the completion of this dissertation possible. First and
foremost, I would like to thank the chairman of my
dissertation committee, Professor Paul Rubin. It has been an
honor and a privilege to have such an accomplished scholar as
a mentor, not only for this project but throughout my graduate
career at Michigan State University. His insights,
suggestions and patient tutoring have made the completion of
this research possible. The other members of my dissertation
committee, Professor Soumen Ghosh, Professor Gary Ragatz and
Professor Shawnee Vickery have all made crucial contributions
to this research as well as my development as a scholar.

Special appreciation is extended to Priya Campbell and
the Simpson Plainwell Paper Company. Other important
contributions to this research have been made by Mr. Michael
Wach, Mr. Scott Twaroski, Dr. Lou Adler and Mr. David Rish.
I also wish, to thank the many' members of the Eli Broad
Graduate School of Business who have helped throughout my
graduate career, most notably the management department

secretarial staff and my fellow doctoral students.

I would also like to acknowledge the important
contributions and sacrifices made by my wife Suzanne and
children, Jason and Jessica, while I pursued my graduate
degrees. I wish to thank my parents for their efforts and
example in spurring my interest in advanced education. I am
grateful for the support and encouragement offered by all of
my family members throughout this process.

Finally, I would like to thank Professor John Hoagland,
the members of the Institute of Water Research and The Center
for Remote Sensing as well as the many others at Michigan

State University for their support and camaraderie.

vi

TABLE OF CONTENTS

LIST OF TABLES . . . . . . . . . . . . . . . . . . .

LIST OF FIGURES . . . . . . . . . . . . . . . . . . .

CHAPTER 1:

CHAPTER 2:

CHAPTER 3:

CHAPTER 4:

CHAPTER 5:

CHAPTER 6:

CHAPTER 7:

APPENDIX A:

APPENDIX B:

REFERENCES

INTRODUCTION . . . . . . . . . . . . . .

LITERATURE REVIEW . . . . . . . . . . . .

CONCEPTUAL FRAMEWORK AND FORMULATION . .

HEURISTIC SOLUTION PROCEDURE . . . . . .

MODEL CALIBRATION AND BENCHMARK PROCEDURE

COMPUTATIONAL RESULTS . . . . . . . . . .

CONCLUSIONS AND FUTURE DIRECTIONS . . . .

EXAMPLE PROBLEM AND SOLUTION . . . . . .

COMPUTER CODE . . . . . . . . . . . . .

viii

ix

16
25
38
52
61
81
85
99

149

LIST OF TABLES

Table 1-1: Machine Dependence for Sequence Dependent

Production Scheduling . . . . . . . . .
Table 5-1: Representative Problem Size . . . . . .
Table 6-V-1: Problem Characteristics Report

Validation Problems . . . . . . . . . .

Table 6-V-2: Performance Summary
Validation Problems . . . . . . . . . .

Table 6-V—3: Problem Solution Report
Validation Problems . . . . . . . . . .

Table 6-I-1: Problem Characteristics Report
Independent Problems . . . . . . . . . .

Table 6-I-2: Performance Summary
Independent Problems . . . . . . .

Table 6—I—3: Problem Solution Report

Independent Problems . . . . . . . . . .
Table 6-S-l: Problem Characteristics Report

Scale Problems . . . . . . . . . . . . .
Table 6-S-2: Performance Summary

Scale Problems . . . . . . . . . . .
Table 6-S-3: Problem Solution Report

Scale Problems . . . . . . . . . . . . .

viii

 

Figure
Figure
Figure

Figure

LIST OF FIGURES

Heuristic Procedure . .

Solution With a Valid Sequence .

Solution With an Invalid Sequence

Sequence with Sub-Touring

ix

 

39

43

44

47

CHAPTER 1: INTRODUCTION

Paper manufacturers and other companies in the process
industries are currently experiencing production pressures
similar to those that batch or repetitive manufacturing
managers have dealt with for the last two decades. Customers
now expect suppliers to operate in just-in—time (JIT)
environments with shorter delivery lead times even as they
request higher quality and broader product lines. Better
production scheduling can increase responsiveness to these new
demands. Scheduling production to minimize waste enables
management to streamline operations making them more efficient
and profitable.

Efficient production sequencing is often an essential
component of production scheduling in the process industries.
The cost of configuring a manufacturing station can be
independent of the sequence of products produced at the
station. However, there are situations where the cost of
setting up for manufacturing is sequence dependent. This is
common in process industries and in the paper industry in
particular, where residual materials from prior manufacturing
can affect the current production.

When the produced materials are cut into finished
products, the production scheduling problem becomes more
complex, a situation common in the paper and process
industries. An optimal solution to a stock cutting problem

determines the best way to cut lengths or shapes from a stock

2

of raw materials. The raw materials have value, and the
objective is to minimize the amount of raw materials used
while satisfying demands for each length or shape. A wide
range of production technologies such as textiles, paper and
steel would benefit from more effective production scheduling
techniques that consider a subsequent stock cutting problem.
In many situations that require the solution of a. stock
cutting problem, the composition of the available stock is
beyond the control of the decision maker. For example, in the
lumber industry the sizes of the trees dictate the composition
and dimensions of the raw material, the stock set the cutting
priorities are based on. In contrast, manufactured products
offer the decision maker more control over the composition of
the cutting stock.

Although there are substantial bodies of literature
addressing both sequence dependent scheduling and stock
cutting problems, to date there has not been a formulation to
address these concerns simultaneously. This situation is
common in the paper industry where paper machines produce
stock of varying widths that will be cut into finished
products. The cost of changing the type of paper is dependent

on both the current and planned production.

EXAMPLE PROBLEM
The relationship between production scheduling and stock
cutting is illustrated by the following example. For the

purpose of simplicity, the example assumes all demand is for

3
one roll of each type of paper in each requested width. A
scheduler plans to produce two types of paper, A and B, where
paper A is required in a roll that is 100" wide and paper B is
needed in widths of 50" and 60" rolls. The company has two
paper machines, 1 and 2, with the capability of producing
widths of 60" and 100" respectively. At the beginning of the
planning horizon both are producing paper C. Table 1-1
describes the cost, in dollars, of switching between papers on
each machine, where the rows show the paper currently under
production, and the columns describe the status of the machine

after the switch. The incidence matrix shows the cost of that

transition.
Table 1-1: Machine Dependence for Sequence Dependent
Production Scheduling
Machine 1 Machine 2
Paper A E Q A E Q
0 4 6 0 4
B 3 O 9 6 O 12
2 9 0 2 11 O

Trim loss can be assumed to be $1 per wasted linear inch.
There are several approaches to scheduling these orders.

A first approach might be to have both orders scheduled
on machine 2 leaving machine 1 free to process other orders.
This would require switching to product A and then to product
B. The transition cost would be 2 + 4 = $6. The trim loss

for paper A would be zero, and the trim loss for the first

4
reel of paper B would be 50" and the second 40". Making the
total cost of producing both orders on machine 2 96 dollars.

Producing paper B on machine 1 and paper A on machine 2
generates a total cost of $21. In this situation the
transition costs would be higher, $11, while the cutting
expenses would be lower, $10.

This simple example illustrates the benefits of
considering the sequence dependent transition costs and
trimming costs simultaneously. As in ‘many ‘manufacturing
situations, production capability in the paper industry
depends on both the type of machinery available and the mix of
products demanded. Setup times for products are sequence
dependent, indicating opportunities for more efficient
operations through better production scheduling methods.
Additional complexity arises when the output will be cut into
finished products. Most paper mills have several machines
with varying capabilities, the most obvious is different
widths. Therefore, a production scheduler in a paper mill
must consider many factors simultaneously when devising
schedules for the production of cutting stock: due dates,
sequence dependent setups, inventory levels, proportion of
fiber recycled and the manufacturing and cutting capability of
each machine.

A rolling planning horizon reflects the dynamic nature of
most production planning systems where the product is demanded

at varying intervals and plans are frequently revised.

5
According to Baker and Peterson (1979), a rolling schedule
results from solving a multi-period production schedule and
then implementing part of it. A rolling planning horizon
introduces an additional level of flexibility, allowing the
scheduler to build inventory levels to meet future demands or
revise a production plan as more information becomes

available.

PROBLEM STATEMENT

Baker (1974) contends that production scheduling should
take place after three fundamental managerial decisions have
been made: the product or service to be provided, the amount
of product or service and the quantity of resources to be made
available. After implementation of these fundamental
managerial decisions, the scheduling problem should be solved
by answering two questions: determining the allocation of
resources to each task and the time when the task will be
performed. This research seeks to answer these two questions
for a situation in which manufactured materials are cut into
products. The manufacturing environment is a multi-product
facility that has several production lines with sequence
dependent setups.

The objective of the research is to develop a method for
devising schedules that meet customer time constraints and
minimize the total cost of four components: production of the
raw stock, trim waste from the stock cutting procedure, lost

sales and the cost of scrapping excess product. The developed

6
algorithm accomplishes this by first determining an efficient
production sequence for each machine, and then setting the
number of reels (form of the paper leaving the paper machines)
to produce. Inventory levels for each period of the planning
horizon as well as the quantity and period that unsatisfied
demand will occur are also specified by the algorithm.

Final products, rolls, are cut from the reels of paper
according to available cutting patterns. A cutting pattern
represents one possible method of dividing a reel into rolls.
For this research, only non-dominated patterns are considered.

A cutting pattern is dominant if no other pattern exists
where it is possible to cut more rolls in any one width while
producing an equal or greater number of rolls in all other
widths cut from the reel. For example, if paper is demanded
in 50" rolls and 20" rolls and the paper machine has a width
of 100" there are several combinations of useful widths that
could be cut from the 100" reel. For instance two 50" rolls
could. be cut, or one 50" roll and two 20" rolls. Both
patterns would be dominant because it is not possible to have
a pattern that will produce more rolls in one of the widths
while producing as many in the other width of interest. A
pattern where there was one 50" roll and one 20" roll cut
would be dominated by the pattern that produces one 50" roll

and two 20" rolls.

7
DESCRIPTION OF THE MANUFACTURING ENVIRONMENT

Stock cutting together with production scheduling is most
commonly encountered in process industries.
Process Industries

Process industries such as paper, food and chemical
manufacturing can be defined by two important facets, what
they' produce and the nature of the material flow in the
process (Taylor, 1979). Manufacturing facilities in process
industries are characterized by large capital expenditures
that require high levels of utilization to achieve economic
viability. Furthermore, products must meet a host of quality
measures. Equipment operating under steady state conditions,
which typically occur well after a start-up or transition
period, are most likely to produce a consistent product.
Achieving this steady state frequently requires considerable
time and expense. Therefore, such facilities usually operate
twenty-four hours per day, year round, except for scheduled
and unscheduled maintenance.

Process industries share some common manufacturing
characteristics. Raw materials represent nearly 80% of the
cost of production. Capital investment is high, and profit
margins are narrow, making efficient production crucial to
achieving profitable operations (Rice & Norback, 1987).

other features also differentiate process industries from
those involved in assembly or fabrication. One is an inverted

bill of materials. Unlike repetitive manufacturing, where

8

final products require many components, the process industries
produce large numbers of final products from relatively few
raw materials. Typically, products are sold at several stages
in the production process, manufacturing equipment is highly
specialized and delivery lead times are short (Nelson, 1983).

This research deals primarily with the paper industry,
which is representative of process industries. The production
scheduling procedures developed here should be applicable to
an assortment of manufacturing situations in other process
industries.
Paper Production

Paper is sold as reels, rolls, or cut sheets. Reels are
the direct result of converting pulp to paper. They are
usually the width of the paper machine. Their defining
characteristics are the type of paper, the basis weight and
quality grade. The type is often a brand name and denotes a
paper of a specific color, composition and texture. Basis
weight is a measure of the mass of the paper, the weight of
500 sheets. Quality grades can be expressed by many measures
such as defects per unit of surface area, gloss, and
discoloration. Rolls are the result of slitting the reels
width-wise immediately after the production of the reel. For
the purposes of this research the slitting procedure is
assumed to be unconstrained since it is fast compared with
other operations, and any number of rolls can be cut from one

reel of paper.

9

Procedures following the slitting operation are called
conversion. Examples are placing an embossing mark on the
paper, finishing one or both sides of the paper and producing
cut sheets. Cut sheets are made by cutting the roll width-
wise as it is unrolled. The result is rectangular sheets that
are usually trimmed to precise dimensions. All cuts, whether
lengthwise or width-wise, are from edge to edge and are called
"guillotine" in the stock cutting literature.

Typically, orders are for a specified tonnage of paper of
a particular type, cut dimensions, basis weight and grade.
For this research, requests for cut sheets will be specified
in terms of the number of rolls of paper required to produce
the order. The scheduling algorithm will then determine the
quantity of reels needed and the most efficient schedule for
producing and then cutting them into rolls.

In the paper mill used as a setting for this research,
four primary raw materials, Northern and Southern hardwoods
and softwoods, are blended to form several types of paper
pulp. The facility has three paper machines, and any given
order could be produced on one or more of them. The type of
paper produced influences the production rate, and a paper
machine can produce approximately one reel every forty-five
minutes. The output of reels can be characterized by the type
of paper and other parameters such as the basis weight and
width. In addition, each kind of paper may be one of several

quality grades as well as different colors. Reels of paper

10
weighing approximately 7000 pounds are usually produced to the
maximum width of the machine. Narrower reels can be produced,
but the practice represents lost production capacity and is
avoided.

After the rolls are produced, manufacturing ceases to be
continuous, and the process is more characteristic of batch
production, making it possible to schedule subsequent
operations independently. Therefore, this is an appropriate
point to segment the process for analysis. This is the method
used in all prior work on the paper trim problem known to this
researcher.

The effectiveness of a given schedule can be defined by
the total cost of producing the demanded material. Although
the effect of inadequate management on two of the primary
costs, over— and under-production, is obvious, poor management
of the transitions between papers or the creation of a large
amount of trim waste (the result of material left over after
reels are cut into rolls) can have other implications. As a
rule, the greater the similarity in the types of paper, the
shorter the time required to convert from one paper to another
of acceptable quality. Paper waste, whether from low quality
or trimmings, called "broke," is recycled. Minimizing the
amount of broke has several advantages. Most important, broke
represents lost production capacity. Second, paper pulp is
nearly 99% water, and the energy required to remove it is

substantial. Finally, excessive recycling lowers the quality

11
of the paper as the fibers break down from repeated
reprocessing. Although this feature would be difficult to
capture in a formulation, it does show that better scheduling
procedures may increase product quality in some circumstances.

Paper is manufactured in both make-to-stock, an approach
where firms attempt to meet customer orders from inventory,
and.make-to-order, a strategy where production is scheduled to
meet specific customer orders. The most common types of
paper, such as the brown wrapper that becomes grocery bags,
are commodities and are usually produced using a make-to-stock
strategy. On the other hand, specialty papers such as
facsimile or computer paper are normally produced in make-to-
order planning systems because of the wide range of production
specifications.

Although some orders are shipped from inventory, it is
more common to have production assigned upon customer request.
This is also the case for much of the make-to-stock items
where an intermediary will frequently order in large volumes
and perform part of the warehousing and distribution function.
Market conditions dictate that paper will be shipped on time,
or the customer is not obligated to accept the order.
Therefore, backordering is unusual for much (u? the paper

industry.

12
IMPORTANCE OF THE PROBLEM

The variety of manufacturing environments in the paper
industry offers many opportunities to demonstrate substantial
improvements in production efficiency through this research.
The high levels of production in the paper industry translate
small (1 or 2 percent) increases in efficiency into
substantial cost savings (Haessler, 1988). Paper sales in the
United States were estimated at $42 billion in 1980 and
represent nearly Al percent of the gross national product
(Noble, 1973).

other companies in the process industries must schedule
cutting production. These businesses account for nearly 50
percent of all production in the United States (Novitsky,
1983), suggesting that this research might have broad
applicability.

Representatives of the paper and steel industries have
described a distinct shift in management emphasis in recent
years. Advances in automation have made achieving the
required technical specifications for the product nearly
routine. Therefore, the focus of managerial attention has
shifted to scheduling and control issues. The paper industry
is moving toward a long term goal of a completely automated
production facility, called mill-wide control. Higher levels
of automation will make it possible to implement schedule

changes much more effectively (Routledge, 1988).

13

Although this research is aimed at designing production
planning tools for the plant level, another important
application would be as a component of a larger model
describing the entire production planning process. This
larger model could be used to address business questions at
the corporate level. These questions might include when and
which machines should be shut down for maintenance or which
facility should be assigned a given job. For these purposes
the algorithm could be used in a Bax-Meal hierarchical
production planning framework (Hax & Meal, 1975).

Most paper is produced by large companies with several
manufacturing facilities. Firms currently apply a
hierarchical planning framework to allocate jobs first to
individual facilities and then to the appropriate machines.
A scheduling algorithm describing job allocations at the plant
level would play an important part in developing an effective
company-wide production planning procedure. The scheduling
algorithm also could be used to schedule maintenance or
vacations and to establish appropriate policies for responses

to changing market conditions.

DESCRIPTION OF SUBSEQUENT CHAPTERS
The following is a brief description of each of the

subsequent chapters.

14

Chapter 2: Literature Review

This chapter provides a review of previous related
research. The primary focus of the review is the two bodies
of literature that provide the foundation for this research,
the stock cutting problem and sequence dependent production
scheduling. Literature describing the process industries with
particular emphasis on production scheduling and the paper
industry is also reviewed.
Chapter 3: Conceptual Framework and Formulation

This chapter describes the manufacturing environment and
the assumptions used in the formulation. An explicit
formulation for scheduling cutting stock in a multi-period,
multi-process facility with sequence dependent setups is
presented.
Chapter 4: Heuristic Solution Procedure

Chapter 4 presents a heuristic procedure for selecting a
near—optimal production plan. The heuristic is an iterative
procedure that sequences all the machines for one period and
then specifies the number of reels of paper to produce.
Chapter 5: Model Calibration and Benchmark Procedure

In this chapter the problem generator, used to create
test problems, is described, as is the procedure used to
determine a bound on the optimal cost of each problem.
Chapter 6: Computational Results

Three groups of problems were solved. The first set of

problems was used to develop the heuristic. The second

15

demonstrates the heuristic’s performance on an independent set
of problems, while the third set of problems shows that the
heuristic is practical for life size problems. Results
obtained using the heuristic are compared to bounds provided
by the benchmarking procedure. In addition, several methods
to reduce the problem size are introduced in Chapter 6.
Chapter 7: Conclusions and Future Directions

The final chapter of this document summarizes the results
and contributions of this research and outlines several

avenues for future research.

CHAPTER 2: LITERATURE REVIEW

Substantial bodies of literature describe production
scheduling with sequence dependent setups and stock cutting
problems independently. Research addressing situations in
which these problems must be considered simultaneously is
lacking. Three important characteristics should be noted.
First, less complex production scheduling procedures that do
not consider the cutting problem have been shown to be NP-Hard
(Hsu, 1983). Second, obtaining an optimal solution to a stock
cutting problem may require generating an inordinate number of
cutting schedules. Finally, due to difficulty associated with
solving these kinds of problems, both bodies of literature

rely heavily on heuristic solution procedures.

SEQUENCE DEPENDENT PRODUCTION SCHEDULING

Among the first authors to review the production
scheduling literature were Day and Hottenstein (1970) who
provided a classification of the research describing it in
terms of a three part classification scheme: the number of
parts composing a job, the production factors possessed by the
shop, and the nature of the availability of jobs for
processing. The authors go on to describe sequencing methods
such as combinatorial approaches, general mathematical
programming, heuristic procedures, and Monte Carlo sampling.

In their recent review of the literature for parallel
machine scheduling, Cheng and Sin (1990) characterized
scheduling problems using four parameters: size of the job

16

 

17

set; number of machines; a system parameter, which gives
information about the job characteristics and the performance
criteria used to determine the better schedules. The
literature base described by Cheng and Sin does not include
investigations of parallel machines that are not identical, as
is the case for this research. Although paper production
facilities can have identical processors, similar machines
with overlapping capabilities are more common. For instance,
Machine A can produce paper with basis weights varying between
20 and 40 pounds. Machine B can produce paper of between 15
and 30 pounds. In this situation 25 pound paper could be
produced on either machine.

The detailed description of the relevant portion of the
production scheduling literature base will begin by
subdividing the research into two categories, single stage and
multi-stage production processes. Because this research deals
with single stage production processes we analyze it in
greater detail by partitioning the literature still further by
defining two subgroups, single and parallel machine scheduling
procedures.

Glassey (1968) and Mitsumori (1972) are among several
early writers who considered single stage, single product
production environments without changeover times. LaRobardier
and Filak (1972) then extended this work by including
production switch-over costs and inventory holding costs.

Driscoll and Emmons (1977) modified the formulation proposed

18
by LaRobardier and Filak for production scheduling on a paper-
making machine. In this formulation inventory costs were
ignored, additional bounds were added, and a backward-time
dynamic programming approach was applied to solve the problem.
The solution procedure minimized the total changeover cost
while meeting customer demand schedules.

Lockett and Muhlemann (1972) researched the single
machine problem in which the setup time was a function of the
prior job and introduced several heuristics for solving the
problem. Barnes and Vanston (1981) addressed the scheduling
of jobs (mm a continuously available machine ix: which the
individual jobs had linear delay penalties as well as sequence
dependencies.

As pointed out by Cheng and Sin (1990), the vast majority
of research on parallel machines has assumed that the
processors are identical. There are large bodies of research
considering both sequence dependent and non-sequence dependent
scheduling on identical parallel processors.

Production scheduling on parallel processors with
independent tasks was addressed by Rothkopf (1966). The
author derived a procedure for determining optimal schedules
for a single machine and a dynamic programming approach to be
applied to an important class of scheduling problems involving
parallel processors. A branch-and-bound procedure was
developed by Elmaghraby and Park (1974) to schedule jobs on

several identical machines. This model assumed that all jobs

19

are available to be processed at time zero and that there are
no a priori precedence relationships between the jobs. Barnes
and Brennan (1977) improve the Elmaghraby and Park algorithm
by proposing three refinements to the algorithm. These
refinements, a bounding procedure and two methods of
establishing natural precedence relationships, are combined in
a more efficient fathoming procedure.

Dearing and Henderson (1984) used a mixed integer linear
program to describe a scheduling problem in the weaving
industry and solved it by relaxing the integer requirements.
Frendewey and Sumichrast (1988) extended this work with a
formulation that considered setup costs, lost production and
overtime costs. Through an alternative problem structure, the
authors reduced the number of variables that are required to
take integer values, making it possible to solve problems of
a realistic size.

Oliff and Burch (1985) implement a production planning
procedure that considers sequence dependencies in the context
of hierarchical production planning. Their approach uses the
production switching rule developed by Mellichamp and Love
(1978) to determine aggregate production levels. A heuristic
is then applied to sequence production within the constraints
imposed by the aggregate production plan. In related
research, Burch, Oliff and Sumichrast (1987) propose a pair of
heuristics in a hierarchical context to derive schedules for

the production of fiberglass cloth.

20

In one of the earliest articles dealing with production
sequencing in the process industries Prabhakar (1974)
described a manufacturing environment in which chemicals were
produced in reactors with subsequent processing. The author
presented a mixed integer program designed to determine
production quantities for parallel processors. The
formulation also considers sequence dependency and limitations
on the storage of intermediate products.

Geoffrion and Graves (1976) propose two versions of the
quadratic assignment method to solve the problem introduced by
Prabhakar. The first approach is the direct application of a
quadratic method, while the second involves applying a linear
program to augment the procedure. Joint application of the
quadratic assignment method in conjunction with linear
programming proved to be more effective for solving the
general formulation of this type of scheduling problem. The
works by Geoffrion, Graves and Prabhakar are complete
descriptions of production sequencing in the process
industries but do not consider a linkage with a cutting
problem.

Smith-Daniels and Smith-Daniels (1986) introduced a lot-
sizing and sequencing formulation for packaging lines in the
process industries. In this formulation the authors minimize
the sum of inventory holding, back-order, setup and tear-down
costs while considering sequence dependencies of items within

a part family. This capacity-constrained formulation assumes

 

21
that one part family is produced on each packaging line in any
period. Later, Smith-Daniels (1988) introduced a heuristic
procedure to solve the problem.

In an important extension of the Economic Lot Scheduling
Problem, Singh and Foster (1987) considered situations with
sequence dependent setup costs for a single machine. They
proposed a heuristic procedure that breaks the problem into
three stages. Although the approach is for a single-machine
facility, an application is described where the algorithm is
embedded in a multi-machine, multi-product environment.

An alternative method of modeling dynamic production
planning problems with sequence dependent setup costs is by
redefining the sequencing variables and adding a set of
logical variables (Bruvold and Evans, 1985). This reduces the
number of 0-1 variables, increasing the likelihood that the
problem can be solved with conventional mixed integer program
solution procedures. Models incorporating this approach can
easily be extended to consider alternative objective functions

or losses due to start-up or shut-down.

THE STOCK CUTTING PROBLEM

Several characteristics can be used to describe the stock
cutting literature. These are: the dimensions of the problem,
the solution procedure (heuristic or optimizing), and whether
the cuts are guillotine (edge to edge). Dyckhoff (1990)
recently described cutting and packing problems and the

connection that exists between them.

22

Although references to cutting and packing problems date
from the late 19305 and early 19405, the first research to
propose an effective algorithm for solving larger problems was
introduced by Gilmore and Gomory (1961). This linear
programming based approach describes an efficient method of
solving the stock cutting problem by limiting the number of
variables using auxiliary problems. In a companion article
Gilmore and Gomory (1963) consider three practical limitations
on the problem, the number of cutting knives, machine
balancing concerns and multiple orders for several machines.

Gilmore and Gomory (1965) also considered higher dimension
problems that can be decomposed into a series of guillotine
cuts performed in stages. More recent research on the two-
phase approach has incorporated an automatic sequential search
procedure to obtain a compromise between computational cost
and trim loss (Ferreira, Soeiro, Neves, & Fonseca e Castro,
1990).

The computational effort can be lessened at the expense
of larger trim loss by reducing the number of cutting patterns
considered. Several authors besides Gilmore and Gomory have
introduced procedures to select better cutting patterns.
Haessler (1975) developed a format for a one-dimensional trim
problem that uses a fixed charge to control the size of the
problem by limiting the number of cutting patterns considered
for the basis. This mechanism optimizes the tradeoff between

minimizing trim waste and computation effort. Pierce (1964)

 

23
proposed the selection of dominant schedules to be considered
for the solution basis.

Stock cutting questions dealing explicitly with the paper
industry include work by Pierce, Johns (1966), and Haessler
(1977). Pierce addressed the paper trim problem for both
single and multiple machines and proposed heuristic solution
procedures. Pierce's formulation minimizes the variable
production costs while disregarding the capacity constraints
and sequence dependence of paper production. Johns suggested
heuristic procedures for solving the paper trim problem while
Haessler described the single machine roll trim problem.

Haessler (1988) proposes two heuristics to solve the roll
trim problem. The first involves relaxing the integer
requirement for the pattern usage. Then additional patterns
can be selected for entry into the basis by considering the
reduced cost of a nonbasic pattern. This procedure reduces
trim loss by increasing the number of cutting patterns
considered for the basis. The second approach proposed by
Haessler is a sequential method where new cutting patterns are
added according to a selection criterion. The two approaches
work at cross purposes; the linear programming based approach
offers more opportunities to lower trim loss by increasing the
number of patterns considered, while the sequential approach,
by contrast, minimizes the number of schedules at the expense
of increased trim loss. The author uses several examples to

illustrate where and how each approach, or a combination of

 

24
the two, would be applied in actual practice. Sweeney and
Haessler (1990) propose solution methods for one-dimensional

cutting problems with rolls of multiple quality grades.

SUMMARY AND CONCLUSIONS

The literature bases are extensive for both the stock
cutting and production scheduling problems. However, all
prior researchers have viewed the problems separately, while
in many manufacturing situations, and the paper industry in
particular, they are clearly linked. Therefore, the most
important contribution of this research is the linkage of

these two problems in the context of paper production.

 

CHAPTER 3: CONCEPTUAL FRAMEWORK AND FORMULATION

Existing formulations of the paper trim problem describe
the number of reels of paper scheduled to be produced on each
machine as well as how the reels should be cut into rolls to
meet customer demand. This research extends the existing
formulations to consider multiple products and the associated
sequence dependency of production in a time-phased

environment.

ASSUMPTIONS OF THE MODEL

The analytical model presented describes the interaction
between the production schedule and the stock cutting
decisions with the objective of optimizing the use of
production facilities and raw materials. While the current
model is designed for the paper industry, the basic concepts
can be applied to other industries as well. The decision
tools developed in this research will enable managers to
specify more precisely when and to which product a machine
should switch in order to utilize the production facilities
most efficiently while meeting customer demands.

The first important assumption is that production
scheduling on the paper machines can be modeled independently
from the preceding pulp production and subsequent conversion
processes. Based on operations at an active mill, pulp
production process can be assumed to have sufficient capacity
and flexibility to meet the needs of most production
schedules. Subsequent processing of the reels of paper, the

25

 

26

conversion process, is effectively decoupled from the
production scheduling on the paper-making machines because its
requirements can be specified in terms of rolls of paper.
For all practical purposes, the conversion process is treated
as another customer, specifying time phased demands that are
inputs for the model. This is consistent with work done by
earlier researchers.

The formulation assumes that paper is cut immediately
after production, and that this process is not bound by
capacity. The slitting operation, conversion of reels to
rolls, is accomplished by rewinding the paper with knife
blades placed at measured intervals in the paper’s path. The
time required to perform this procedure is negligible,
compared with the time required to produce the paper. The
model further assumes that paper type is not a function of the
machine it is produced on.

There are two other features of the paper industry that
play an important part in the formulation of the model.
First, paper machines operate continuously and require that
the transition to the following paper begin when the
production goals for the current paper are completed. Second,
backordering is not an acceptable practice in the industry and
demands are required to be delivered on their requested due

dates.

27

MODEL DESCRIPTION

The formulation determines the number of reels of paper
of each type to be cut according to each pattern, the order in
which the reels should be produced on each machine in the
facility, the level of inventories throughout the horizon and
the composition of lost sales. This is done in a way that
minimizes the sum of production cost, the cost of over-
production, the cost of lost sales and the transition costs
incurred as machines are switched between types of paper.

Constraints on the decision maker are: the specific
capabilities of each.machine, particularly the width; the time
available for production on each machine; and the demand
requirements. The formulation must also recognize the need to
specify a unique, uninterrupted sequence of transitions
between types of papers on each. machine over the entire

planning horizon.

MODEL FORMULATION

The objective of the formulation is to minimize the total
of the production and cutting costs, transition costs, the
cost of over-production and the cost of lost sales. This
objective is met while subject to two major classes of
constraints. The first deals with the usual requirements and
limitations of meeting demand with inventory and production
subject to capacity limitations. The second class of

constraints assure that the sequencing relationships are met.

28
The following formulation is organized in three parts:
definitions, an explicit statement of the formulation, and
then a detailed explanation of the objective function and each

constraint in the formulation.

Definitions

Indices:

i labels the state (type of paper currently produced)
of the machines, i = 0,1,...,N, where i = 0

indicates the communicating state that each machine
must visit at the beginning of a period and N is
the number of products produced;

j labels the state (type of paper) to which a machine
will be switched, (j = 0,1,...,N), where j = 0 is
the communicating state that each machine must
visit at the end of each period;

k indexes the widths that a paper of a given type may
be demanded by customers, k = 1, 2,..., K”
m enumerates the machines, m = 1,2,...,M, and M is

the number of machines in the facility;

q indexes the cutting patterns for machine m and
product i, q = 1,2,...,Qm;

t enumerates the planning periods, t == 1,2,...,T,
where T is the number of planning periods;

Parameters:
bmt the time available on machine m in period t;
cim the time required to produce a reel of product i on

machine m;

Cimq the sum of production cost and trim waste, for a
reel of paper of type 1 produced on machine m and
cut according to pattern q;

Dan the demand for rolls of paper of width k, type i in
period t;
am the time required to switch from paper 1 to paper j

on machine m;

 

E

ijm

lfln

PM“

rm

m0

ViOmO

Wik
Variables:

d

unit

It:

ml

V

ijm!

29

the transition cost for changing from paper i to
paper j on machine m;

a large number;

the cost of lost sales, per roll, for paper of type
1, width k in period t;

the width of paper-making machine m;

the number of rolls of width k to be cut from each
reel using pattern q on machine m when producing
product i, subject to the condition

N K1 91m
EEZWikPl-kmqsLm m=l,2,...,M (3-1)
1=1 =1 q=1

the recycling cost for a roll of paper of type i
and width k;

the initial slack (unused time) on each machine at
the beginning of the planning horizon, assumed to
be zero;

the initial conditions for each machine, where me
is a binary variable indicating a transition from
paper i to paper j;

the k“1 physical width (in inches) of paper type i;

a real number associated with state i on machine m
for period t, used to prevent sub-touring;

the total cost of the production schedule;

the inventory, in rolls, of width k, product type i
at the end of period t;

the number of rolls of width k of product i
produced in period t;

the slack (accumulated unused time) on machine m at
the end of period t;

a zero one variable showing a switch between paper
types i and j (i¢j) on machine m in period t;

 

3O

Ximqt the number of reels of type 1 produced on machine m
and then cut according to pattern q in period t;
Zikt the demand, in rolls, of width k and type i not
satisfied in each period (lost sales).
Formulation
Minimize:
T N M 9111! T N
G: 2 2 E Cimgimqt + Z:
t=1 1=1 m=1 q=1 t=1 1=1

Subject to:

M pin
2 2 Pikquimqt _ Rikt = O
m=1 q=1

N Om N N
E CimXimqt + Z 2 eijmvijmt '
q=1 1=1 1=1

1‘1

1=1

S

m(t—1) + Smt = bmt

Iik(t-1) + Rik: ‘ Iikt + Zikt = Dikt

Qjm

N
1=0 q=l

r???“

n n
HHH
Hgfiz

n3

n WK
H H

H

n 5 L‘-
II II II
HEIZ

\-

(3’2)

(3-3)

(3-4)

(3-5)

(3-6)

V.

g.
'1 Z
1..
§
('1'

2

V0

Ll
II
p

ViOm(t-1)

jmt =

=1

._ V0

imt

=0

31

N N
Z Vijmt ‘ Z Vjimt = 0
1=o 1=o

dimt _ djmt
Zikt' Ximqt'

Smt'

Riktl Iikt’ dimtl

Ximqt

V

ijmt

20

m = 1,2,
t: = 1,2,
m = 1,2,.
t = 1,2,.
1 = 1,2,.
m = 1,2,.
t: = 1,2,.
j = 1,2,.
m = 1,2,.
t = 1,2,.
1:0,1,
1 = 9:1:
ms” £7312“
t = 1,2,
1' = 1,2,
k = 1,2,
m = 1,2,
g: ll2! '
t: = 1,2,.

integer V 1,171, g,

“-~

"39::
3

‘~

'93 22

s~

iVZ

and t

Oor1Vi,j,m, andt

(3'7)

(3'8)

(3'9)

(3-10)

(3-11)

(3-12)

(3-13)

(3-14)

32
EXPLANATION OF THE FORMULATION
The following section is divided into two parts, a
description of the objective function and a detailed
explanation of the function of each of the constraint sets.

Objective Function

Minimize:
T N M Qim T N N M
G‘ E )3 2 Ciquimqt + 2 Eiijijmt +
t=1 1=1 m=1 q=1 t=1 1=1 1=1 m=1
1‘1 (3—2)
N Ki T N K1
IiinkT + E 11kt Zikt

H
II
p
7?
II
H
V?
II
p.)

[.-
II
P
a.
II
p.»

The objective function has four major components:
production and cutting charges, setup charges, the cost of
lost sales and a recycling charge for any excess rolls
produced. The possibility of producing extra rolls results
from limiting the available cutting patterns to those that are
non-dominated. Although the formulation stipulates that the
rolls would accumulate until the final period, in actual
practice extra production would be recycled as it is produced.
Constraints

Conversion of reels to rolls in each period:

M 9m 1 = 1,2,. ,
Z: Z: Pikquimqt _ Rikt : 0 k = 1’2' ' 1K1 (3-3)
m=1q=1 t : 1,2

The first set of constraints, equation (3-3), relates the

number of reels of paper, XMm, of type i cut according to

 

33
schedule q in period t on machine m to the number of rolls,
Rm, of width k of paper type i produced during period t, where
Kiis the number of widths paper 1 is demanded in. The number
of rolls of width k cut for product i on machine m must be
selected such that the total of the roll widths does not
exceed the width of the paper machine. Each cutting pattern
has elements PM“ (the number of rolls of width k cut from a
reel of type i produced on machine m and slit according to
pattern q). Cutting patterns are generated such that
N K1 01..

2;:wikPimqsLm m=l,2,...,M (3—1)
1=1 =lq=1

where Wik is the dimension of the km width for product i and Lm
is the width of paper machine m.

Machine capacities:

N Om N N
Z: 2 CimXimqt + eijmvijmt _
1=1 q=1 1=1 g: (3-4)
_ 1n= l 2 ... M
Sm(t—1)+Smt_bmt t=l:2:...lT

I

Equation (3-4) describes the capacity constraints for
each machine in each period. Machine capacity not required
for the current period can be reassigned through the slack
variables, Sm, to the following period. The time required to
switch from paper type i to type j on machine m is em” and the

time to produce a reel of paper type i on machine m is cm.

 

34
The total amount of time available on machine m in period t is
bmt (usually 24 hours).

Production, inventory, and demand balance:

(3-5)

1

1 .. N
Iik(t—1) + Rikt ' Iikt + Zikt = Di,“ 1; ”I;

I"'I

Equation (3-5) describes the relationship between demand,
production and inventory for each of the roll widths for each
type of paper.

Setup before paper production can begin:

N Qjm j
no r1 t

~

€22

(3'6)

II II II
HFJH

,2,...
,2,...
,2,...

The i index shows the type of paper for which a given
machine is configured. The zero state indicates a null state
used to transmit the status of the machines at the end of the
prior period to the current period. The formulation relies on
the assumption that the ending condition of each machine in
period t is identical to the beginning condition of the same
machine in the next period. To achieve this there must be a
theoretical transition to an artificial ending state (0) for
each machine in each period, (Violm = 1). In addition, each
machine must begin each period by leaving the initial state,
(me = 1). These requirements are enforced by equations (3-7)

and (3—9).

 

35

Required transition to the ending state:

.. M' _
iOmt = l .I (3 7)

‘bflz
<

...
II
P

Equation (3-7) ensures that there is a transition to the
ending state for each machine in each period. Equation set
(3-8), included earlier for clarity, is omitted when solving
the model. It ensures a transition out of the initial state
and is made redundant by equation (3-9).

Continuity of machine state:

v30m(t—1) _ L’oimt = 0

chH
II II II
H+4H

,2,---,N
,2,...,14 (3-9)
,2,...,T

Equation (3—9) ensures that each machine begins each
period producing the same type of paper it was producing at
the end of the previous period.

Setup restrictions:

1=O i=0

N N j = 1,2,... N
Z Vijfllt - E Vjimt = O Ig:%:§: : z - [L141 (3—10)
Constraint set 3-10 requires that if there is a
transition into a state on a machine in a period, then there

must be a transition out of that same state.

Sub—tour restrictions:

— djm, + (N+1) v. (3-11)

1

jmtS‘N

~ ~

rr§l<$<.H.
II II M» II II
hi3 2:2

 

36

The Tucker (1960) constraints, equation set (3—11),
assure that each state (paper to be produced) for each machine
in each period will be assigned a unique real value, dmv
Through this requirement there will be one complete production
sequence for each machine in each period. This sequence will
begin and end with the null state, and make one visit to each
state in which there will be production.

Equations (3-12), (3-13) and (3-14) define the domains of
the variables. It is important to note that the inventory and
lost sales variables, although defined as continuous, will be
integer in any solution through additivity and the integrality
requirements of Xma-

Equation (3-15), not shown in the complete formulation,
is an auxiliary set of restrictions that can be used to limit
the number of changes on each machine at the paper mill by
requiring that a minimum number of reels, g, are produced if

there is a transition to a new paper.

91ml:

i
Z: Ximqt _ 961mg,- + gViOmt Z O m
q=1 t

..,N
.,M (3-15)

' I

II II II
HFJH
NBJN

The binary indicator variables, &

um!

represent the
decision to switch to paper type i on machine m in period t.
A strict implementation of this policy is difficult because of
situations where production overlaps periods. Relaxing the

minimum production amount for the last paper produced in the

37
period is a practical way to meet this type of managerial

priority with the current formulation.

SUMMARY

Chapter 3 presents the formulation for the stock cutting
problem iJ1 conjunction with sequence dependent production
scheduling within the specific context of a facility producing
specialty papers. The objective is to minimize the sum of
production, cutting, transition, lost sales and over-
production costs, subject to demand, capacity and sequencing
constraints. Paper-making machines operate continuously,
requiring that the transition to the next paper scheduled
begin immediately after production of the current paper
ceases. This is addressed through the slack variables and a
null production state. The formulation, as presented, is a
rigorous description of the production planning problem that
confronts managers in the paper industry. However, the large
number of integer and binary variables make conventional
solution procedures impractical. The following chapter

describes an effective heuristic solution procedure.

 

CHAPTER 4: HEURISTIC SOLUTION PROCEDURE

This description of the heuristic procedure used to
obtain solutions for the formulation presented in Chapter 3 is
presented in four sections. The first is an introduction
describing the operation of the heuristic in broad terms. The
next two sections describe in detail the two major steps of
the heuristic, determining the production sequence and then
specifying the number of reels to produce. The fourth section

summarizes the chapter and presents important conclusions.

INTRODUCTION

The heuristic, shown in Figure 4-1, is an iterative
procedure that produces a production plan by adding
constraints and a limited number of binary variables to the
formulation described in Chapter 3. After the additions, the
heuristic proceeds by repeatedly solving the enhanced
formulation with the binary requirements enforced on selected
transition variables and the domain restricted on a few of the

integer variables, X used. to determine the production

mm!
quantities.

Enforcement of the integrality requirements is based on
two major objectives. The first is determination of a
suitable production sequence, that the different papers will

be produced on each machine. Second, after the production

sequence is set, production quantities can be specified.

38

V

Complete
Production Schedule

 

Figure 4-1:

39

 

Determine First
Incomplete Period
(1)

 

 

 

     
      

Specify Vwm, Nona,” and
VIOmGH) Binary

Set CNT‘ ofthe Vuml
Variables as Binary

 

 
   

Fix Existing Sequence

 

 

Set Domain for Production
Quantities in Prior
Periods (leq, to ximqfl-ll)

 

 

 

Heuristic Procedure

 

 

40
PRODUCTION SEQUENCING
Production sequencing is accomplished by determining the
last period with an incomplete production sequence on one of
the machines and then selectively enforcing some binary
requirements for variables describing transitions on machines
in the incomplete period. Once enough of the transition

variables, V are specified as binary (or are fixed), to

mm:
determine a production sequence for the period, the heuristic
then sets the production quantities. Although the procedure
for setting the production quantities is straightforward and
will be described later, it is necessary now to describe some
additional details associated with determining the production
sequence.

Figure 4-1 shows the sequence of steps used to solve
problems using the heuristic procedure. Initially the
formulation is solved with all of the integrality restrictions
relaxed. Subsequently, selected sets of the transition and
production variables are forced to take integer values
requiring the solution of a Mixed Integer Program (MIP) at the
start of each iteration.

The results from the solved. MIP are then tested to
determine the first period with an incomplete production
sequence, period I, a situation where some transition
variables in the period are not binary. Completely binary
transition variables guarantee, through the Tucker

constraints, a unique and uninterrupted production sequence

41

for all machines in periods prior to period I. Transition
variables for those periods are fixed at their current values.
Transition variables for period I that are a part of completed
sequences are fixed up to but not including the transition to
the null state at the end of the period. This leaves the
possibility of adding another paper to the sequence during the
next iteration.

Three mechanisms are incorporated in the formulation to
encourage the complete sequencing of period I in the next
iteration. Specifying all variables describing the transition
from the endpoints of the completed sequences is the first and
most important mechanism. Second, all transitions to the null
state at the end of period I are required to be binary.

A final mechanism assures that period I will be correctly
sequenced. This is done by rapidly increasing the number of
transition variables in period I required to be binary. Each
time there is an attempt to sequence the same period the
counter, CNT, is incremented. This count is then raised to
the sixth power to determine the number of transition
variables, in addition to those already described, that will
be specified binary.

Transitions out of the null state in period I+1 are also
required to be binary. This assures that once period I is
sequenced, there will be continuity between the ending state
of each machine in period I and the initial states of the

machines in period I+1.

42
Several mechanisms are employed to assist the formulation
in completing the production sequence in a given period. The
first is the addition of binary variables, 6mv Their purpose
is to show whether there will be production of paper j on
machine m in period t. The delta variables work with an

additional constraint set, (4-1), in two ways.

N j
2 Vijmt ‘ 5m: = 0 ’g

4%
1i]

l2l"'l
,2,...,M' (4-1)
,2,... T

I

ll ll ||
HFJH

First, if there is a transition to paper j these constraints
require that the sum of all transition variables describing
entrances to state j will equal one. This is an important
property because it forces the formulation to recognize, in
some sense, one complete setup. Second, the constraints
require that transition variables for papers that will not be
produced are set to zero.

Solving this formulation with the binary requirements of

the V variables relaxed frequently produces a valid

ijmt
production sequence, however, this is not guaranteed.
Incomplete sequences can occur when the transition variables
are allowed to take fractional values. This results in

solutions that suggest several simultaneous partial

transitions to the next paper scheduled on a machine. Figures

 

ixi‘l“

43
4-2 and 4-3 give examples of solutions with and without

complete sequences for a single period on one machine.

V =1
02 V23=1

A
@®@@®@
V

mm

V1o=1
Figure 4-2: Solution With a Valid Sequence

Figure 4-2 is a complete production sequence, with a
clear ordering of the papers, beginning in the null state and
then entering state 2, proceeding to paper 3, then paper 1 and
finally returning to the null state. In the example where the

transition variables are allowed to take fractional values,

 

44
Figure 4-3, it is not possible to establish the production

sequence .

Qi/
Vu=06

vm=os

Figure 4-3: Solution With an Invalid Sequence

When transition variables do take fractional values the
formulation must be constrained further by stipulating that
the variables describing the transitions to the null state are
binary. However, there is an additional practical
consideration that further limits the number of transition
variables that take fractional values.

As first pointed out in Chapter 3, the transition to
different types of papers can be costly and may introduce
difficulties in achieving consistent quality. For this and
other managerial reasons, many production planners would

prefer to produce at least a minimum number of reels, 9, once

 

45
production begins. This operating priority is carried out

with a set of constraints of the form:

out

i
E Ximqt _ 961ml: + gViOmt 2 O m
q=1 t

1,2,...,N
1,2,...,M’ (3’15)
1,2, .,

In this research the minimum number of reels of
production for each setup was set at four. This addition to
the formulation plays an important role in the performance of
the heuristic for several reasons. First, the minimum
production size limits the number of papers produced on a
machine in a period to three or four. Second, higher
production volumes in constraint set (3-6)

N Qjm j
F0 r1 t

1,2, . ..,N
1,2,...,M (3-6)
1,2, .. T

° I

will encourage each of the transition variables associated
with paper types that are produced to take larger values.

Once the values of the V-

ljmt

variables are large enough, the

Tucker constraints, equation (3-11), become binding.

dimt _ djmt 1 (3-11)

+ (N+1) V7,,t s N

or S KL“. H.
II ll ‘0‘ ll ll
H i: 2 2

-
ax

Finally, and most important, constraint set (3-15) allows
the enumeration process to rule out sequences with small

production quantities. In some cases this can assist the

 

46
fathoming of the branch-and-bound tree considerably with a
minimal impact on solution quality.
It is useful to note that the minimum required production
quantity, constraint set (3—15), provides enough structure to
make it practical to require that the variables describing the

decision to produce a paper on a machine in a period, 6 are

jmt r
binary for the period with the incomplete sequence. However,
this requirement and the added constraints are not sufficient
to guarantee that the production sequence will be complete
through the current period. For this reason, it is necessary
to restrict further the values of the transition variables,
anI in the incomplete period.

The most confining restriction is a requirement that
transition variables whose solution values were equal to one
in the prior iteration and are part of a complete sequence are
set at one for all subsequent iterations, fixing established
production sequences. An exception is made for variables
describing transitions to the null state. These transitions
are kept binary until the sequencing is complete on all
machines for the current period, providing the heuristic with
additional opportunities to schedule different papers on a
machine by assuring an endpoint for each production sequence
in the first incomplete period.

The next restriction stipulates that all transitions from
the endpoints of the production sequence on each machine are

binary, forcing a complete transition to the next paper or the

 

47

null state, should the solution procedure chose to do so.
Three other sets of transition variables are also restricted
to binary status. These are the required transition to the
null state in the current period and the transitions from and
to the null state in the subsequent period. These are
important because they require continuity between periods and
they constitute class 1 specially ordered sets (Beale &
Tomlin, 1969) whose structure is exploited.

Allowing some transition variables to take real values
can generate solutions with sub-tours. Sub-touring, shown in
Figure 4-4, occurs when complete transition sequences can be

devised that do not pass through the null state.

VM=L° va=m4

A
@ooooo

Vm=LO
Figure 4-4: Sequence with Sub-Touring

Sub—touring is prevented by increasing the number of
transition variables (besides the specially ordered sets and
the required transition from the endpoints) required to take
binary values on each successive attempt to sequence a period.
For instance, the first time there is an attempt to sequence
period one, no variables beyond the previously mentioned sets

would be set binary. On subsequent attempts to sequence

 

48
period one, the number of transition variables specified as
binary would be increased rapidly. This approach allows the
procedure to establish the production sequence and then assure
that no sub-touring has taken place. Once the sequencing,
without sub-touring, is complete for the first period, the
sequencing procedure can begin for the second and the

production quantities can be set for the first.

SPECIFYING PRODUCTION QUANTITIES

As mentioned earlier, determination of the production
quantities is straightforward. The approach is to establish
an integer set of possible production quantities around the
truncated Xm¢ values found during the sequencing step of the
heuristic. For this research the set of integers had a range
spanning from two below to three above the truncated Xmm
values.

Two sets of variables are given ranges. The first set
consists of any production variables for papers scheduled to
be produced in the period immediately preceding the period
currently being sequenced. The ranges of production
quantities provide additional flexibility for the sequencing
procedure while assuring that the integrality requirements
will be met in the earlier period. Production quantities are
fixed at their integer values for periods more than one

period before the one undergoing sequencing.

 

49

In most situations the procedure just described would be
sufficient to guarantee production in integer quantities.
However, if more than one period happens to be sequenced in an
iteration it would be possible to have fractional production
variables in more than just the prior period. Therefore,
ranges are also applied to any of the production variables in
periods more than one period before the current period that
take fractional values.

In situations where the total demand level is low and the
bulk of the demand is required late in the horizon, the
heuristic is faced with a host of identical solutions (from a
cost perspective). Most production schedulers would prefer to
keep the machines operating in the short term in hopes that
additional work would come available later. This priority is
implemented by attaching a small profit (when compared to the
transition costs) to the slack variables, where that profit
increases in the later planning periods. This incentive
motivates the formulation to minimize slack in the earlier
periods.

At modest levels of utilization the heuristic, as just
described, solves most of the scheduling problems
expeditiously. However in situations where there is the
possibility of substantial slack early in the horizon or lost
sales late in the horizon, an additional concession can be
made to limit the number of alternative solutions and increase

the performance of the heuristic procedure.

 

50

Large numbers of essentially equivalent solutions are
generated because each lost sale has the same cost despite
type or timing. As a practical matter, there is no indication
that the decision to reject a sale occurs at the plant level.
In situations where it is likely that a plant will be unable
to meet all of its scheduled demand, steps are taken to reduce
the demand level or change its timing. For this research,
lost sales are an important part of the planning criteria and
a mechanism that introduces minor differences in those costs
can improve the solution procedure. In this case, an
additional charge is attached to the cost of a lost sale,
where the cost varies as a function of the paper type and time
period. This provides the branch-and-bound. procedure an
additional mechanism to differentiate nodes for fathoming.

Although these enhancements were developed to increase
the solution speed of the heuristic, they also assure that
work is completed as soon as possible, making it obvious when
the machines will be out of work. There might be situations
where it would be useful to reverse the incentive structure,
forcing slack time to the beginning of the planning horizon.
Solutions generated this way would provide :3 conservative
estimate of the earliest time that capacity can be made

available for unscheduled orders.

SUMMARY AND CONCLUSIONS
Chapter 4 has described a practical heuristic solution

procedure for obtaining good solutions to the formulation

 

51
presented in Chapter 3. The two part procedure seeks to
determine a production sequence, the order of production of
paper on each machine, by adding new papers to an existing
production sequence one period at a time.

After the production sequence is established in a period,
the second step of the heuristic specifies the number of reels
of paper to produce in the periods up to and including the
period sequenced most recently.

Although this procedure will produce a production
schedule, it is not an optimizing procedure; consequently, the
quality of those schedules cannot be guaranteed. The next
chapter describes the procedure used to benchmark the

performance of this heuristic procedure.

 

CHAPTER 5: MODEL CALIBRATION AND BENCHMARK PROCEDURE

As mentioned earlier, the production scheduling of stock
that will be cut has applicability in a broad spectrum of
manufacturing environments. While limiting the discussion to
the paper industry diminishes that spectrum considerably,
there are many different market and manufacturing environments
in the paper industry alone. Within the paper industry the
simplest manufacturing, from a production scheduling
perspective, is the production of large volumes of essentially
the same product, such as newsprint. Production is
standardized, not only in terms of the type of paper produced
but also with respect to the widths. This type of application
requires machine widths of up to 300 inches, the largest
machines in the industry. At the other end of the spectrum
are plants that produce a ‘wide assortment of colors and
textures as in the production of construction paper. Plant
operations in these facilities are more consistent with batch
or semi-batch procedures where the machines are stopped
between production runs so that they can be thoroughly
cleaned before the production of the next paper begins.

A market that is experiencing considerable growth is
specialty papers, such as facsimile, most types of writing or
computer papers (uncoated freesheet), and the backings for
pull away stickers (release paper). Plants that produce for
these markets typically have smaller machines producing

several different types of paper, and each paper can be

52

 

53
produced in several grades and/or basis weights. The problems
used for testing the performance of the heuristic were
generated based on nwnufacturing and marketing situations
consistent with this type of production. The information used
to specify distribution functions for the parameters of the
test problems was obtained through interviews with individuals

in the paper industry.

PROBLEM GENERATION

Problem generation begins by entering parameters that
characterize the size of the problem. These are the number of
periods (T), the number of machines (M), the number of
distinct papers (N), and the level of expected demand. Based
on this general description of the problem size, a complete
manufacturing scenario is generated using probability
distributions described below.

Generation of the cutting patterns for each machine
depends on the capabilities of the machines, primarily the
width, as well as the widths of paper required. For the
specialty papers on which this research is based, the machines
are toward the narrow end of the spectrum, 100 to 140 inches
in width. Further, there is considerable diversity in the
paper widths demanded; at a representative mill, the narrowest
width demanded is 20 inches while the largest of the common
widths is nearly 80 inches. Machine widths and paper widths
are uniformly distributed between the largest and smallest

width, and rounded to the nearest inch.

 

54

Cutting patterns and the associated cutting charges are
generated by enumerating all possible patterns. While this is
tedious, it is only necessary to do it once. For the purposes
of the simulation all rolls in an order are for the same type
and width of paper. Each kind of paper could he demanded in
a maximum of four different widths over the planning horizon.

To limit the number of possible patterns to those that
are most likely to be useful, only dominant patterns (defined
in Chapter 1) are considered. Although this assumption will
tend to cause over-production, the amount is small when
compared to the total production volume, and in actual
practice the cost of storing or disposing of excess production
is quite small.

Simulated demand, measured in rolls of paper of a given
width and type, is generated for each day of the planning
horizon. The number of rolls demanded depends on two
parameters, the largest number of orders expected in a day,
and the maximum number of rolls required per order. All
parameters are drawn from the appropriate uniform
distributions.

Transition times and costs for each machine vary
considerably depending on the degree of difference between the
types of paper. Changing between basis weights is relatively
easy and requires only a few minutes, while transitions
between papers of different colors may require stopping the

machine and completely rinsing it clear, a procedure that can

 

 

55
take several hours. This research assumes that several
similar kinds of paper are produced, each in an assortment of
basis weights. Based on similarities in the paper types,
anticipated transition times are assumed to be uniformly
distributed between 0.1 and 2.0 hours. The expected time
required to produce a reel of paper is drawn from a uniform
distribution over 0.5 to 1.0 hours. Similarly, the transition
costs are uniformly distributed between $40 and $800. The
production cost has three components, energy, fiber and labor.
Trim waste and the recycling costs are adjusted to reflect the
value of the fiber that can be recovered. Fiber represents
one half the total cost, while labor and energy cost each
represent one quarter (D. Rish, personal communication, August
10, 1993). Finally, the cost of a lost sale was set at a

large arbitrary number.

BENCHMARK PROCEDURE

This research is based on operating conditions at two
mills that produce a fairly wide assortment of specialty
papers that are sold in bulk form and then used as a raw
material in subsequent manufacturing operations. This type of
paper mill may have three paper-making machines and will try
to establish a production plan for the next two to four weeks.
On this horizon, the production planner may have to schedule
five distinct kinds of paper with five to seven different
basis weights. Each type, paper kind and basis weight

combination, can have as many as four different widths. Table

 

56
5-1 shows the dimension of a small formulation describing a

production scheduling decision at the example mill.

Table 5-1: Representative Problem Size

Problem Parameter

Planning horizon 14 days
Number of product types 10
Number of machines 3
Maximum number of widths per product type 4

Variable Type

Cutting patterns, Ximql 3206'
Transition, an 4620'
Rolls of paper, Rikt 434
Inventory, Im 465
Lost sales, Zm 434
Slack, Smt 42
Sub-tour restriction, dm, 504
Total number of variables 9705

' Integer or binary

Besides the 9705 variables described in Table 5-1, this
formulation would have 6832 constraints. Although, by modern
standards, a problem of this dimension might be characterized
as a medium-sized linear program, the large number of integer
and binary variables precludes any opportunity of obtaining an
optimal solution within an acceptable amount of time.
Therefore, much of this research effort has focused on finding
an effective heuristic solution procedure.

The most appropriate mechanism for conclusively
establishing the effectiveness of a heuristic procedure would

be to compare the quality of the solution obtained using the

 

57

heuristic to the optimal solution. However, the computational
cost of obtaining optimal solutions for relatively small
problems can be prohibitive. Various methods were considered,
most notably Lagrangean.Relaxation (Fisher, 1981), to overcome
the intractable nature of solving the formulation to
optimality. Failure to find. a relaxation that produced
subproblems with an exploitable structure dictated the use of
a more direct means to demonstrate the effectiveness of the
heuristic procedure. This procedure entailed using branch-
and-bound (Land and Doig, 1960) to solve the optimal
formulation whose objective function is given by equation (3-
1). The solution obtained from the heuristic was employed to
assist the pruning of the enumeration tree. The performance
of the branch-and-bound procedure was further enhanced by
using specially ordered sets and a specified branching order.

Additional binary variables, 6m” whose purpose and usage
are the same as in equation set (4-1), will not compromise the
integrity of the formulation and offer opportunities to
improve the fathoming of the enumeration tree. Branching on
these variables first usually decreases the time required to
find an optimal solution. In a sense, addressing these
variables early in the fathoming process is analogous to a
manager first deciding whether to produce, and then later
deciding how much to produce. As mentioned in the description
of the heuristic in Chapter 4, the entrances and exits to and

from the null state constitute a specially ordered set whose

 

58

exploitation further improves the performance of the branch-
and-bound procedure. It is interesting to note that the
variables describing the switch to any given paper on each
machine in each. period are ordered sets also. However,
empirical trials showed that only the sets describing the
entrances and exits to and from the null state offered any
significant change in computation effort.

The objective of the benchmark procedure is to provide a
measure of the suboptimality of the solution obtained with the
heuristic. The best method is to use the objective value
obtained from the heuristic solution as an upper bound during
the fathoming of an enumeration tree for the optimal
formulation. If the enumeration tree is completely fathomed,
the solution obtained using the heuristic is shown to be
optimal; if the heuristic is suboptimal, then the actual
optimal solution will be uncovered.

The computation effort required to fathom the enumeration
tree using the heuristic’s objective value made this approach
impractical for most of the problems. The number of binary
and integer variables, as well as the availability of
alternate optima, generates enumeration trees that require
considerable amounts of memory and exorbitant amounts of
computer time to solve. Problems that have more than one
optimal solution could require the algorithm to consider a
large proportion of the enumeration tree. Further, if the

alternative optima were from disparate parts of the tree, the

 

59
algorithm would have to consider more of the tree
simultaneously. This may be the case when there is a capacity
limitation in the last part of the planning horizon and the
procedure is indifferent to which period the sale is lost in.

Given that the purpose of the benchmark procedure is to
estimate the error associated with the heuristic solution, it
is possible to use a series of proposed incumbents to
establish a bound on the optimal solution. For the purposes
of this discussion an incumbent is a bound on the actual
objective used to dissect the search space and increase the
chances that the pruning procedure will fathom the entire tree
or encounter a feasible solution.

A binary search procedure was applied in which the first
incumbent was a value halfway between the objective obtained
from the relaxed formulation and the objective obtained from
the heuristic. If the enumeration tree was completely
fathomed without encountering a feasible solution, a lower
bound was established on the optimal solution and another
incumbent was generated halfway between this bound and the
upper limit. The initial upper limit was the objective value
of the heuristic. Upper limits were updated if one of two
circumstances occurred: a better feasible solution was
encountered, or an incumbent value was too large to prevent
the enumeration tree from outgrowing the computer’s
capabilities. This binary search procedure continued until an

incumbent was proven optimal, or the value of the largest

 

6O
incumbent capable of fathoming the entire tree was
established. From this incumbent, a maximum error was

calculated for the objective found using the heuristic.

 

CHAPTER 6: COMPUTATIONAL RESULTS
The objectives of the research were to develop a solution
procedure for problems of a practical size and demonstrate the
quality of those solutions. For this purpose three classes of
problems, Validation, Independent and Scale, were developed

and solved.

INTRODUCTION

Each of the problem classes had a distinct function in
proving the effectiveness of the heuristic procedure. The
primary use of the Validation problems was to develop and
validate the heuristic, while the Independent problems
provided an unbiased evaluation of the heuristic's
performance. Scale problems were designed to give some
insight into how the computation cost would grow as the length
of the planning horizon increases, and to ShOW’ that the
heuristic could be used to solve problems of a practical size.

The Validation and Independent problems each have three
different sizes, (T), (M) and (P), with (T) the smallest. For
each size there are three different utilization levels, low,
medium and high. Each Scale problem has three machines, four
products and is specified at the low utilization level making
it possible to consider a broader range of values when
evaluating the effect of increasing the length of the planning
horizon on computation costs.

The next three sections provide descriptions of the
Validation, Independent and Scale problems along with their

61

 

62
accompanying results. Three tables illustrate these
descriptions. The fifth section of the chapter describes
several procedures that could reduce the problem size and/or
the computation effort. The chapter concludes with a summary

of the important results.

VALIDATION PROBLEMS

The first class of problems, shown in Tables 6-V-1
through 6-V-3, was used to develop and validate the heuristic
solution procedure. Parameters for this set of problems were
selected to ease ‘the coding’ and 'validation procedure and
provide insights into how the heuristic would perform in
diverse situations, such as large machine widths with small
paper widths, which induces a large number of possible cutting
patterns and substantial diversity in the manufacturing
environment.
Description of the Tables

Three sets of three tables, nine in all, describe the
problem sets. The first table in each set provides
information about the size of the problems. The first five
columns contain the name of the problem, the number of periods
(T), the number of distinct types of paper (N), and the number
of machines (M). Demand, the sixth column, is specified at
three levels. '

The best description of the problem size is found in the

last two columns of the first table of each set, the number of

63
variables and columns necessary to describe the optimal
formulation presented in Chapter 3.

Demand levels for the Validation problems are somewhat
arbitrary, with the average demand per day for the medium case
twice as large as for the low, and for the high demand case
three times as large as for the low. Demand patterns for the
validation problems were generated with limited regard for the
machine widths. Although it is assured ‘that there is a
machine with sufficient width to produce the paper, it is
possible to have lost sales in a wide paper while a narrow
machine is idle.

All paper widths generated for the Independent and Scale
problems are less than the widths of the narrowest machines.
This requirement is consistent. with the situation at the
example mill and produces much more realistic production
schedules, where, in actual practice, large mismatches between
the market demands and the production environment would be
addressed higher in the production planning hierarchy, before
the paper mill is committed to delivering the order.

The Independent problems were generated for three
different demand levels, where the low level of demand is
designed to operate the machines at 50 percent of capacity,
the medium demand level to operate at 70 percent of capacity,
and the high level case to operate at 90 percent of capacity.
Estimated utilization was the ratio of the expected number of

reels demanded to the expected shop capacity (in reels per

 

64

day). Calculation of the expected shop capacity is
straightforward, and is the product of the expected machine
capacity (reels per hour), the length of the production day
(twenty-four hours), the number of machines and the number of
days in the planning horizon. Expected demand is the product
of the expected number of orders per day, the expected number
of rolls per order, the number of days in the planning horizon
and the inverse of the estimated number of rolls per reel of
paper. The rolls per reel factor is estimated as the ratio
of the expected machine width (averaged across machines) to
the average expected paper width (averaged across paper
types).

Determining meaningful bounds for solutions obtained
using a heuristic is always difficult in this type of
research. The second table for each data set provides two
comparison measures for the objective obtained using the
heuristic. The first is a "Best Incumbent" category where the
best feasible solution that is obtained either with the
heuristic or optimizing procedure is recorded. The second
category used to interpret the quality of the heuristic
solution is the "Best Bound" column. This value represents
the best bound, without regard to feasibility, achieved
through the binary search procedure described in Chapter 5.

There are some important observations to be made about
the use of these tables. First, for the purposes of this

discussion, "incumbent" will refer to any feasible solution as

 

65

good as or better than the solution obtained using the
heuristic procedure. If the solution obtained using the
heuristic procedure is proven to be optimal, the values in all
three columns of the second tables will be identical, and the
value in the "Maximum Error" column will be zero. In
situations where a feasible solution is better than the
solution obtained using the heuristic and proven to be
optimal, then values in the "Best Incumbent" and "Best Bound"
columns will be identical and the maximum error will be
calculated based on the value of the incumbent value. In
situations where it is not possible to prove that either the
incumbent or the heuristic solution value is optimal, the
maximum error is calculated using the best bound.

Tables 6-V-3, 6-I-3 and 6-S—3 contain the objective
values of each problem solved with the integrality
restrictions relaxed, and the solution time required to solve
the heuristic using an Intel 486/66 microprocessor. Solutions
for the mixed integer programs in both the optimizing and
heuristic procedures were solved using version 2.1. of the
CPLEX Mixed Integer-Optimizer (CPLEX Optimization Inc., 1993).
The objective value quoted in the "Linear Relaxation
Objective" is obtained by solving the formulation with all of
the integrality requirements relaxed in the optimal

formulation.

 

66

Table 6-V-1: Problem Characteristics Report

Validation Problems

 

Problem
Name I _ _ Demand Variables Constraints
T-A-l 2 2 2 low 119 76
T-A-2 2 2 2 medium 104 72
T-A-3 2 2 2 high 83 68
M-A-Z 3 3 3 low 345 261
M-A-Z 3 3 3 medium 362 273
M-A-3 3 3 3 high 286 249
P-A-l 5 4 3 low 705 590
P-A-2 5 4 3 medium 655 590
P-A-3 5 4 3 high 782 610
Table 6-V-2: Performance Summary
Validation Problems
Problem Best Best Heuristic Maximum
Name Incumbent Bound Objective Error(%)
T-A-1 40,826.1 40,826.1 40,826.1 0.00
T-A~2 33,899.1 33,899.8 34,120.9 0.65
T-A-3 40,647.2 40,647.2 40,674.2 0.07
M-A-l none 171,245.2 171,597.6 0.21
M-A-2 953,024.9 950,080.1 953,401.8 0.35
M-A-3 727,453.2 727,453.2 728,265.3 0.11
P-A-l 501,052.4 501,052.4 501,052.4 0.00
P-A-2 531,733.0 531,733.0 532,423.2 0.13
P-A-3 none l,637,565.0 1,645,936.0 0.51

Tables 6-V-2 and 6-V-3 show that both the heuristic and
the optimizing procedure performed well on this type of
problem. In addition, the solution speed for the heuristic is
very good. Although these problems were not intended to give
insights into the relationship between the problem size and
the computation effort, the information in Tables 6—V—1 and 6-

V-3 make it appear that computation effort is increasing in

 

67
nearly the same proportion as the problem size. It should be
pointed out that the heuristic reads and writes several files,
and a substantial portion of the time used to solve the
problem is spent in these activities. The question of the

relationship between computation effort and problem size will

be addressed in the section dealing with the Scale problems.

Table 6-V-3: Problem Solution Report

Validation Problems

 

Linear Heuristic
Problem Relaxation Best Bound Solution
Name Objective Usinq CPLEX Objective Time
T-A-l 40,826.1 40,826.1 40,826.1 1
T-A-2 34,120.9 33,899.8 34,120.9 1
T-A-3 40,674.2 40,647.2 40,674.2 < 1
M-A-l 171,597.6 171,245.2 171,597.6 2
M-A-Z 953,401.8 950,080.1 953,401.8 3
M-A-3 728,265.3 727,453.2 728,265.3 3
P-A-l 501,052.4 501,052.4 501,052.4 6
P—A-Z 532,423.2 531,733.0 532,423.2 5
P-A-3 1,645,936.0 1,637,565.1 1,645,936.0 21

It should be noted that problems where the benchmark
procedure demonstrated that it was not possible to obtain a
schedule without lost sales were discarded. This procedure is
consistent with current practices. The authority to reject a
sale is not held at the plant level and when a situation does
arise where it is obvious that the plant will not meet the
demand requirements, arrangements are made to lessen the
demand or reschedule it. The interesting question of
determining which order to reject or reschedule is left for

further research.

 

68

INDEPENDENT PROBLEMS

The second set of problems, whose characteristics are
shown in Tables 6-I-1 through 6-I-3, demonstrate the
performance of the heuristic with an independent set of
problems. The number of machines in this class was kept at
three, a number quite common in practice. The parameters for
these problems were randomly generated using distribution
functions based on the operating conditions at an actual mill.
Detailed descriptions of the environment and the derivation of
the parameters used in the generation of the problems are

given in Chapter 5.

Table 6-I-1: Problem Characteristics Report
Independent Problems

Problem
Name T N M Demand Variables Constraints
T-B-l 2 low 151 104
T-B-2 2 2 3 medium 154 108
T-B-3 2 2 3 high 168 108
M-B-l 3 3 3 low 306 243
M-B-2 3 3 3 medium 351 243
M-B-3 3 3 3 high 361 267
P-B-l 5 4 3 low 751 600
P—B-2 5 4 3 medium 908 620
P-B-3 5 4 3 high 892 610

 

69

Table 6-I-2: Performance Summary

Independent Problems

 

Problem Best Best Heuristic

Name Incumbent Bound Objective
T-B-l 137,418.3 137,418.3 138,433.3
T-B-Z 120,910.6 120,910.6 120,910.6
T-B-3 160,068.0 160,068.0 160,676.0
M-B-l 108,651.7 108,651.7 108,990.6
M-B-Z none 271,194.0 272,819.9
M-B-3 none 301,023.4 303,095.2
P-B-l none 222,791.0 223,400.4
P-B-2 none 988,326.2 996,658.8
P-B-3 none 920,083.3 923,957.4
Table 6-I-3: Problem Solution Report

Independent Problems
Linear Heuristic

Problem Relaxation Best Bound Solution
Name Objective Using CPLEX Objective
T-B-l 136,836.2 137,418.3 138,433.
T-B-2 120,532.6 120,910.6 120,910.
T-B-3 159,825.4 160,068.0 160,676.
M-B-l 107,760.2 108,651.7 108,990.
M-B-2 270,687.3 271,194.0 272,819
M-B-3 300,174.6 301,023.4 303,095.
P-B-l 221,960.3 222,791.0 223,400.
P-B-2 987,529.3 988,321.2 996,658.
P-B-3 919,253.8 920,083.3 923,957.

* Time is in minutes on a 486/66.

Maximum

Error 1 g i

0.74
0.00
0.38

0.31
0.60
0.69

0.27

0.84
0.42

3
6
0

A
haHla

6 2

.9 3

2 3

4 3
8 35
4 13

The tables clearly show that the heuristic produces very

good solutions on these problems.

accomplished with minimal computation effort.

As mentioned earlier,

facing the possibility of missing a sale,

reschedule the demand requirements.

steps are

Further, these results are

when the production planner is

taken to

This observation makes it

 

7O
unlikely that the demand properties for the early part of the
horizon are the same as for later in the horizon. These two
features are addressed by lowering the maximum order size by

two-thirds in the first period and one-third in the second.

SCALE PROBLEMS

To give some insight into how the heuristic procedure
would perform as the problem size is increased, a third set of
problems was developed. The Scale problems only consider the
effect of an increasing planning horizon. For these problems
the number of products is fixed at four and the number of
machines is fixed at three. The following three tables

summarize the solution results.

Table 6-S-1: Problem Characteristics Report
Scale Problems

Problem
Name 1 M Variables Constraints
S-1 2 4 3 289 232
S-2 3 4 3 600 360
S-3 4 4 3 805 512
S-4 5 4 3 797 610
S-5 6 4 3 951 720
S-6 7 4 3 950 812
S-7 8 4 3 1311 992
S-8 9 4 3 1444 1098
S-9 10 4 3 1673 1240
S-lO 11 4 3 1713 1276
S-ll 12 4 3 2208 1512
S-12 13 4 3 1565 1482
S-13 14 4 3 2134 1652

 

 

 

71

Table 6-S-2: Performance Summary
Scale Problems

Problem Best Best Heuristic Maximum
Name Incumbent Bound Objective Error(%)
S-l 56,842.1 56,842.1 56,842.1 0.00
S-2 32,543.9 32,543.9 32,543.9 0.00
S-3 467,143.7 464,681.2 467,352.6 0.58
S-4 335,314.2 333,670.3 335,702.9 0.61
S-5 625,037.9 622,477.4 625,482.6 0.48
S-6 659,336.1 656,836.8 660,249.9 0.52
S-7 445,635.1 444,070.5 445,697.2 0.37
S-8 818,749.9 813,457.8 818,784.1 0.65
S-9 none 800,714.2 803,497.9 0.35
S-lO 1,060,281.9 1,056,315.8 1,061,013.6 0.44
s—11 1,765,124.9 1,753,409.5 1,765,925.6 0.71
s-12 none 1,620,498.7 1,622,933.l 0.15
S-13 none 1,340,279.9 1,346,557.5 0.47

The results shown in Tables 6-S-1 and 6-S-3 give insights
into how the computation burden increases as the planning
horizon is lengthened. Although computing costs are rising
much faster than the problem size it is clear that problems of
a practical size can be solved using a modest platform. Table
6-S-2 presents the most encouraging results, showing that the
solution quality remains very high even for the longest

planning horizons.

 

 

72

 

Table 6—S-3: Problem Solution Report
Scale Problems
Linear Heuristic
Problem Relaxation Best Bound Solution
Name Objective Using CPLEX Objective Time*
S-l 56,062.7 56,842.1 56,842.1 2
s-2 32,156.7 32,543.9 32,543.9 2
S-3 463,575.8 464,681.2 467,352.6 5
s-4 332,567.0 333,670.3 335,702.9 9
S-5 621,054.4 622,477.4 625,482.6 7
S-6 655,073.4 656,836.8 660,249.9 22
S-7 442,758.3 444,070.5 445,697.2 7
S-8 812,033.2 813,457.8 818,784.1 30
S-9 800,040.1 800,714.2 803,497.9 25
s-1o 1,055,102.4 1,056,315.8 1,061,013.6 17
s-11 1,752,656.6 1,753,409.5 1,765,925.6 97
s-12 1,619,417.4 1,620,498.7 1,622,933.1 15
s-13 1,339,188.4 1,340,279.9 1,346,557.5 66

* Time is in minutes on a 486/66.

Although it is appropriate to define the problem size in
terms of the parameters presented in Table 6-V-1 and 6-I-1
(number of machines, number of days in the planning horizon
and the number of paper types), it should be pointed out that
problem size is also a function of the widths and number of
different widths of the papers demanded and the widths of the
paper-making machines. Wide machines producing a broad
assortment of narrow papers will be prone to generating a
large number of cutting patterns, increasing the problem size.
The heuristic procedure significantly reduces the number of
binary and integer variables in any one formulation. This is
done by selectively relaxing integrality requirements and/or
restricting the domain of the integer and binary variables.

However, within any one of the subproblems in the heuristic

 

73
procedure, there are still enough binary and integer variables
that solution times would be unacceptable even (n1 a very
powerful platform. The results of this research would have
wider applicability if the size of the formulations could be

reduced even slightly.

PROBLEM SIZE REDUCTION TECHNIQUES

When considering the possibility of solving problems of
a practical size, there are at least two methods that could be
used to condense the problems. Both approaches are based on
aggregation methods.

Although single days are the preferred planning period,
it has been suggested (S. Twaroski, personal communication,
October 15, 1991) that there are situations where the planning
increment could be extended to forty-eight hours. It is clear
that in situations where this approach could be applied, the
problem size could be reduced by 50%. A less dramatic but
still useful approach would be to aggregate the weekend into
a single planning period. This is appropriate in situations
where the subsequent conversion processes are shut down over
the weekend.

The second aggregation approach is based on the similar
nature of many of the products produced. Although demand will
be for a specific kind of paper in a specific basis weight,
the most formidable transitions are those between kinds of
paper. Once a machine is producing, it is a much simpler

problem to change basis weights than it is to switch to an

 

74
entirely different kind of paper. At the example mill, the
practice is to have a fairly definite schedule with respect to
the kind of paper produced, while the timing of the production
of the different basis weights is left open.

The production planner at the example mill prefers to
schedule on a 21-day planning horizon, giving vendors adequate
notice for required materials. Although the planner will
consider twenty or more paper kind/basis weight combinations,
there will only be approximately five distinct paper types.
At first inspection, it might seem reasonable to aggregate
demand for the basis weights. Then the production scheduling
issue might be analogous to many hierarchical production
planning situations where part-families (paper kinds) are
scheduled first and then the individual part (basis weight) is
scheduled within the time allocated for the particular part
family. The issue of the cutting patterns confounds this
approach when considering production scheduling le a paper
mill. This is because the demands are for a specific weight
and width for each kind of paper, and aggregating the basis
weights would introduce ambiguity, at best, into the meaning
of the cutting patterns. The following proposed modifications
would make it possible to reformulate the problem with
significantly fewer transition variables, the most important
consideration for the most time-consuming step of the

heuristic.

75
Grouping products into part families is another method
that could be used to reduce the problem size. Paper kind,
characterized by a name brand, Kashmir Natural for instance,
will denote the name of the part family. The different basis
weights represent the individual members of the part family.
The production scheduling problem can now be reformulated,

where the sequencing variables, V represent a switch

um,
between different part families rather than individual paper
types. This requires introducing an index, f, to identify the
individual members of the part families.

The production variables still describe the number of
reels of a type of paper to produce on a machine m in a period
t and then cut according to pattern q, but it must be
recognized that the i index now represents the paper family.
Production variables are now expressed with the form: Xfimv

Except for the i and j indexes and the N parameter, most
of the variables and parameters used in the following
formulation have the same meaning as when they were first
presented in Chapter 3. Only new or different indices,

parameters or variables will be described in the following

definitions.

Definitions

Indices:

f enumerates the individual paper types in each part
family, f = 1,2,...,E, where E is the number of

paper types in part family i;

 

 

q

76

labels the state (paper family) on the machines,

i = 0,1,...,N, where i = 0 indicates the
communicating state that each machine must visit at
the beginning of a period and N is the number of
paper families produced;

labels the state (paper family) to which a machine
will be switched, (j = 0,1,...,N), where j = 0 is
the communicating state that each machine must
visit at the end of each period;

denotes the width of paper of a given type demanded
by customers, k = 1,2,...,Kfi;

enumerates the cutting patterns for machine m and
paper type f from family i, q = 1, 2,...,Qm“

Parameters:

cfin

Cfimq

the time required to produce a reel of product f
from family i on machine m;

the sum of production cost and trim waste, for a
reel of paper of type f from family i produced on
machine m and out according to pattern q;

the demand for paper of width k, type f, family i
in period t;

the time required to switch from paper family i to
paper family j on machine m;

the transition cost for changing from paper family
i to paper family j on machine m;

the cost of lost sales, per roll, for paper of type
f in family i, width k in period t;

the number of different paper families;

the number of rolls of width k to be cut from each
reel using pattern q on machine m when producing
product f from family i, subject to the condition

N F1 Kti inm

122: quwfikpfikmqs‘z‘m m=l,2,...,M (3'1)

f==1k1

the recycling cost for a roll of paper of type f,
family i in width k;

 

77

Violuo the initial conditions for each of the machines,
where me is a binary variable indicating a
transition from part family i to part family j;

Wfik the k‘h physical width (in inches) of paper type f
from family i;

Variables:

db, a real number associated with state i on machine m
for period t, used to prevent sub-touring;

Ifin the inventory of roll. width ‘k, product type f,
family i at the end of period t;

Rfikt the number of rolls of width k of product f from
family 1 produced in period t;

an a zero one variable indicating a switch between
states i and j (i¢j) on machine m in period t;

Xmm the number of reels of type f from family 1
produced on machine m and then cut according to
pattern q in period t;

Z“, the demand, in rolls, of width k and type f, family

i not satisfied in each period.

 

78
w
Minimize:

9n.

2

EM»
5..
H
MS"
31"]:

G=
q=1

z
E
a"

I :11: I £110 +
‘1

I.
II
....
M
u
...
7:-

Subject to:

Qtim

M
21 Zip fikqufimqt - Rfikt = 0
m=1

q=1

N F1 Qfim N N
Z: ; 2 Cfimeimqt +
1=1 =1 q=1 1=1

i-‘J-‘

a

ll HJN

S

Ifik(t—1) + Rfikt ‘ Iiikt + Zfikt =

iOmt = 1

T
Cfiqufimqt + 2:

[ill
my
sin
l-.H

2 Z: f2: 2 1511:: Zfikt

.4

un
l—‘i—‘HH
NNNN

fik‘i‘hh.

”Kt—1) + Smt b

(wk-Hue.

tram

o3
II II
"3

(3‘2)

(3'3)

(3‘4)

(3‘5)

(3'6)

(3‘7)

 

79

N
_ n1= 1 2 .. M _
Vbjmt ' 1 t = 1.:2: . :1‘ (3 8)
1=1
i = 1,2,. ,

ViOIMt-l) — Voimt = 0 $1: i:§:: :M (3-9)
N N = 1,2, . I, (3 10)
V = = 1 2 . M' ‘

1:0 ijmt _Z;V jimt = 1:2:. .:
1 = 0,1, .,N
1 = Q11! - IN
dimt - djm + (N+1) V133": s N J at 1 (3-11)
n7: 1,2,. .,M
t = 1,2, ,T
1 = 1,2, .,N
z . X . s f’= 1,2, . ,F}
f1kt' f1mqt' mtI k = 1,2, . . ' lei (3_12)
n7= l 2 ..M
Rfikt' Ifikt' djmt Z 0 q = 1:2: .:Qfim
32mm: integer V f,i,m,q, and t (3-13)
Vfimt 0 or'l V i,j,m, and t (3-14)

The problem reduction technique just described has the
potential to offer substantial improvements in the usability
of the procedure by reducing the number of paper types (N) for
which the transition variables must be generated. In one
example mill this approach would reduce the number of types by
nearly 80%.

Although the formulation does ignore transition costs
between basis weights, these times are small compared with the

effort required to switch between paper kinds. In facilities

 

80
where it is common to maintain some flexibility in the
production schedule, this time is small enough that it would
not compromise the scheduling process. This approach
addresses the most time consuming and important aspect of the

heuristic procedure, generation of the production sequence.

SUMMARY AND CONCLUSIONS

For problems where there is sufficient capacity to meet
all of the demand requirements, it has been shown that the
heuristic procedure can obtain good solutions with an
acceptable computation effort. This chapter also suggests a
problem size reduction technique that will increase the
applicability of the heuristic. This technique relies on the
part family concept, where the various basis weights represent

the members of the paper brand family.

 

CHAPTER 7: CONCLUSIONS AND FUTURE DIRECTIONS

The heuristic developed here, together with the problem
reduction techniques discussed earlier, is an efficient method
of scheduling in environments where manufacturing is sequence
dependent and produced. materials are cut into finished
products. Many businesses operate in this manner, typically
with a large investment in both capital equipment and raw
materials. This research provides production managers with a
tool that can significantly improve profitability. While this
is a significant contribution itself, it also suggests other
areas for future research.

There are modifications that could improve the
effectiveness of the current heuristic, either by decreasing
the time it takes to obtain a good solution or, if considered
useful, by forcing the heuristic to develop schedules more
consistent with those obtained using an optimizing procedure.
The most direct approach to lowering the over-production
charge would be the selective consideration of dominated
cutting patterns. Selective inclusion of more patterns in
this application may not have a noticeable impact on solution
time given the small amount of time the heuristic now directs
to this effort and could reduce the amount of material
recycled.

The most pressing speed issue is the large increase in
computing cost as the heuristic is forced to reject some

sales. As mentioned in Chapter 4, this problem is

81

 

82
particularly acute if demand is heavy late in the planning
period.

A. related, and 'very interesting, question is how ‘to
respond to situations where demand will not be satisfied. In
an actual production environment, the scheduler would have a
variety of options available to respond to schedule
infeasibility. These might' include negotiating" with the
customer to reschedule all or part of the order, shipping by
an alternate carrier, diverting some work to another plant or,
if it is an in-house delivery, allowing the order to be late
under the assumption that the time could be made up later. It
should be pointed out that many feasible schedules could be
improved if the right constraint could be relaxed. In these
cases, frequently, the biggest difficulty is deciding which
constraint to relax.

Among the most obvious methods of relieving production
pressure on one mill is the diversion of some work to another
facility. This could be addressed by incorporating the
current formulation into a comprehensive company—wide model
that can consider several plants in dispersed geographic
areas. This approach is appealing because it offers an
opportunity to determine production schedules for each.machine
in the organization that are optimal not just with respect to
each plant, but also with respect to the entire organization.
It will also allow the company to investigate other issues

that affect the production scheduling at the plant level, such

 

83

as the selection and timing of a machine shutdown for
maintenance, engineering changes, or the management of excess
production capacity. Finally, a company-wide model could be
used to investigate plant expansion or modernization efforts.

Another interesting research question is the relationship
between product quality and the production schedule. To a
large extent this is implied in the changeover costs and times
already considered in the model. However, this could be
extended by incorporating the proportion of virgin materials
as an additional set of constraints. Different papers require
different ratios of virgin raw materials to recycled fiber.
This would give the production scheduler two mechanisms to
control the recycled material, the cutting patterns selected
and product composition. Production could be scheduled in a
way that best utilizes these components. This implementation
could be used. with various inventory' policies to provide
management with another tool for managing product quality.

In summary, the research meets its objective of
developing an effective method of obtaining good production
schedules for environments that face a stock cutting problem
in conjunction with sequence dependent setups, and does so
with a reasonable amount of computation effort. The results
of this research have applicability in the paper industry as
well as other, primarily process, industries at the plant
level. Given the large production volumes associated with the

process industries, the results of this research can achieve

 

substantial dollar savings through modest

 

production efficiency.

 

 

APPENDICES

 

 

 

85

 

86
PROBLEM (Prototype 3)
Random Number Seed: 32

The Number of Periods is: 5

The Number of Machines is: 3

Number of Types of Paper: 4

Maximum Machine Width: 300

Minimum Machine Width: 75

Maximum Paper Width: 300

Minimum Paper Width: 20

Expected Level of Utilization: high

Maximum Number of Orders/Day: 4

Minimum Number of Rolls/Order: 13

Maximum Number of Rolls/Order: 88
Variable Legend

Type of Variable Number

Cutting Patterns:

Roll variables:

Inventory Variables:

Rejected Rolls variables:
Machine Initialization Variables:
Transition Variables:

Slack Variables:

Tucker Variables:

Setups on Each Day:

Total Number of Variables
Constraint Legend
Type of Constraint

Reel to Roll Conversion
Capacity Constraints

Demand Constraints

Required Setups

Exit From the Null State
Matched Exit to Entrance
Exit from Each State Visited
Tucker Constraints

Required Trans. Before Prod.
Minimum Production

Total Number of Constraints

Number

783

205
255
315
365
377
677
692
782
842

End
50

115
175
190
250
310
610
670
730

 

87

PROBLEM SPECIFICATIONS

Machine Number Machine Width
1 136
2 214
3 207

Time Required to Produce a Reel of Paper (in hours)

Machine Number 1

paper(x) Production Time
1 0.8
2 0.9
3 0.9
4 0.7

Machine Number 2

paper(x) Production Time
1

DUN
0000

.6
.7
.6
Machine Number 3

paper(x) Production Time
1 0.5
2 1.0
3 1.0
4 1.0
Cost of Transferring Between Papers (in Ss)

Machine Number 1

paper(x) paper(y) Transition Cost
1 0.9
48.9
401.4
63.2
112.1
150.5
265.3
202.0
352.5
370.1
391.3
419.0

bb-bwaNNNI—‘H
UMP-bNHbUH-bww

 

88

Machine Number 2

paper(x) paper(y) Transition Cost

1 2 2 .

1 3 362.8
1 4 550.0
2 1 94.3
2 3 181.8
2 4 369.0
3 1 356.7
3 2 262.4
3 4 187.2
4 1 279.3
4 2 185.0
4 3 366.8

Machine Number 3

paper(x) paper(y) Transition Cost
2 .
l 3 262.6
1 4 554.5
2 1 83.9
2 3 117.8
2 4 638.4
3 1 474.4
3 2 758.5
3 4 607.5
4 1 422.7
4 2 796.3
4 3 685.4

Time to Transfer Between Papers (in hours)
Machine Number 1

paper(x) paper(y) Transition Time

whoop-Awooxxioooomw

bbbwwwNNNl-‘HH
wMH-bMo-Ihwl-Ipw
HHHHOHHHOI—IHO

 

 

89

Machine Number 2

paper(x) paper(y) Transition Time
1 2 1.9
1 3 1.7
1 4 1.2
2 1 1.0
2 3 1.3
2 4 0.9
3 1 1.2
3 2 1.6
3 4 0.9
4 1 1.8
4 2 0.1
4 3 0.5

Machine Number 3

paper(x) paper(y) Transition Time
1 0.6
1 3 1.9
1 4 1.7
2 1 1.6
2 3 0.8
2 4 1.0
3 1 0.8
3 2 1.9
3 4 1.4
4 1 1.5
4 2 0.3
4 3 1.1

DEMAND SCHEDULE
Requirements for Day 1
Paper Type 1 is needed in the following width(s):
22 Rolls of Width 88 Inches
Paper Type 3 is needed in the following width(s):
17 Rolls of Width 41 Inches
Requirements for Day 2
Paper Type 2 is needed in the following width(s):

15 Rolls of Width 95 Inches
45 Rolls of Width 120 Inches

Paper Type 4 is needed in the following width(s):
25 Rolls of Width 129 Inches
Requirements for Day 3
Paper Type 2 is needed in the following width(s):

38 Rolls of Width 58 Inches
59 Rolls of Width 120 Inches

 

90

Requirements for Day 4

Paper Type

1 is needed in the following width(s):

135 Rolls of Width 88 Inches

Paper Type

3 is needed in the following width(s):

74 Rolls of Width 142 Inches

Requirements for Day 5

Paper Type

2 is needed in the following width(s):

118 Rolls of Width 95 Inches

Paper Type

3 is needed in the following width(s):

41 Rolls of Width 142 Inches

Paper Type

4 is needed in the following width(s):

78 Rolls of Width 129 Inches

Widths Demanded for Each Paper

Paper 1 Widths
Paper 2 Widths
Paper 3 Widths
Paper 4 Widths

Initial Configuration
Machine 1: paper

Machine 2: paper
Machine 3: paper

88

95 58 120

41 107 142 56
129 131

2

2

4

 

CUTTING PATTERNS

Patterns for Machine
Paper Widths (in

88
1

Patterns for Machine
Paper Widths (in

95 58
O 0
0 2
1 0

Patterns for Machine
Paper Widths (in

41 107

wi-‘OO
OOI—‘O

Patterns for Machine
Paper Widths (in

129 131
0 1
1 0

Patterns for Machine
Paper Widths (in

88
2

Patterns for Machine
Paper Widths (in

95 58
0 1
O 3
1 2
2 O

1 and Paper
inches):

1 and Paper
inches):

120
1
0
0

1 and Paper
inches):

142

0000

l and Paper
inches):

2 and Paper
inches):

2 and Paper
inches):

120

OOOI—I

91

OHON

Cost($)

2219.20

Cost($)
2680.00

2622.40
2320.00

Cost($)
2564.80
2492.80

2348.80
2723.20

Cost($)

2838.40
2809.60

Cost($)

4032.40

Cost($)

4061.20
4003.60
4536.40
4234.00

 

92

Patterns for Machine 2 and Paper 3
Paper Widths (in inches):

41 107 142 56 Cost($)
0 0 1 1 4349.20
0 2 0 0 4579.60
1 O O 3 4507.60
1 O 1 0 4133.20
1 1 0 1 4435.60
2 0 0 2 4291.60
2 1 O 0 4219.60
3 0 O 1 4075.60
5 0 0 0 4450.00
Patterns for Machine 2 and Paper 4
Paper Widths (in inches):
129 131 Cost($)
0 1 3384.40
1 0 3355.60
Patterns for Machine 3 and Paper 1
Paper Widths (in inches):
88 Cost($)
2 3983.40
Patterns for Machine 3 and Paper 2
Paper Widths (in inches):
95 58 120 Cost($)
0 1 1 4012.20
0 3 0 3954.60
1 l 0 3652.20
2 O 0 4185.00
Patterns for Machine 3 and Paper 3
Paper Widths (in inches):
41 107 142 56 Cost($)
O O 0 3 3868.20
0 O 1 1 4300.20
1 O 1 0 4084.20
1 l O 1 4386.60
2 O O 2 4242.60
2 1 O 0 4170.60
3 0 O 1 4026.60
5 0 O 0 4401.00
Patterns for Machine 3 and Paper 4
Paper Widths (in inches):
129 131 Cost($)

0 1 3335.40
1 0 3306.60

 

REEL PRODUCTION

Day 1
Machine 1

Paper

Machine 2
Paper

Paper

Machine 3
Paper

Paper

Day 2
Machine 1

Paper

Machine 2

Paper

Machine 3

Paper

Day 3
Machine 1

Paper

Machine 2
Paper

Paper

Type

Type

Type

Type

Type

Type

Type

Type

Type

Type

Type

M

N

U

N H

J)

5..

M

N

w

Reels
24

Reels

Reels
17

Reels

Reels

Reels
35

Reels
32

Reels
51

Reels
25

Reels

Reels

Wl=

Wl=

W1=

i-'

W

W1=

93

95

95

NO

129
1

H0

88

1
W2= 107
O

OH

W2= 131
0

W2= 107

CO

1
W2= 107

00

W3= 120
1

W3= 120
l

W3= 142
l

W3= 120

W3= 142

W3= 120
l

W3= 120

W3= 142

W4=

OHII

Oi—‘ll

56

56

56

 

Machine 3
Paper Type

Paper Type

Day 4
Machine 1

Paper Type

Machine 2
Paper Type

Paper Type

Machine 3
Paper Type

Paper Type

Day 5
Machine 1

Paper Type

Machine 2
Paper Type
Paper Type

Paper Type

Machine 3

Paper Type

ROLL RECORD
Day 1
Paper Type 1

Width
88

...

(a

h

..1

w

M

(a)

b

N

U

.h

M

Reels

Reels

Reels
32

Reels
l

Reels
3O

Reels

Reels
3
Reels

32

Reels
24

Wl=

Wl=

Wl=

W1:

W1=

Wl=

(in rolls)

Prod.
22

94

88

H0

129
1

88

41

95

41

129

B.Inv.
0

W2= 107

CO

W2= 131
0
W2= 107
O
0
W2= 107
O
W2= 131
O
0
W2= 107
0

W2= 131
0

W2= 58

E.Inv.
0

W3= 142
1
1
W3= 142
1
W3= 120
W3= 142
1
W3= 120
0
W3= 142
1
W3= 120
Dmd.
22

W4= 56
1
0

W4= 56
1

W4= 56
O

W4= 56
l

Rej.
0

 

Paper Type 2

Paper Type 3
Width

41

107

142

56
Paper Type 4
Width

129
131

Day 2
Paper Type 1

Width
88

Paper Type 2

Paper Type 3
Width

41

107

142

56
Paper Type 4
Width

129
131

Day 3
Paper Type 1

Width
88

Paper Type 2

(in

(in

(in

(in

(in

(in

(in

rolls)

rolls)
Prod.
0

0

rolls)

Prod.
102

rolls)
Prod.
0

O
0

rolls)
Prod.
2
O
34
32
rolls)
Prod.

35
0

rolls)

Prod.
32

rolls)

Prod.
0
30
55

95

B.Inv.

B.Inv.

0000

B.Inv.

B.Inv.
0

B.Inv.
102

E.Inv.
6

49

E.Inv.
102

E.Inv.
1

E.Inv.

51
32

E.Inv.
l

E.Inv.
134

Dmd.

Dmd.

Dmd.
O

D

00003

Dmd.
25

Dmd.
0

Dmd.

38
59

Rej.

Rej.

0000

Rej.

Rej.
0

Rej.

Rej.

0000

Rej.

Rej.
O

Rej.

 

Paper

Paper

Day 4

Paper

Paper

Paper

Paper

Day 5

Paper

Paper

Paper

Type 3

Width
41

107

142
56

Type 4
Width
129
131

Type 1

Width
88

Type 2

(in

(in

(in

(in

(in

(in

rolls)
Prod.
7
0
20
13
rolls)
Prod.
0

0

rolls)

Prod.
2

rolls)
Prod.
38
O
0
rolls)
Prod.
3
0
33
30
rolls)
Prod.

32
0

rolls)

Prod.
0

rolls)

Prod.
80
O
0

rolls)
Prod.
0
0

11
ll

96

B.Inv.

51
32
B.Inv.

B.Inv.
134

B.Inv.
10

B.Inv.

B.Inv.

17

B.Inv.
1

3O
75

E.Inv.

71
45

E.Inv.

E.Inv.
1

E.Inv.

E.Inv.
42

Dmd.

0000

Dmd.

Dmd.
135

U
Q.

0003

Dmd.

Dmd.

Dmd.
118

Rej.
0

Rej.

Rej.

0000

Rej.
0

Rej.

27

0000
u.

Rej.

0000

 

Paper Type

4 (in rolls)

Width

129
131

COST SUMMARY

SLACK INFORMATION

Day

Cost of Transitions:
Cost of Production:
Cost of Lost Sales:

Value of Recyled Inven.:

Prod.
36
0

97

B.Inv.

42

4311.34
1734893.38

0.00
-93268.80

Total Cost of the Schedule is:

1

Accumulated
Accumulated
Accumulated

2

Accumulated
Accumulated
Accumulated
3

Accumulated
Accumulated
Accumulated

4

Accumulated
Accumulated
Accumulated

5
Accumulated

Accumulated
Accumulated

Slack
Slack
Slack

Slack
Slack
Slack

Slack
Slack
Slack

Slack
Slack
Slack

Slack
Slack
Slack

PRODUCTION SEQUENCE

Day

1

Machine 1

Sequence is:

Machine 2

Sequence is:

Machine 3

Sequence is:

on
on
on

on
on
on

on

on

on

0
:1

on

on
O
on

:1

Machine
Machine
Machine

Machine
Machine
Machine

Machine
Machine
Machine

Machine
Machine
Machine

Machine
Machine
Machine

*§k* *de* *de* *§§*
(”Mi-4

WNH

51:31:54
u:Mla

WNH

(DMD-l

E.Inv.
0

1645936.00 Dollars

000 000 000 H00
Nth-b 0‘00

(JO-b

Ol—‘N
come

0 0

HM!»

Hours
Hours
Hours

Hours
Hours
Hours

Hours
Hours
Hours

Hours
Hours
Hours

Dmd.

Rej.

 

Day 2

Machine
Sequence

Machine
Sequence

Machine
Sequence

Day 3

Machine
Sequence

Machine
Sequence

Machine
Sequence

Day 4

Machine
Sequence

Machine
Sequence

Machine
Sequence

5

Machine
Sequence

Machine
Sequence

Machine
Sequence

1
is:

98

 

 

99

 

19

20

21

22

23

52

550

505

100

PROGRAM Exx
REAL MAXTT,MINTT,MAXPT,MINPT
DOUBLE PRECISION Ix1
CHARACTER*12 NAMSTR

INTEGER STATUS
DIMENSION IWI(15,4)
WRITE(*,19)

FORMAT(5(/),5X,'What is the name of the problem ?’)

NAMSTR = ’TEMP'
WRITE(*,20)
FORMAT(3X,’How many periods are there? '
READ(*,*) IT
WRITE(*,Zl)
FORMAT(3X,’How many products are there?
READ(*.*) N
WRITE(*,22)
FORMAT(3X,’How many machines are there?
READ(*,*)
WRITE(*,23)
FORMAT(3X,’What is the level of demand?
’ (l = Low; 2 = Medium; 3 = High)’)
READ(*,*) IOPT3
WRITE(*,52)
FORMAT(3X,'Random Number Seed? ’)
READ(*,*) IXl

NWMAX

= 4
DAYL = 24
MAXMW = 300
MINMW = 75
MAXPW = 300
MINPW = 20
AMTCST = 14.4
ALABOR = 7.0
CUTCST = 7.0
MAXTC = 2 * 400
MINTC = 0.1 * 400
CLOST = l40000.0

OPEN(88,FILE='EXX.FLE’)
WRITE(88,550)IT,N,M,IOPT3,IX1,ITOYS
FORMAT(I2,/,I2,/,I2,/,12,/,D15.8,/,I2)

CLOSE(88)
IPERIOD = 1
ITERAT = -1

OPEN(88,FILE=’INSUR.FLE')
WRITE(88,505)IPERIOD,ITERAT
FORMAT(5X,I3,5X,I3)

CLOSE(88)
MAXTT = 2.0
MINTT = 0.1
MAXPT = 1.0
MINPT = 0.5
Y = 500.0

HPR = (MAXPT+MINPT)/2.0
AVPW= (MAXPW+MINPW)/2.0
AVMW= (MAXMW+MINMW)/2.0
ROPRE = AVMW/AVPW
MINSIZ = 8
MINR = NINT((MINSIZ/HPR)*ROPRE)
IF(IOPT3.EQ.1)THEN
MAXO = 4
UTIL = 0.3
ENDIF

)

I

I

I

)

)

I

 

101

IF(IOPT3.EQ.2)THEN
MAXO = 4
UTIL = 0. 6

IF(IOPT3. EQ. 3)THEN

MAXR = NINT(

x 4. 0*UTIL*(M/1. 0)*24. 0*(1/HPR)*

x (1. 0/MAXO)*ROPRE) - MI

CALL PRO(IT, N, M, IOPT3, 1x1 ,NWMAX,

x DAYL,MAXO,MINR,MAXR,MAXMW,MINMW,MAXPW,MINPW,

x MAXTC,MINTC,CUTCST,MAXTT,MINTT,

x MAXPT,MINPT,IWI,NAMSTR,ALABOR,AMTCST)
OPEN(88,FILE='DELTA.FLE’)
OPEN(90,FILE=’DELTAP.FLE’)

Do 64, J=1, IT

DO 64, J1=1, M

Do 64, J2=l, N
WRITE(88,5511)INUM
WRITE(90,5511)INUM

5511 FORMAT(5X,I1)
64 CONTINUE
CLOSE(88)
CLOSE(90)
OPEN(88,FILE='ICON.FLE')
Imm=o
WRITE(88,5511)INUM
CLOSE(88)
IOPTl — 1
LOLIM = 0
IUPLIM = 1
CALL HETEST( IWI,
x Y,CLOST,
x FACTOR,LOLIM,IUPLIM,
x AMTCST)
STOP
C LAST CARD OF Exx
END
c
SUBROUTINE HETEST( IWI,
x Y,CLOST,
x FACTOR,LOLIM,IUPLIM,
x AMTCST
DIMENSION IIW(15),IQK(15,7),OBJ(15,15,7),
x TRANS(15,15,7
x ANIT(7,15),OBJX(200,15,7),IDOT(4,200,7,15),CAP(15,7),
x IDMD(14,15,4), IV(0:15,16,7,0:14),MW(7),
x ININV(15,4),RHS(14,7)
INTEGER IWI(15,4),INV(0:14,15,4),
x ROLLPR(14,1S,4,2),
x FACTOR,FLGDAY

REAL IPROD(200,15,7,14)
CALL READIT(DAYL,IT,N,M,IIW,IQK,
1 OBJ,TRANS,ANIT,OBJX,IDOT,CAP,IDMD,MW)
D0 6, I=0,IT
6 L=1,M
6, J=o,N
DO 6, K=l,N+l
6, Kl=l,IQK(J,L)

 

102

IF((J.GT.O).AND.(I.GT.0))

x IPROD(K1,J,L,I) = o
IV(J,K,L,I) = o
6 CONTINUE

OPEN(12,FILE=’IV.TEM’)
DO 8, I=0,IT
DO 8, Il=1,M
DO 8, 12=0,N
, I3=1,N+1

IF(I2.EQ.I3)GOTO 8
IF((12.EQ.O).AND.(I3.EQ.N+1))GOTO 8
ATEM = IV(I2,I3,I1,I)/1.0
WRITE(12,4455)ATEM

4455 FORMAT(F8.5)

8 CONTINUE
CLOSE(12)
FACTOR = 4

WRITE(*,1212)
1212 FORMAT(3X,'What is the FACTOR value 2 ')
READ(*,*) FACTOR

OPEN(88,FILE='FACTOR.FLE’)
WRITE(88,9011)FACTOR

9011 FORMAT(I3)
CLOSE(88)
FLGDAY = 1
MIPCNT = o

50 CONTINUE

MIPCNT + 1

s-u-I

U

0

m

01
~ f‘QHXII
rooms-01.x:-

clil II II II
HIIVi-‘l-Ii-‘H
C40"

~ 0

RHS(K,I)
ROLLPR(K,
55 CONTINUE
Do 57, 1:0, IT
DO 57, Il=1,N
DO 57, 12=l,IIW(Il)
INV(I,11,12) = 0
57 CONTINUE
INCOMP = 0
CALL IP(IV,IPROD,ININV,Y,CLOST,
FACTOR,IWI,LOLIM,IUPLIM,
AMTCST,INCOMP)
RETURN
c LAST CARD OF HETEST
END
CsINCLUDE READ_WRT.FOR
CsINCLUDE PRO.FOR
c

ii
0

I

><><

SUBROUTINE READIT(DAYL,IT,N,M,IIW,IQK, ‘
1 OBJ,TRANS,ANIT,OBJX,IDOT,CAP,IDMD,MW)
DIMENSION IIW(15),IQK(15,7),
1 OBJ(15,15,7), TRANS(15,15,7), ANIT(7,15),OBJX(200,15,7),
2 IDOT(4,200,7,15),CAP(15,7),IDMD(14,15,4),MW(7)
OPEN(96,FILE='PROB_ST.FLE’)
READ(96,SOOO)DAYL
5000 FORMAT(F7.4)
READ(96,5001) IT,N,M
5001 FORMAT(I10
READ(96,5002) (IIW(I),I=1,N)

 

5002

5003
5004
2
5005
3
5006
5007
5008

5009
7

5010
8

5011

5000

5001

><

waH

103

FORMAT(I2)

DO 1,I=1,N

READ(96,5003) (IQK(I,J),J=1,M)
FORMAT(I2)

CONTINUE

DO 2 I=1, M

READ(96,5004) (ANIT(I,J),J=1,N)
FORMAT(F6.1)

CONTINUE

DO 3 I=1, N

READ(96,5005) (CAP(I,J),J=1,M)
FORMAT(F6.3)

CONTINUE

DO 4, I=1, N

DO 4, Il=1, M

DO 4, I2=1, IQK(I,I1)
READ(96,5006) (IDOT(J,IZ,Il,I),J=1,4)
FORMAT(12)

CONTINUE

DO 5, I=1, N

D0 5, 11:1, N

READ(96,5007) (OBJ(I,I1,J),J=1, M)
FORMAT(F8.2)

CONTINUE

DO 6, I=1, N

Do 6, Il=l, M

READ(96,5008) (OBJX(J,I,Il),J=l, IQK(I,I1))
FORMAT(F8.2)

CONTINUE

DO 7, I=1, N

DO 7, 11:1,

READ(96,5009) (TRANS(I,I1,J),J=1, M)
FORMAT(F7.1)

CONTINUE

ICNT = o

ICNT = ICN
READ(96,5010)IDMD(I,I1,12)
FORMAT(I6)

CONTINUE
READ(96,5011)(MW(I),I=1,M)
FORMAT(I5)

CLOSE(96)
RETURN

LAsT CARD OF READIT
END

SUBROUTINE WRITIT(DAYL,IT,N,M,IIW,IQK,CAP,IDOT,
OBJ,OBJX,TRANS,IDMD,ANIT,MW)

DIMENSION ANIT(7,15),OBJ(15,15,7),
TRANS(15,15,7),OBJX(200,15,7),
IDOT(4,200,7,15), CAP(15,7),IIW(15),
IQK(15,7),MW(7),

IWI(15,4),IDMD(14,15,4)

OPEN(96,FILE='PROB ST.FLE’)
WRITE(96,5000) DAYL_

FORMAT(F7.4)
WRITE(96,5001) IT,N,M

FORMAT(IlO)

WRITE(96,5002) (IIW(I),I=1,N)

 

 

 

 

5002

5003
5004
2
5005
3
5006
5007
5
5008
6

5009
7

5010

5011

><><><

104

FORMAT(I2)

DO l,I=1,N

WRITE(96, 5003) (IQK(I,J),J=1,M)

FORMAT(I2)

CONTINUE

DO 2 I=1, M

WRITE(96,5004) (ANIT(I,J),J=1,N)

FORMAT(F6.1)

CONTINUE

Do 3 I=1,

WRITE(96,5005) (CAP(I,J),J=1,M)

FORMAT(F6.3)

CONTINUE

Do 4, I=1, N

Do 4, I1= 1,

DO 4, 12: 1, IQK(I1)

WRITE(96, 5006) (IDOT(J, 12, 11, I), J= L 4)

FORMAT(I2)

CONTINUE

DO 5, I=1, N

DO 5, 11:1, N

WRITE(96,5007) (OBJ(I,I1,J),J=1, M)

FORMAT(F8.2)

CONTINUE

Do 6, I=1, N

DO 6, 11=1, M

WRITE(96,5008) (OBJX(J,I,Il),J=l, IQK(I,I1))

FORMAT(F8.2)

CONTINUE

DO 7, I=1, N

DO 7, 11:1, N

WRITE(96,5009) (TRANS(I,I1,J),J=1, M)

FORMAT(F7.1)

CONTINUE

Do 8, I=1, IT

Do 8, 11:1, N

DO 8, I2=1, IIW(I1)

WRITE(96,5010)IDMD(I,Il,I2)

FORMAT I6

CONTINUE
WRITE(96,5011)(MW(I),I=1,M)
FORMAT 5)

CLOSE(96)I

RETURN

LAST CARD OF WRITIT

END

SUBROUTINE PRO(IT,N,M,IOPT3,IX,NWMAX,

DAYL,MAXO,MINR,MAXR,MAXMW,MINMW,MAXPW,MINPW,

MAXTC,MINTC,CUTCST,MAXTT,MINTT,

MAXPT,MINPT,IWI,NAMSTR,ALABOR,AMTCST)

REAL MAXTT,MINTT,MAXPT,MINPT

DOUBLE PRECISION Ix

CHARACTER*12 NAMSTR

DIMENSION ANIT(7,15),OBJ(15,15,7),

TRANS(15,15,7),OBJX(200,15,7),

IDOT(4,200,7,15), CAP(15,7),IIW(15),

IQK(15,7),MW(7), IWI(15,4),IDMD(14,15,4)
IXSAVE = IX

CALL MACHW(M,IX,MW,MAXMW,MINMW)

CALL PAPW(M,MW,IX,IIW,N,IWI,NWMAX,MAXPW,MINPW)

CALL INITl(ANIT,OBJ,TRANS,IQK,IIW,

20

41

48

1

105

IDOT,IT,N,M,OBJX,IDMD)

CALL DEMAND(IT,IIW,MAXO,MINR,MAXR,IX,N,
NWMAX,IDMD)

CALL PATRNS(N,M,IWI,MW,IDOT,IQK)

CALL OBJF(OBJ,N,M,IX,MAXTC,MINTC)

CALL MASSAG(OBJ,N,M)

CALL OBJFX(OBJX,IDOT,MW,IQK,IWI,M,CUTCST,
ALABOR,AMTCST)

CALL TRANSF(TRANS,M,IX,MAXTT,MINTT)

CALL CAPF(CAP,N,M,MAXPT,MINPT,IX)

CALL ICOND(M,N,ANIT,IX)

SCREEN(M, MW, N, IIW, IT, IDMD, IWI,

fl

CAL
XOBJ, TRANS, OBJX, CAP, IQK, IDOT, ANIT)

X
X

X

UNH

L REPORT(M, N, IIW, IT,
IQK, MAXO, MINR, MAXR, MAXMW, MINMW,MAXPW,
MINPW,IXSAVE,IOPT3,NAMSTR)
CALL WRITIT(DAYL,IT,N,M,IIW,IQK,CAP,IDOT,
OBJ,OBJX,TRANS,IDMD,ANIT,MW)
RETURN
END

SUBROUTINE INIT1(ANIT,OBJ,TRANS,IQK,IIW,
IDOT,IT,N,M,OBJX,IDMD)

DIMENSION ANIT(7,15),OBJ(15,15,7),

TRANS(15,15,7),OBJX(200,15,7),
IDOT(4,200,7,15), CAP(15,7),IIW(15),
IQK(15,7),IDMD(14,15,4)

D0 1, I=1, M

DO 1, Il=l,N

ANIT(I,I1) = 0.0

CONTINUE

Do 20, I=1, N

DO 20, Il=1, N

DO 20, I2=1, M

OBJ(I,I1,12) = 0.0
TRANS(I,Il,I2) = 0.0
CONTINUE

DO 41, I=L 4

D0 41, 11: 1, 200

Do 41, 12: L 7

DO 41, I3=1, 15
IDOT(I,I1,I2,I3) = 0
OBJX(Il,I3,I2) = 0.0
CAP(I3,I2) = 0.0
IQK(I3,I2) = o
CONTINUE

DO 48, I=1, IT

DO 48, I1=1, N

Do 48, 12:1, IIW(Il)
IDMD(I,I1,I2) = 0.0
CONTINUE

TURN
LAST CARD OF INITl
END

SUBROUTINE MACHW(M,IX,MW,MAXMW,MINMW)
DOUBLE PRECISION DRAND,
DIMENSION MW(7)
DO 5,I=1,10
R = DRAND(IX)
CONTINUE
DO 10, I=1, M

 

10

10

21

23

25

26

28

29
30

106

R = DRAND(IX)
MW(I) = (MAXMW-MINMW) * R + MINMW
CONTINUE
RETURN
LAST CARD OF MACHW
END

DOUBLE PRECISION FUNCTION DRAND(IX)
DOUBLE PRECISION A,P,IX,BlS,816,XHI,XALO,LEFTLO,FHI,K
DATA A/16807.DO/,BlS/32768.DO/,
1 Bl6/65536.D0/,P/2147483647.D0/
XHI=IX/316
XHI = XHI—DMOD(XHI,1.D0)
XALO = (IX-XHI*Bl6)*A
LEFTLO = XALo/Ble
LEFTLO = LEFTLO - DMOD(LEFTLO,1.DO)
FHI = XHI*A+LEFTLO
K=FHI/BlS
K = K-DMOD(K,1.DO)
IX = (((XALO-LEFTLO*816)-P)+(FHI-K*Bl$)*Bl6)+K
IF(IX.LT.O.D0)IX=IX+P
DRAND = IX*4.656612875D—10
RETURN
LAST CARD OF DRAND
END

SUBROUTINE PAPW(M,MW,IX,IIW,N,IWI,NWMAX,MAXPW,MINPW)
DOUBLE PRECISION DRAND, Ix
DIMENSION MW(7), IIW(15),IWI(15,4)

DO 10,I=1,N

DO 10,J=1,NWMAx

IWI(I,J) = 0

CONTINUE

DO 23,I=1,N

R = DRAND(IX)

DO 21, K=1,NWMAX

IF((R.GT.(K—1.0)/(NWMAX+0.0))
X.AND.(R.LE.K/(NWMAX+0.0)))THEN

IIW(I) = K

GOTO 22

ENDIF

CONTINUE

CONTINUE

CONTINUE

DO 30, I=1, N

DO 30, I1=1, IIW(I)

CONTINUE

R = DRAND(IX)

IWI(I,I1) = R * (MAXPW—MINPW) + MINPW
Do 26, 12:1,11—1
IF(IWI(I,12).EQ.IWI(I,Il))GOTO 25
CONTINUE

Do 28, 12:1, M

IF(IWI(I,Il).LE.MW(12)) GOTO 29

CONTINUE

GOTO 25

CONTINUE

CONTINUE

RETURN

LAST CARD OF PAPW

END

 

107

SUBROUTINE SCREEN(M,MW,N,IIW,IT,IDMD,IWI,
X OBJ,TRANS,OBJX,CAP,IQK,IDOT,ANIT)
DIMENSION ANIT(7,15),OBJ(15,15,7),
TRANS(15,15,7),OBJX(200,15,7),
IDOT(4,200,7,15), CAP(15,7),IIW(15),
IQK(15,7),MW(7), IWI(15,4),IDMD(14,15,4)
0PEN(44,FILE=’SCREEN.FLE’)
2 FORMAT(SX)
WRITE(44,6000)
6000 FORMAT(ZOX,’ PROBLEM SPECIFICATIONS’,/)
WRITE(44,6001)
6001 FORMAT(lSX,’Machine Number Machine Width')
DO 200, I=1, M
WRITE(44,6002) I, MW(I)
6002 FORMAT(lBX,12,20X,I3)
200 CONTINUE
WRITE(44,7015)
7015 FORMAT(5X,/,
X’Time Required to Produce a Reel of Papers (in hours)’,/)
Do 210,I=1, M
WRITE(44,7016)I
7016 FORMAT(5X, 'Machine Number’,I2,/,
X12X,’paper(x)’,lOX,’Production Time’)
D0 210,I1=1,N
WRITE(44,7017)I1,CAP(I1,I)

WNH

7017 FORMAT(13X,I2,19X,F3.1)
210 CONTINUE
WRITE(44,7001)
7001 FORMAT(/,15X,'Cost of Transferring Between Papers (in $s)’)

DO 230,I=1, M
WRITE(44,7002)I

7002 FORMAT(5X, ’Machine Number',I2,/,

X12X,’paper(x)’,5X,'paper(y)’,5X,’Transition Cost’)

Do 229,I1=1,N
Do 229,I2=1,N
IF(Il.NE.I2)THEN
WRITE(44,7003)Il,I2,0BJ(I1,12,I)

7003 FORMAT(14X,Il,12X,I2,14X,F6.1)
ENDIF
229 CONTINUE
WRITE(44,2)
230 CONTINUE
WRITE(44,7010)
7010 FORMAT(15X,’Time to Transfer Between Papers (in hours)')

00 240,I=l, M
WRITE(44,7011)I

7011 FORMAT(5X, ’Machine Number’,I2,/,

X12X,’paper(x)’,5X,’paper(y)',5x,'Transition Time')

Do 239,Il=1,N
DO 239,I2=1,N
IF(I1.NE.I2)THEN
WRITE(44,7003)I1,12,TRANS(I1,I2,I)

ENDIF
239 CONTINUE
WRITE(44,2)
240 CONTINUE
WRITE(44,6003)
6003 FORMAT(25x,'DEMAND SCHEDULE’,/)

DO 300, J=1, IT
WRITE(44,6007)J

6007 FORMAT(5X,’Requirements for Day ’,12,/)
DO 300, I=1, N

 

6004

6008
289
290
6009
300
9670
9671
302
9672
9673
304
9674
9675
306

8000

8011

9000

8012

9001
350

9002

108

ITEM = 0
Do 250, I4U=1,IIW(I)
ITEM = IDMD(J,I,I4U) + ITEM
CONTINUE
IF(ITEM.EQ.0)GOTO 299
WRITE(44, 6004) I
FORMAT(10X,’Paper Type ’
’ is needed in the following width(s):')
DO 290, I1= 1, IIW(I)
IF(IDMD(J,I,I1).EQ.O)GOTO 289
WRITE(44,6008)IDMD(J,I,Il),IWI(I,Il)
FORMAT(lSX,I4,’ Rolls of Width ’,I3,’ Inches’)
CONTINUE
CONTINUE
WRITE(44, 6009)
FORMAT ( 5X)
CONTINUE
CONTINUE
WRITE(44,9670)
FORMAT(/,5X,’IIW = (number of widths for each paper)',/)
Do 302,I=1,N
WRITE(44,9671)I,IIW(I)
FORMAT(lOX,’Paper type ',IZ,’ Has ',I2,’ Widths’)
CONTINUE
WRITE(44,9672)
FORMAT(/,5X,'IWI = ',/)
Do 304,I=1,N
WRITE(44,9673)I,(IWI(I,Il),I1=l,IIW(I))
FORMAT(5X,'Paper ',12,' Widths ’,4(I4,4X))
CONTINUE
WRITE(44,9674)
FORMAT(/,5X,’INITIAL CONDITIONS')
Do 306, I=1,M
WRITE(44,9675)(ANIT(I,I1),Il=1,N)

FORMAT(5X,’MACHINE ’,5(F5.2,3X))
CONTINUE

WRITE(44,8000)M,N

FORMAT(5X,'CUTTING PATTERNS: ,/,

15x, 'NUMBER OF MACHINES = M = ’,I2,/,
15x, ’NUMBER OF PRODUCTS = N = ',12,/)

WRITE(44,8011)
FORMAT(5X,’NUMBER FOR ROLLS OF THE FOLLOWING WIDTHS:’)
DO 355, I=1,M
DO 355, Il=1,N
WRITE(44,9000)I,I1,MW(I)
FORMAT(/,5X,’PATTERNS FOR MACHINE',I2,

' AND PAPER ',12,//,

15X,’Machine Width = ’,13,' inches',/,

15X,’With Paper Widths (in inches):’,
WRITE(44,8012)(IWI(Il,12),I2=l,IIW(Il))
FORMAT(9X,I3,8X,I3,

12X,I3,12X,I3)

IF(IIW(I1).EQ.4)THEN
D0 350,I3=1,IQK(Il,I)
WRITE(44,9001)(IDOT(I4,I3,I,Il),I4=1,4),OBJX(I3,I1,I)
FORMAT(9X,I2,9X,I2,13X,I2,13X,I2,7X,’COST = ’,F7.2)
CONTINUE
ENDIF
IF(IIW(Il).EQ.3)THEN
DO 351,I3=l,IQK(I1,I)
WRITE(44,9002)(IDOT(I4,I3,I,I1),I4=l,3),OBJX(I3,Il,I)
FORMAT(9X,I2,9X,12,13X,I2,13X,2X,7X,’COST = ',F7.2)

9003
352

9004
353

10

21
22

31
32

46

109

CONTINUE
ENDIF

IF(IIW(I1).EQ.2)THEN
Do 352, 13= L IQK(IL I)
WRITE(44, 9003)(IDOT(I4, 13, I, I1), I4= 1, 2), OBJX(I3, I1, I)
FORMAT(9x, 12, 9x, 12, 13x, 2x,13x, 2x, 7X,'COST = ',F 72)
CONTINUE
ENDIF

IF(IIW(Il).EQ.1)THEN
DO 353,I3=1,IQK(Il,I)
WRITE(44,9004)(IDOT(I4,I3,I,I1),I4=1,l),OBJX(I3,I1,I)
FORMAT(9X,12,9X,2X,13X,2X,13X,2X,7X,'COST = ',F7.2)
CONTINUE
ENDIF

CONTINUE

CLOSE(44)

RETURN

LAST CARD OF SCREEN
END

SUBROUTINE DEMAND(IT,IIW,MAXO,MINR,
MAXR, Ix, N, NWMAx, IDMD)

DOUBLE PRECISION DRAND IX

DIMENSION IIW(15), IDMD(14, 15, 4)

IDMD(I,I1,I2) = o

CONTINUE

DO 100, I=1,IT

R = DRAND(IX)

DO 21, K=1,MAXO + 1
IF((R.GT.(K-1.0)/(MAXO+1)).AND.(R.LE.K/(MAXO+1.0)))THEN
NORDER = K-l
IF((I.EQ.1).AND.(NORDER.GT.2))NORDER = 2
IF((I.EQ.2).AND.(NORDER.GT.3))N0RDER =
GOTO 22

ENDIF

CONTINUE

CONTINUE

IF(NORDER.EQ.O)GOTO 100

Do 50, Il=1, NORDER

R = DRAND(IX)

Do 31, K= 1,

IF((R. GT. (K-l. 0)/(N)). AND. (R. LE. K/(N+O. 0)))THEN
IPICK =

GOTO 32

ENDIF

CONTINUE

CONTINUE

R = DRAND(IX)

IROLLS = NINT(((MAXR-MINR)/l.0)*R)+MINR
IF(I.EQ.1)IROLLS = NINT(IROLLS/3.0)
IF(I.EQ.2)IROLLS = NINT(IROLLS*(2.0/3.0))

R = DRAND(IX)

Do 46, K=1,IIW(IPICK)
IF((R.GT.(K-1.0)/(IIW(IPICK)))

X.AND.(R.LE.K/(IIW(IPICK))))THEN
IWIDTH — K

GOTO 47
ENDIF
CONTINUE

 

110

47 CONTINUE
IDMD(I,IPICK,IWIDTH) = IDMD(I,IPICK,IWIDTH) + IROLLS
so CONTINUE

100 CONTINUE
RETURN

C LAST CARD OF DEMAND
END

C

SUBROUTINE OBJF(OBJ,N,M,IX,MAXTC,MINTC)
DOUBLE PRECISION DRAND,
DIMENSION OBJ(15,15,7)

Do 100, =1,M

Do 100, 11=1,N

DO 100, 12=1,N

IF(I1.N .I2)THEN

R = DRAND(1X)

OBJ(I1,I2,I) = R*(MAXTC-MINTC)+MINTC

1

END F
100 CONTINUE
RETURN
C LAST CARD OF OBJF
END
C

SUBROUTINE TRANSF(TRANS,M,IX,MAXTT,MINTT)
REAL MAXTT,MINTT
DOUBLE PRECISION DRAND, 1x
DIMENSION TRANS(15,15,7)
DO 100, I=1,M
DO 100, 11=1,15
DO 100, 12=1,15
IF(Il.NE.I2)THEN
R = DRAND(IX)
TRANS(Il,I2,I) = R*(MAXTT-MINTT)+MINTT

ENDIF
100 CONTINUE
RETURN
C LAST CARD 0F TRANSF.FOR
END

SUBROUTINE OBJFX(OBJX,IDOT,MW,IQK,IWI,
x M,CUTCST,ALABOR,AMTCST)
DIMENSION OBJX(200,15,7),IDOT(4,200,7,15),
1 IQK(15,7),MW(7), 1w1(15,4)
DO 1000, 1=1, 15
DO 1000, J=1, M
DO 500, K=l, IQK(I,J)
LENGTH = 0
DO 400, L=1,4
LENGTH = LENGTH + IDOT(L,K,J,I)*IWI(I,L)
400 CONTINUE
OBJX(K,I,J) = (MW(J)-LENGTH)*CUTCST
OBJX(K,I,J) = OBJX(K,I,J) +
x (LENGTH/1.0)*(ALABOR+AMTCST)
IF(OBJX(K,I,J).LT.0.0)THEN
WRITE(*,SSSS)

5555 FORMAT(5X,’ERROR IN OBJFX.FOR',/,
X5X,'PATTERN LONGER THAN THE WIDTH OF THE MACHINE')
PAUSE
ENDIF
500 CONTINUE
1000 CONTINUE

RETURN

 

111

LAST CARD OF OBJFX
END

SUBROUTINE ICOND(M,N,ANIT,IX)
IX

DOUBLE PRECISION DRAND,

D

21
100

9674

IMENSION AN1T(7,15)

DO 100, J=1,M

R = DRAND(IX)

DO 21, x=1, N

IF((R. GT. (K—l. 0)/(N)). AND. (R. LE. (K)/(N)))THEN
ANIT(J, K) = 1.0

GOTo 22

ENDIF

CONTINUE

CONTINUE

CONTINUE

WRITE(44,9674)
FORMAT(/,5X,'INITIAL CONDITIONS’)

RETURN
LAST CARD OF ICOND.FOR
END

SUBROUTINE CAPF(CAP,N,M,MAXPT,MINPT,IX)

REAL MAXPT,MINPT

D
D

CSINCLUDE
CSINCLUDE
CSINCLUDE
C

><><

555

1000

1001

1010

XX

1002

><><><

OUBLE PRECISION DRAND, 1x
IMENSION CAP(15,7)

Do 100, 1=1,N

Do 100, J=1,M

R = DRAND(1X)

CAP(I,J) = R*(MAXPT-MINPT) + MINPT
CONTINUE

RETURN

LAST CARD OF CAPF.FOR
END

REPORT.FOR
PATRNS.FOR
MASSAG.FOR

SUBROUTINE REPORT(M,N,IIW,IT,
IQK,MAXO,MINR,MAXR,MAXMW,MINMW,MAXPW,
MINPw,IXSAVE,10PT3,NAMSTR)
CHARACTER*12 NAMSTR,UTIL

DIMENSION IIW(15),IQK(15,7)
WRITE(*,SSS)

FORMAT(5X,’ENTERING REPORT')
OPEN(43,FILE=’REPORT.FLE’)
WRITE(43,1000)NAMSTR

FORMAT(25X,'PROBLEM — ',A12)
WRITE(43,1001)IXSAVE
FORMAT(/,5X,’Random Number Seed = ’,I3,/)

WRITE(43,1010)IT,M,N

FORMAT(5X,’The Number of Periods is: 12, /,
5X,’The Number of Machines is: ’,Il ,/,
5X,'Number of Types of Paper: ,I2, /)

WRITE(43,1002)MAXMW,MINMW,MAXPW,MINPW

FORMAT(5X,’Maximum Machine Width = ',I4,/,
5X,'Minimum Machine Width = ’,I4,//,
5X,’Maximum Paper Width = ’,I4,/,

X,'Minimum Paper Width = ’,I4)

IF(IOPT3.EQ.1)UTIL = ’low’

 

1003

1005

2000

2001

210

2002

230

2003

240

2004

2005

260

2006

112

IF(IOPT3.EQ.2)UTIL = ’medium’
1E(IOPT3.EQ.3)UT1L = ’high’
WRITE(43,1003)UTIL,MAXO,MINR,MAXR

FORMAT(/,5X,'Expected Level of Utilization = ',A12,/,
5X,'Maximum Number of Orders/Day = ',I3,/,
5X,'Minimum Number of Rolls/Order = ’,I3,/,
5X,’Maximum Number of Rolls/Order = ',I3)

CALL SIZE(IT,N,M,IQK,IIW)
WRITE(43,1005)
FORMAT(/,35X,’Constraint Legend’)
IBEG = l
IEND = 0
WRITE(43,2000)
FORMAT(/,5X,'Type of Constraint’,15X,’Beginning ’,
0 ’Ending ’,/)
DO 200, I=1,IT
D0 200, I1=1,N
DO 200, 12=1,IIW(Il)
IEND = IEND+1
CONTINUE
WRITE(43,2001)IBEG,IEND
FORMAT(5X,’Reel to Roll Conversion’,1OX,I5,14X,IS)
IBEG = IEND + 1
Do 210, I=1,IT
DO 210, Il=1,M
IEND = IEND + 1
CONTINUE
WRITE(43,2002)IBEG,IEND
FORMAT(5X,'CapaCity Constraints’,13X,IS,14X,IS)
IBEG = IEND + 1
Do 230, I=1,IT
DO 230, 11=1,N
Do 230, IZ=1,IIW(Il)
IEND = IEND + 1
CONTINUE
WRITE(43,2003)IBEG,IEND
FORMAT(5X,’Demand Constraints’,15X,15,14X,IS)
IBEG = IEND + 1
D0 240, I=1,IT
Do 240, Il=1,M
Do 240, 12=1,N
IEND = IEND + 1
CONTINUE
WRITE(43,2004)IBEG,IEND
FORMAT(5X,’Required Setups',18X,I5,14X,IS)
IBEG = IEND + 1
DO 250, I=1,IT
D0 250, Il=1,M
IEND = IEND + 1
CONTINUE
WRITE(43,2005)IBEG,IEND
FORMAT(5X,’Exit From the Null State ’,8X,IS,14X,IS)
IBEG = IEND + 1
DO 260, I=1,IT
DO 260, Il=1,M
DO 260, I2=1,N
IEND = IEND + 1
CONTINUE
WRITE(43,2006)IBEG,IEND
FORMAT(5X,'MatChed Exit to Entrance’,9X,I5,14X,I5)
IBEG = IEND + 1
DO 270, I=1,IT

270

2007

280

2008

285

2208

2209

2009

6000

113

DO 270, Il=1,M

CONTINUE
WRITE(43,2007)IBEG,IEND
FORMAT(5X,’Exit from Each State Visited’,5X,I5,l4X,IS)
IBEG = IEND + 1
DO 280, I=1,IT
DO 280, Il=1,M
DO 280, 12=O,N
D0 280, I3=1,N+1
IF(IZ.NE.I3)THEN
IF((12.EQ.0).AND.(I3.EQ.N+1))GOTO 280
IEND = IEND + 1
ENDIF
CONTINUE
WRITE(43,2008)IBEG,IEND
FORMAT(5X,’Bread Crumb Constraints’,1OX,15,14X,IS)
IBEG = IEND + 1
DO 285, I=1, IT
DO 285, I1=1, M
D0 285, I2=1, N
IEND = IEND + 1
CONTINUE
WRITE(43,2208)IBEG,IEND
FORMAT(5X,’Required Trans. Before Prod.’,
5x,15,14x,15)
IBEG = IEND + 1
Do 345, I=1,IT
DO 345,I1=1,M
D0 345,I2=1,N
IEND = IEND + 1
CONTINUE
WRITE(43,2209)IBEG,1END
FORMAT(5X,’Minimum Production’,15X,IS,14X,I5)
WRITE(43,2009)IEND
FORMAT(/,5X,’Total Number of Constraints’,25x,15)
CLOSE(43)
WRITE(*,556)
FORMAT(5X,’LEAVING REPORT’)
RETURN
LAST CARD OF REPORT
END

SUBROUTINE SIZE(IT,N,M,IQK,IIW)
DIMENSION IQK(15,7),IIW(15)

WRITE(43,6000)

FORMAT(/,SX,’For Integer Number of Reels’,/,
56X, ' Beg. End’,/)

ICUT = 0

Do 200, I=1,IT

DO 200, I1=1,N

DO 200,I2=1,M

ICUT = ICUT + IQK(Il,I2)
CONTINUE

IBG = 1

IED = ICUT

DO 220,I=O,IT
DO 220,11=1,N
IF(I.GT.O) IROLLS = IROLLS + IIW(Il)

 

5788
X

6002

6003

6004

6005

6006

6007

720

6008

114

INV = INV + IIW(Il)
CONTINUE
WRITE(43,5788)ICUT,IBG,IED
FORMAT(5X,’Number of Cutting Patterns: ',17X,I4,
3X,I4,3X,I4)
IBG = IBG + ICUT
IED = IED + IROLLS
WRITE(43,6002)IROLLS,IBG,IED
FORMAT(5X,’Number of Roll variables:’,18X,16,
3x,14,3x,14)
IBG = IBG + IROLLS
D = IED + INV
WRITE(43,6003)1NV,IEG, IED
FORMAT(5X,’Number of Inventory Variables:',l3X,I6,
3x,14,3x,14)
IBG = IBG + INV
IED = IED + IROLLS
WRITE(43,6004)IROLLS,IBG, IED
FORMAT(5X,'Number of Rejected Rolls variables: ',7X,I6,

3x,14,3x,14)
IBG = IBG + IROLLS
IED = IED + N*M

WRITE(43,6005)N*M,IBG, IED
FORMAT(5X,’Number of Machine Initialization Variables: ’,IS,

3x,14,3x,14)
ITRAN = 0
Do 630, I=1,IT
DO 630, Il=1,M

D0 630, I2=O,N
DO 630, I3=1, N+1
IF(I2.E .13)GOT0 630
IF((I2.EQ.O) .AND. (I3.EQ.N+1)) GOTO 630
ITRAN = ITRAN + 1
CONTINUE
IBG = IBG + N*M
IED = IED + ITRAN
WRITE(43,6006)ITRAN,IBG,IED
FORMAT(5X,'Number of Transition Variables: ’,12X,IS,
3X,I4,3X,I4)
IBG = IBG + ITRAN
IED = IED + IT*M
ISLACK = IT*M
WRITE(43,6007) ISLACK,IBG,IED
FORMAT(5X,’Number of Slack Variables:’,l8X,I5,
3x,14,3x,14)
ITOUR = 0
DO 720, I=1, IT
D0 720, Il=1, M
DO 720, 12=0, N+1
ITOUR = ITOUR + l
CONTINUE
IBG = IBG + ISLACK
IED = IED + ITOUR
WRITE(43,6008)ITOUR,IBG,IED
FORMAT(5X,'Number of Bread Crumb Variables:’,l2X,IS,
3x,14,3x,14)
IGCONS = 0
DO 750, 1=1, IT
DO 750, Il=1, M
D0 750, I2=1, N
IGCONS = IGCONS + l
CONTINUE

6200
X
X

6009

C

C
X
X
X

C

C

C

C
X
X
X
X
X

53

54

115

IBG = IBG + ITOUR
IED = IED + IGCONS
WRITE(43,6200)IGCONS,IBG,IED
FORMAT(5X,’Number of Setups on Each Day:’,15x,15,
3x,14,3x,14)
ITOTAL = ICUT+2*IROLLS+IGCONS+
INV + M*N+ITRAN+ISLACK+ITOUR
WRITE(43, 6009) ITOTAL
FORMAT(/, 5X,’Tota1 Number of Variables: ’,30X,I6)
LA ST CARD OF SIZE

END

SUBROUTINE ELAPSE(1YEAR,IMONTH,IDAY, IHRS,IMINS,
ELHRS,ELMINS)

INTEGER*2 IYEAR,IMONTH,IDAY,IHRS, IMINS,
YEAR, MONTH, DAY, HRS, MINS,
SECS,HSECS

INTEGER ELHRS,ELMINS

CALL GETDAT(YEAR,MONTH,DAY)

CALL GETT1M( HRS, MINS, SECS, HSECS )

1F(1YEAR.EQ.YEAR)THEN

IF(IMONTH.EQ.MONTH)THEN

IF(IDAY.EQ.DAY)THEN

MINTOT= (HRs*6o.o+M1NS) — (IHRS*60.0+IMINS)

CALL MNTOHS(MINTOT,ELHRS,ELMINS)

RETURN

ELSE

MINTOT = 24.0*60.0-(IHRS*60.0+IMINS)

MINTOT = (DAY-IDAY-1)*24.0*60.0 + MINTOT

MINTOT = MINTOT+HRS*60.0+MINS

CALL MNTOHS(MINTOT,ELHRS,ELMINS)

RETURN

ENDIF

PAUSE’ELAPSE...CHANGE 1N MONTH, ELAPSED TIME INVALID'

ENDIF

PAUSE’ELAPSE...CHANGE 1N YEAR, ELAPSED TIME INVALID'

ENDIF

RETURN

LAST CARD 0F ELAPSE

END

SUBROUTINE MNTOHS(MINTOT,ELHRS,ELMINS)
INTEGER ELHRS,ELMINS

ELHRS = INT(MINTOT/60.0)

ELMINS = INT(MINTOT-ELHRS*60.0)

RETURN

LAST CARD OF MNTOHS

END

SUBROUTINE TIMING(ELHRS,ELMINS,EDHRS,EDMINS,
ETHRS,ETMINS,IYEAR,IMONTH,
IDAY,IHRS, IMINS,MIPCNT)
INTEGER*2 IHRS, IMINS, SECS, HSECS,
LHRS, LMINS,
IYEAR,IMONTH,IDAY,
LYEAR,LMONTH,LDAY
INTEGER ELHRS,ELMINS,EDHRS,EDMINS,ETHRS,ETMINS
OPEN(86,FILE=’TIME.FLE’)
WRITE(86,53)
FORMAT(//,25X,’Solution Time Summary’)
WRITE(86,54)1YEAR,1MONTH,1DAY,IHRS, IMINS
FORMAT(///,5X,’Solution Procedure began on: ’,/,

55

56

57

58

59

10

X 5X,’Year: ',I4,/,
X 5X,’Month: ’,I2,/,
X 5X,’Day: ’,I2,’ at:',/,
X 5X,’Hour: ’,I2,/,
X 5X,'Minutes: ’,12)
WRITE(86,55)ELHRS,ELMINS
FORMAT(///,SX,’Time required to solve the LP: ',I3,
X ’ Hours and ',12,’ Minutes’)

WRITE(86,56)EDHRS,EDMINS
FORMAT(/,5X,’Time required to solve dailies: ',I3,
X ' Hours and ’,I2,’ Minutes’)
WRITE(86,57)ETHRS,ETMINS
FORMAT(//,5X,'Total time to solve the heuristic: ’,I3,
X ’ Hours and ’,I2,’ Minutes')
CALL GETDAT(LYEAR,LMONTH,LDAY)
CALL GETTIM( LHRS, LMINS, SECS, HSECS )
WRITE(86,58)LYEAR,LMONTH,LDAY,LHRS, LMINS

FORMAT(///,5X,’Solution Procedure ended on: ’,/,
X 5X,'Year: ’,I4,/,
X 5X,'Month: ’,12,/,
x 5X,’Day: ',12,' at:’,/,
x 5X,’Hour: ',12,/,
X SX,’Minutes: ’,I2)
WRITE(86,59)MIPCNT
FORMAT(//,5X,’The large LP/MIP was solved:',I3,
X ’ times')
CLOSE(86)
LAST CARD OF TIMING

END

SUBROUTINE PATRNS(N,M,Iw1,Mw,IDOT,IQK)
DIMENSION IWI(15,4),MW(7),
XIDOT(4,200,7,15),NROLLS(4),IQK(15,7)

Do 1000, 1=1,N

DO 1000, J=1,M

IQ = 0

DO 5,L4=1,4

NROLLS(L4) = 0

CONTINUE

1TER1 = 0

Do 10, L=1,4

IF(IWI(I,L).EQ.0)GOT0 10

ITER = MW(J)/(IWI(I,L) - 0.000001) + 1

IF(ITER.GT.ITER1)ITER1 = ITER

CONTINUE

ITER = 1TER1

Do 800, K1=o,1TER

Do 700, K2=0,ITER

DO 600, K3=0,1TER

DO 500, K4=O,ITER

ITOTAL = IWI(I,1)*K1+IWI(I,2)*K2+
x IWI(I,3)*K3+IWI(I,4)*K4

IF(ITOTAL.GT.MW(J))GOTO 499

NROLLS(1) = K1

NROLLS(2) = K2

NROLLS(3) = K3

NROLLS(4) = K4

CALL PATEST(NROLLS,I,J,IDOT,1Q)
CONTINUE

CONTINUE

CONTINUE

CONTINUE

800
999

1000

20

184

185

200

220

4545

117

CONTINUE

CONTINUE

IQK(1,J) = IQ
CONTINUE

RETURN

LAST CARD OF PRTRNS
END

SUBROUTINE PATEST(NROLLS,I,J,IDOT,IQ)
INTEGER

DIMENSION NROLLS(4),1D0T(4,2oo,7,15)
ITEM = o

= 1, 4
ITEM = NROLLS(M1) + ITEM
UE

IQTEM = IQ

IF(ITEM.EQ.0)RETURN

CONTINUE

IQ =IQTEM

Do 200, Q=1, IQ
IF((IDOT(1,Q,J,I).GE.NROLLS(1))
AND.(IDOT(2,Q,J,1).GE.NROLLS(2))
AND.(IDOT(3,Q,J,I).GE.NROLLS(3))
AND.(IDOT(4,Q,J,I).GE.NROLLS(4)))RETURN
IF((IDOT(1,Q,J,I).LE.NROLLS(1))
.AND.(IDOT(2,Q,J,I).LE.NROLLS(2))
AN

AN

. D.(IDOT(3,Q,J,I).LE.NROLLS(3))
D.(IDOT(4,Q,J,I).LE.NROLLS(4)))THEN
Do 185,1WQ=Q,IQ—1

DO 184,L4=1,4

IDOT(L4,IWQ,J,I) = IDOT(L4,IWQ+1,J,I)
CONTINUE

CONTINUE

IQTEM = IQTEM - 1

GOTO 20

ENDIF

CONTINUE

IF(IQ.GT.200)PAUSE'TOO MANY CUTTING PATTERNS'
IQ = IQ + 1

D0 220, L4=1,4

IDOT(L4,IQ,J,I) = NROLLS(L4)

CONTINUE

RETURN

LAST CARD OF PATEST

END

X><X x><x

SUBROUTINE FINISH(REELPR,IT,M,N,IIW,
IQK,IWI,IDOT,ROLLPR,OBJ,IV,
IDMD,TRANS,CAP,OBJX,IOPT1,IPROD,CLOST,
SLACK,AMTCST )

DIMENSION IIW(15),IQK(15,7),OBJ(15,15,7), TRANS(15,15,7),
OBJX(200,15,7),IDOT(4,200,7,15),CAP(15,7),
IV(0:15,16,7,0:14),RHS(14,7),IDMD(14,15,4),SLACK(14,7)
INTEGER REELPR(14,7,15,200),1w1(15,4),1NV(0=14,15,4),

N><X

X#

x ROLLPR(14,15,4,2),SEQ(10)
REAL IPROD(200,15,7,14)
NWMAX = 4
MAXN = 4

WRITE(*,4545)
FORMAT(5X,’ENTERING FINISH’)
DO 5,I=0,IT

Do 5,Il=l,N

 

10

4995

4994

4993

5000

5001

95

5002

5003

100
101

150

>=X

118

D0 5,12=1,IIW(11)
INV(1,11,12) = o
CONTINUE
Do 10, I=1,IT
D0 10, Il=1,M
DO 10, 12=1,N
DO 10, I3=1,IQK(I2,I1)
REELPR(I,I1,12,I3)=REELPR(I,I1,I2,I3)+
NINT(IPROD(13,12,11,I))
CONTINUE
OPEN(44,FILE=’RESULTS.FLE')
WRITE(44,4995)
FORMAT(2OX,’ REEL PRODUCTION')
IF(IOPT1.EQ.0)THEN
WRITE(44,4994)
FORMAT(19X,’(optimizing procedure)’)
ENDIF
IF(IOPT1.EQ.1)THEN
WRITE(44,4993)
FORMAT(2OX,’(heuristic procedure)’)
ENDIF
D0 101, I=1,IT
WRITE(44,5000)1
FORMAT(/,5X,’DAY ',12)
DO 100, Il=1, M
WRITE(44,5001)11
FORMAT(10X,’MACHINE ',11)
DO 100, 12:1, N
ITEl = 1
ITEM = 0
D0 95, 15=1,1QK(12,11)
IF(REELPR(I,Il,I2,IS).GT.0)ITEM=1
CONTINUE
IF((ITEM.EQ.1).AND.(ITE1.EQ.1))THEN
WRITE(44,5002)12,(IWI(12,16),16=1,4)
FORMAT(7X,'PAPER TYPE ',12,' REELS',
’ w1= ',I3,
' W2= ’,I3,’ W3= ',I3,’ W4= ',I3)
ITEl = 0
ENDIF
DO 100, I3=1, IQK(12,11)
IF(REELPR(I,Il,IZ,I3).GT.O)THEN
WRITE(44,5003)REELPR(I,11,12,13),
(IDOT(14,13,11,12),14=1,4)
FORMAT(21X,I2,4(8X,I3))
ENDIF
CONTINUE
CONTINUE
Do 150, I=1,IT
DO 150,11=1,N
D0 150,I2=1,IIW(I1)
ROLLPR(I,11,12,1) =
ROLLPR(I,11,12,2) =
DO 150, J=1,IT
DO 150,Jl=1,M
DO 150,J2=1,N
DO 150,J3=1,IQK(J2,J1)
IF((I.EQ.J).AND.(Il.EQ.J2))THEN
ROLLPR(I,Il,IZ,l)= ROLLPR(I,I1,I2,1)+
REELPR(J,J1,J2,J3)*IDOT(I2,J3,J1,J2)
ENDIF
CONTINUE

00

r"

 

5999

6000

6002

119

D0 160,I=1,IT

DO 160,I1=1,N

DO 160,I2=1,IIW(I1)
IENDIv=INV(I-1,11,12)-IDMD(I,11,12)+ROLLPR(I,II,12,1)
IF(IENDIV.GE.0)THEN

INV(1,11,12) = IENDIv

ENDIF

IF(IENDIV.LT.O)THEN

INV(1,11,12) = o

ROLLPR(I,I1,I2,2) = -IENDIv

ENDIF

CONTINUE

WRITE(44,5999)

FORMAT(/,24X,’ROLL RECORD’,/)

DO 300,I=1,IT

WRITE(44,6000)I

FORMAT(5X,'DAY ',12)

DO 300, 12=1,N

WRITE(44,6002)12

FORMAT(10X,'PAPER TYPE ',12, (in rolls)’, /, 14x,

X’Width Prod. B.Inv. E. Inv. Dmd. Rej.’)

XD<N

6003
300

375

377

382

Do 300,I3=1,IIW(I2)
WRITE(44,6003)

IWI(12,13), ROLLPR(I,12,I3,I),
INV((1-1),12,I3),INV(I,12,13),IDMD(I,12,I3),
ROLLPR(I,I2,I3,2)

FORMAT(14X,I3,8X,I3,7X,I3,8X,I3,7X,I3,7X,13)
CONTINUE

=1,
RHS(I,Il)=RHS(I,Il)+TRANS(IZ,I3,I1)*IV(IZ,I3,I1,I)
CONTINUE

DO 377, I=1, IT

Do 377, 11:1, M

Do 377, 12:1, N

DO 377, I3=1, IQK(I2,I1)
RHS(I,Il)=RHS(I,Il)+CAP(12,I1)*REELPR(I,11,12,13)
CONTINUE

CTRAN = o. o

CLos =

DO 380, I=1, IT

DO 380, Il=1,M

DO 380, 12: 0, N1

DO 380, 13: 1, N+

z = z + OBJ(12,1I3, Il)*IV(I2, I3, 11, 1)

CTRAN = z

CONTINUE

Do 382, I=1, IT

DO 382, Il=1, M

Do 382, 12:1,

D0 382, I3=1, IQK(I2,Il)
z = z + OBJX(13,12,11)
*REELPR(I,Il,I2,I3)

CONTINUE

CPRO = z — CTRAN

DO 384, I=1, IT

 

384

386

5555

6969
X

7005

7001

7000
400

2999

3000

3001

450

3002
500

55

~

120

D0 384, I2=1, N
DO 384, I3=1, IIW(I2)

z = z + CLOST * (ROLLPR(I,I2,I3,
CONTINUE
CLOS = z — CTRAN — CPRO
DO 386, I2=1, N
DO 386, I3=1, IIW(I2)
z=z-(IWI(I2,I3)*AMTCST*INV(IT,12,
CONTINUE
COV = z - CTRAN - CPRO - CLOS
WRITE(44,5555)
FORMAT(/,SX,'COST SUMMARYz',

2)/1.0)

13))/1.o

/)
WRITE(44,6969)CTRAN,CPRO,CLOS,COV

FORMAT(5X,'Cost of Transitions =
5X,’Cost of Production =
5X,’Cost of Lost Sales =
5X,'Cost of Over Prod. =

WRITE(44,7005)z

FORMAT(//,10X,'Total Cost of the

Dollars’,/)

WRITE(44,7001)

FORMAT(5X,'SlaCk Information’)

DO 400,I=1,IT

WRITE(44,6000)I

D0 400,I1=1,M

ANUM = SLACK(I,I1)

WRITE(44,7000)II,ANUM

FORMAT(SX,’Accumulated. Slack on

' is: ',F6.1,' Hours’)

CONTINUE

WRITE(44,2999

’,F14.2,/,
’,F14.2)

Schedule is: ’,F14.2,

Machine #’,IZ,

FORMAT(/,20X,’PRODUCTION SEQUENCE')

D0 500,I=1,IT
WRITE(44,3000)I
FORMAT(/,5X,’DAY ',12)

D0 soo,J=1,M
WRITE(44,3001)J
FORMAT(/,10X,’MACHINE ',12)
ICNT = o

IAM = 0

DO 450, No=1, MAXN

D0 450, N1=1, N
IF(IV(IAM,N1,J,I).EQ.1)THEN
ICNT = ICNT + 1

SEQ(ICNT) = N1

IAM = N1

ENDIF

CONTINUE
WRITE(44,3002)(SEQ(N2),N2=1,ICNT)
FORMAT(10X,’SEQUENCE IS: ',12(3x,
CONTINUE

CLOSE(44)

RETURN

LAST CARD OF FINISH

END

SUBROUTINE MASSAG(OBJ,N,M

12))

)
DIMENSION OBJ(15,15,7),TRAN(15,15)

WRITE(*,SS)
FORMAT(5X,’ENTERING MASSAG')
OBJ(7,7,1) = 0.0

SHORT = 0.0

 

130

150
200

10

175
200
500

121

DO 200, I=1, M
CONTINUE
DO 130, I3=1,N
DO 130, I4=1,N
TRAN(I3,I4) = OBJ(13,I4,I)
CONTINUE
D0 150, Il=1, N
DO 150, 12=1,
IF(I1.EQ.12)GOTO 150
CALL DIJKST(TRAN,N,SHORT,11,I2)
IF(SHORT.LT.OBJ(11,12,I))THEN
OBJ(11,12,I) = SHORT
GOTO 125
ENDIF
CONTINUE
CONTINUE
RETURN
LAST CARD OF MASSAG
END

SUBROUTINE DIJKST(TRAN,N,SHORT,ISOUR,ISINK)
DIMENSION DIST(15),LABEL(15),TRAN(15,15)

N
I
DIST(I) = TRAN(ISOUR,I)

DO 10, I=1,N
LABEL(I) = o
CONTINUE
LABEL(ISOUR) = 1
DO 200, J=1,N
MARK

TDIS = 1000000.0

DO 150, J1=1,N

IF((TDIS.GT.DIST(J1))
.AND.(LABEL(J1).EQ.0))THEN

MARK = J1

TDIS = DIST(J1)

ENDIF

CONTINUE
IF(MARK.EQ.O)THEN
PAUSE'ERROR IN DIJKST...1’
ENDIF

LABEL(MARK) = LABEL(NMARK) + 1
IF(MARK.EQ.ISINK)GOTO 500
NMARK = MARK
DO 175, J3=1, N
TEM = DIST(MARK) + TRAN(MARK,J3)
IF((TEM.LT.DIST(J3)
.AND.(LABEL(J3).EQ.0)) THEN
DIST(J3) = TEM
ENDIF
CONTINUE
CONTINUE
CONTINUE
SHORT = DIST(ISINK)
RETURN

LAST CARD OF DIJKST.FOR
END

SUBROUTINE HETEST( IWI,IOPT1,
X Y,CLOST,

 

50

55

57

><>< ><><><>< ><>¢

H

><>< XX

XXX

><>< ><>< ><><

122

EACTOR,LOLIM,IUPLIM,
AMTCST )
DIMENSION IIW(15), IQK(15, 7), OBJ(15, 15, 7),
TRANS(15,15 7),
ANIT(7, 15), OBJX(200, 15, 7), IDOT(4, 200, 7, 15), CAP(15, 7),
IDMD(14, 15,4), IV(0: 15, 16, 7,0: 14),m ),
ININV(15, 4), RHS(14, 7L SLACK(1L 7)
INTEGER REELPR(14,7,15,200),IWI(15,4),INV(O:14,15,4),
ROLLPR(14,15,4,2),
FACTOR,LOST(14,15,4,2),INCOMP
REAL IPROD(200,15,7,14)
CALL READIT(DAYL,IT,N,M,IIW,IQK,
OBJ,TRANS,ANIT,OBJX,IDOT,CAP,IDMD,MW)
D0 6, I=o,IT
DO 6, L=1,M
DO 6, J=0,N
DO 6, K=1,N+1
IV(J,K,L,I) = o
CONTINUE
INCOMP = IT+1
CONTINUE
DO 55, K=1, 14
D0 55, I=1, 15
D0 55, J=1,
D0 55, L=1,

ROLLPR(K,
CONTINUE
DO 57, I=0, IT
Do 57, 11=1,N
DO 57, I2=1,IIW(I1)

INV(I,I1,I2) = o

CONTINUE

CALL READZO(IT,N,M,

IIW,IQK,INV,IPROD,IV,IFLG,LOST,

SLACK,INCOMP )
IF(IFLG.EQ.0)THEN
CALL IP(IV,IPROD,ININV,Y,CLOST,
FACTOR,IWI,LOLIM,IUPLIM,
AMTCST,INCOMP)

STOP

ENDIF

CALL FINISH(REELPR,IT,N,N,IIW,
IQK,IWI,IDOT,ROLLPR,OBJ,IV,
IDMD,TRANS,CAP,OBJX,IOPTl,IPROD,CLOST,
SLACK,AMTCST)

RETURN
LAST CARD OF HETEST

END

ll
0

SUBROUTINE READZO(IT,N,M,

IIW,IQK,INV,IPROD,IV,IFLG,LOST,

SLACK,INCOMP

DIMENSION IIW(15), IQK(15,7),INV(O:14,15,4),
TIV(O:15,16,7,0:14),SLACK(14,7),
TIVT(O:15,16,7,0:14)

INTEGER IV(O:15,16,7,0:14),
LOST(14,15,4,2),
DELTA(14,7,15),INCOMP

CHARACTER*8 FIELD1,FIELD2,FIELDlA,FIELD2A

CHARACTER*1 L(93)

 

9090

11
500

15

1010

20

4O

50

6O

65

123

REAL IPROD(200,15,7,14),TDELTA(14,7,15)
FIELDl = ' NUMBER'

FIELD2 = ' .....C’

WRITE(*,9090)

FORMAT(5X,'ENTERING READZO')

D0 6, I=0,IT

DO 6, L1=1,M

DO 6, K=1,N+1

TIVT(J,K,L1,I) = 0.0

CONTINUE
DO 11, 10:1, IT
DO 11, I1=1, N
D0 11, 12=1, M
DO 11, I3=1, IQK(11,12)
IPROD(I3,I1,12,IO) = 0.0
CONTINUE
FORMAT(F7.2,5X,F10.2)
OPEN(9S,FILE='KRUS.TXT’)

ICNTT = o

CONTINUE

READ(95,1010)FIELD1A,FIELD2A,L

FORMAT(A8,A8,93A1)

IF(FIELD2A.NE.FIELD2)GOTO 15

READ(95,1010)FIELD1A,FIELD2A,L

EPSILN = 1/1ooooo.o

IFLGC = 0
D0 20, I=1, IT
DO 20, I1=1, N
Do 20, 12:1, M
DO 20, 13:1, IQK(I1,I2)

CALL REDLIN(B)

ITEM = NINT(B)

ATEM = ABS(B—ITEM/1.0)
IF(ATEM.GT.EPSILN)IFLGC = 1
IPROD(I3,I1,I2,I) = B
CONTINUE
DO 40, I=1,IT
D0 40, I1=1, N
DO 40, I2=1, IIW(I1)

CALL REDLIN(B)
CONTINUE

DO 50, I=0,IT

DO 50, Il=1, N

DO 50, 12:1, IIW(I1)

CALL REDLIN(B)

INV(I,I1,I2) = NINT(B)

CONTINUE

D0 60, I=1,IT

DO 60, I1=1,N

DO 60, I2=1,IIW(I1)

CALL REDLIN(B)

LOST(I,I1,I2,1)

)

NINT(B)
LOST(I,I1,I2,2 o
CONTINUE
DO 65, I=1,M
D0 65,I1=1,N
CALL REDLIN(B)
TIVT(I1,N+1,I,0) = B
TIV(I1,N+1,I,0) = B
IV(I1,N+1,I,O) = NINT(B)
CONTINUE

 

 

IFLGT = o
4455 FORMAT(F8.5)
OPEN(12,FILE='IV.TEM’)
DO 70, I=1,IT
Do 70, Il=1,M
DO 70, 12=o, N
DO 70, I3=1, N+1
IF(I2.EQ.I3)GOTO 7o
IF((I2.EQ.O) .AND. (I3.EQ.N+1)) GOTO 70
CALL REDLIN(B)
READ(12,4455)TEMP
IF((B.GT.0.0).AND.(B.LT.1.0))IFLGT=1
TIVT(12,I3,11,I) = B
IF((B.GT.0.0).OR.(TEMP.GT.O))
x TIV(I2,I3,Il,I) = 1.0
70 CONTINUE
CLOSE(12)
OPEN(12,FILE=’IV.TEM’)
DO 71, I=1,IT
D0 71, Il=1,M
D0 71, 12:0, N
D0 71, 13:1, N+1
IF(12.EQ.I3)GOTO 71
IF((I2.EQ.0) .AND. (I3.EQ.N+1)) GOTO 71
WRITE(12,4455)TIV(I2,I3,I1,I)
71 CONTINUE

SLACK(I,Il) = B
80 CONTINUE
DO 120, I=1, IT
DO 120, Il=1, M
DO 120, 12:0, N+1
CALL REDLIN(B)
120 CONTINUE
IFGLD = o
OPEN(89,FILE=’DELTAP.FLE’)
DO 150, I=1, IT
DO 150, I1=1, M
DO 150, I2=1, N
CALL REDLIN(B)
IF((B.GT.0.0).AND.(B.LT.1.0))IFGLD = 1
IF((B.GT.0.0).AND.(B.LE.1.0))DELTA(I,I1,12)=1
TDELTA(I,11,12) = B
READ(89,5511)INUM
5511 FORMAT(5X,I1)
IF(INUM.EQ.1)DELTA(I,I1,I2)=1
150 CONTINUE
CLOSE(89)
CLOSE(95)
OPEN(89,FILE=’DELTAP.FLE')
D0 160, I=1, IT
DO 160, I1=1, M
DO 160, 12:1, N
WRITE(89,5511)DELTA(I,11,12)
160 CONTINUE
CLOSE(89)
IF(IFGLD.EQ.1)THEN
CALL FLAGS(TIV,IV,TIVT,IT,N,M,IPROD,IQK,
x TDELTA,INCOMP)

 

125

IFLG = o
CLOSE(12)
RETURN

ENDIF
IF((IFLGC.EQ.O).AND.(IFLGT.EQ.O))THEN
DO 171, I=1,M
D0 171,I1=1,N
IV(O,I1,I,0)=NINT(TIVT(0,I1,I,O))
171 CONTINUE
Do 172, I=1,IT
DO 172, Il=1,M
DO 172, 12=o, N
DO 172, I3=1, N+1
IF(I2.NE.13)THEN
IF((I2.EQ.O) .AND. (I3.EQ.N+1)) GOTO 172
IV(12,I3,11,I)=NINT(TIVT(I2,13,II,I))
ENDIF
172 CONTINUE
OPEN(88,FILE=’BINARY.FLE')
WRITE(88,7723)
CLOSE(88)
OPEN(88,FILE='INTEGER.FLE’)
WRITE(88,7722)
7722 FORMAT(5X,’INTEGER.FLE’)
CLOSE(88)
IFLG = 1
RETURN

DIF
IF(IFLGT.EQ.O)THEN
D0 175, I=1,M
Do 175,I1=1,N
IF(TIVT(0,I1,I,0).EQ.0)IV(0,I1,I,O)=3
IF(TIVT(0,I1,I,0).EQ.1)IV(O,I1,I,O)=2
175 CONTINUE
D0 180, I=1,IT
DO 180, Il=1,M
DO 180, I2=o, N
DO 180, I3=1, N+1
IF(I2.EQ.I3)GOT0 180
IF((I2.EQ.O) .AND. (I3.EQ.N+1)) GOTO 180
IF(TIVT(I2,13,I1,I).EQ.0)IV(I2,I3,I1,I)=3
IF(TIVT(I2,I3,I1,I).EQ.1)IV(I2,I3,I1,I)=2
180 CONTINUE
OPEN(88,FILE='BINARY.FLE')
WRITE(88,7723)

7723 FORMAT(5X,’BINARY.FLE’)
CLOSE(88)
0PEN(88,FILE=’ICON.FLE')
ICON = 2
WRITE(88,5511)ICON
CLOSE(88)

IFLG = o

RETURN

ENDIF

CALL FLACS(TIv,Iv,TIVT,IT,N,M,IPROD,IQK,
x TDELTA,INCOMP)

OPEN(88,FILE=’ICON.FLE’)

ICON = 1

WRITE(88,5511)ICON

CLOSE(88)

IFLG = o

RETURN

 

126

C LAST CARD OF READZO
END

SUBROUTINE REDLIN(B

CHARACTER*1 L(64),C(13),LDIGIT(15)
CHARACTER*8 FIELD(4)

INTEGER INTG(13),MANT(13)

DATA LDIGIT /’0’,'1',’2',’3’,’4’,’5’,'6’,’7’,
X I+I’I_I’l III.I’IEI/
READ(95,1010)FIELD,C,L
1010 FORMAT(4A8,13A1,64A1)

IF(C(10).EQ.LDIGIT(15))THEN
IF(C(11).EQ.LDIGIT(12))THEN
B = 0.0
RETURN

ENDIF
PAUSE’PROBLEM IN REDLINE - E’
STOP
ENDIF
IPOINT = 0
DO 100,I=1,13
IF(C(I).EQ.LDIGIT(14))THEN
I

IPOINT =
GOTO 101
ENDIF
100 CONTINUE
101 CONTINUE

IF(IPOINT.EQ.0)THEN
DO 200,I=1,13
DO 195,11=1,10
IF(C(I).EQ.LDIGIT(I1))THEN
INTG(I) = I1—1
GOTO 200
ENDIF
195 CONTINUE
IF(C(I).EQ.LDIGIT(13))THEN
INTG(I) = o
GOTO 200

DIF
PAUSE’ERROR IN REDLINE - NON STANDARD INPUT’
200 CONTINUE

DO 3oo,I=1,13
B = B + INTG(I)*(10**(13—I))
300 CONTINUE
ENDIF
IF(IPOINT.NE.0)THEN
DO 4oo,I=1,IPOINT—1
DO 395,11=1,10
IF(C(I).EQ.LDIGIT(I1))THEN
INTG(I) = I1—1
GOTO 400
ENDIF
39s CONTINUE
IF(C(I).EQ.LDIGIT(13))THEN
INTG(I) = o
GOTO 400
ENDIF
PAUSE’ERROR IN REDLINE — NON STANDARD INPUT w'
400 CONTINUE
DO 500,I=IPOINT+1,13
DO 495,Il=1,1o

 

127

IF(C(I).EQ.LDIGIT(11))THEN
MANT I

= 11-1
GOTO 500
ENDIF
495 CONTINUE
IF(C(I).EQ.LDIGIT(13))THEN

MANT(I) = o

GOTO 500

ENDIF

PAUSE’ERROR IN REDLINE - NON STANDARD INPUT W’
500 CONTINUE

= 0.0

DO 600,I=1,IPOINT - 1

B = B + INTG(I)*(10**(IPOINT-1—I))
600 CONTINUE

DO 700,I=IPOINT+1,13

B = a + (MANT(I)*(o.1**(I-IPOINT)))/1.o
700 CONTINUE

ENDIF

RETURN
C LAST CARD OF REDLIN

END

C
CSINCLUDE FLAGS.FOR
C

SUBROUTINE FLAGS(TIV,IV,TIVT,IT,N,M,IPROD,IQK,

X TDELTA,INCOMP)
DIMENSION TIV(0:15,16,7,0:14),IV(0:15,16,7,0:14),
x TIVT(O:15,16,7,0:14),IQK(15,7)

INTEGER DELTA(14,7,15),MSTATUS(7)
REAL IPROD(200,15,7,14),TDELTA(14,7,15)
INTEGER COMPLE,STATUS,INCOMP
WRITE(*,909O
9090 FORMAT(5X,’ENTERING FLAGS')
DO 10, I=0,IT
DO 10, Il=1,M
DO 10, I2=O,N
DO 10, I3=1,N+1
MSTATUS(I1) = o
IV(12,I3,II,I) = o
10 CONTINUE
INCOMP = 0
D0 15, I=1,IT
INCOMP = INCOMP + 1
DO 15, Il=1,M
DO 15, 12=o,N
DO 15, I3=1,N+1
IF(I2.EQ.I3)GOTO 15
IF((I2.EQ.O).AND.(I3.EQ.N+1))GOTO 15
IE((TIVT(I2,I3,11,I).GT.o.0).AND.
x (TIVT(12,I3,11,I).LT.1.0))GOTO 16
15 CONTINUE
16 CONTINUE

WRITE(*,5566)INCOMP

5566 FORMAT(5X,'INCOMP = ', I3)
DO 300, Il=1,M
I = o

115 CONTINUE
I = I + 1
IBEG = o

120 CONTINUE

 

128

D0 200, J1=1,N+1

IF(TIVT(IBEG,J1,I1,I).EQ.1.0)THEN
IF(J1.LT.N+1)THEN
IV(IBEG,J1,Il,I) = 2

IE(I.GE.INCOMP)GOTO 201
IV(IBEG,J1,I1,I) = 2

GOTO 115
ENDIF
ENDIF
200 CONTINUE
201 CONTINUE

D0 250, J1=1,N+1
IV(IBEG,J1,I1,I) = 1

250 CONTINUE

300 CONTINUE
OPEN(88,FILE=’INSUR.FLE’)
READ(88,505)IPERIOD,ITERAT
CLOSE(88)

505 FORMAT(5X,I3,5X,I3)
IF(IPERIOD.EQ.INCOMP)THEN
ITERAT = ITERAT + 1

ELSE

ITERAT = -l
IPERIOD = INCOMP
ENDIF

OPEN(88,FILE=’INSUR.FLE’)
WRITE(88,505)IPERIOD,ITERAT
CLOSE(88)
ITERAT = ITERAT ** 6
WRITE(*,7733)ITERAT

7733 FORMAT(5X,’ITERAT = ’,I4)
DO 350, I=1,INCOMP
D0 350, Il=1,M
DO 350, 12=o,N
DO 350, I3=1,N+1
IF(I2.EQ.I3)GOTO 350
IF((I2.EQ.0).AND.(I3.EQ.N+1))GOTO 350
IF(I.LT.INCOMP)THEN
IF(IV(I2,I3,11,I).EQ.0)IV(12,I3,I1,I)= 3

ENDIF
IF((I.EQ.INCOMP).AND.(ITERAT.GT.O).AND.
x (IV(I2,I3,11,I).EQ.0))THEN

ITERAT = ITERAT - 1

IV(12,I3,I1,I)= 1

ENDIF

350 CONTINUE

IF(INCOMP.LE.IT)THEN
DO 355, Il=1,M
D0 354, 12=1,N
IV(O,12,Il,INCOMP+l)=1
IV(I2,N+1,I1,INCOMP+1)=1
IV(I2,N+1,Il,INCOMP)=1

354 CONTINUE
355 CONTINUE
ENDIF

OPEN(89,FILE=’DELTA.FLE’)
DO 365, I=1, IT
D0 365, Il=1, M
DO 365, I2=1, N

360

5511
365

8989

X><><X ><X

8989

129

DELTA(I,I1,12) = o
IF(I.EQ.INCOMP)THEN
APROD = 0.0
DO 360, J1=1,M
D0 360, J2=1,IQK(I2,J1)
APROD = APROD + IPROD(J2,12,Il,I)
CONTINUE
IF(APROD. GT. 0. o)DELTA(I, 11, 12) = 1
ENDIF
WRITE(89, 5511) DELTA(I,11,12)
FORMAT(5X,I 1)
CONTINUE
CLOSE(89)
WRITE(*,8989)
FORMAT(5X,'LEAVING FLAGS’)
RETURN
LAST CARD OF FLAGS
END

SUBROUTINE IP(IV,IPROD,ININV,Y,CLOST,
FACTOR,IWI,LOLIM,IUPLIM,
AMTCST,INCOMP)
DIMENSION IQK(15,7),ITRAC(15,4),
OBJ(15,15,7), TRANS(15,15,7), ANIT(7,15),OBJX(200,15,7),
IDOT(4,200,7,15),CAP(15,7),IDMD(14,15,4),
IV(O:15,16,7,0:14),
MW(7),ININV(15,4),IWI(15,4)

INTEGER FACTOR,INCOMP,STATUS

REAL IPROD(200,15,7,14)
COMMON ITRAK(15,4),IIW(15)
OPEN(4, FILE=’(C)PRINTER')
0PEN(44,FILE=’KRUS.MPS’)

WRITE(*,8989)

FORMAT(5X,’ENTERING IP.FOR’)

TOL = 0.02
CALL READIT(DAYL,IT,N,K,IIW,IQK,
OBJ,TRANS,ANIT,OBJX,IDOT,CAP,IDMD,MW)

NMAx = 6
ICO
IC1
IC2
IC3
IRo
IR1
IR2

II II II II II II II II

OOOOOOOO

CALL WTRK(1, 1R3, IR2, IR1, IRO)
CALL ROWS(IIW,N,K,IT,
IR3,IR2,IR1,IR0)

CALL INIT(IT,N,K,IIw,ITRAC,IQK)

o

IC1 = 0
IC2 = 0
IC3 = O

OPEN(45,FILE=’TEMP.PRN')
CALL EXQKNT(IR3,IR2,IR1,IRO,

XIC3,IC2,IC1,ICO,IQK,IT,N,K,ITRAC,
XITRAK,IDOT,OBJX,IIW,CAP

CALL RCOEFF(IT,N,IIW,IR3,IR2,IR1,IRO,

XIC3,IC2,IC1,ICO

X

CALL INVEN(IR3,IR2,IR1,IRO,IC3,IC2,IC1,ICO,
ITRAC,ITRAK,IIW,IT,N,

m1

35
227

229

X

130

IWI,AMTCST)
ANUM = 1.0
CALL RTRC(5,IC3,IC2,IC1,IC0,ITRAC)
CALL RTRC(6,IR3,IR2,IR1,IR0,ITRAK)
Do 227, I=1,
DO 227, Il=1,
CALL INCR(IC3, IC2, ICl, ICO)
CALL INCR(IR3, IR2, IR1, 1R0)
WRITE(45,35) IC3,IC2,IC1,IC0,IR3,IR2,IR1,IR0, ANUM
FORMAT(’ ’,4(Il),’ ',4(Il),' ’,F10.1)
CONTINUE
CALL RTRC(1,IR3,IR2,IR1,IR0,ITRAK)
CALL RTRC(6,IC3,IC2,IC1,IC0,ITRAC)
DO 228, I=1,IT
DO 228, Il=1,K
DO 228, I2=O,N
DO 228, I3=1, N+1
IF(I2.NE.I3)THEN
IE((12.EQ.0) .AND. (I3.EQ.N+1)) GOTO 228
CALL INCR(IC3,IC2,IC1,ICO)
IF(12 .EQ. 0) GOTO 228
IF(I3 .EQ. N+1) GOTO 228
IF(OBJ(IZ,I3,Il) .EQ. 0.0) GOTo 228
WRITE(45,35) IC3,IC2,IC1,Ico,IR3,IR2,IR1,IRo,
OBJ(I2,I3,I1)
ENDIF
CONTINUE
CALL RTRC(2,IR3,IR2,IR1,IRO,ITRAK)
DO 229, 15:1, IT
DO 229, I6=1, K
CALL RTRC(6,IC3,IC2,IC1,ICO,ITRAC)
CALL INCR(IR3,IR2,IR1,IRO)
D0 229, I=1,IT
D0 229, Il=1,K
DO 229, 12=o,N
DO 229, 13:1, N+1
IF(12.NE.I3)THEN
IF((12.EQ.0) .AND. (I3.EQ.N+1)) GOTO 229
CALL INCR(IC3,IC2,IC1,ICO)
IF(12.EQ.O) GOTO 229
IF(I3.EQ.N+1) GOTO 229
IF(TRANS(I2,I3,I1) .EQ. 0.0) GOTO 229
IF((I5 .EQ. I) .AND. (16.EQ. Il))THEN
WRITE(45,35)
IC3,IC2,IC1,Ico,IR3,IR2,IR1,IR0,TRANS(12,I3,11)
ENDIF
ENDIF
CONTINUE
CALL RTRC(4,IR3,IR2,IR1,IR0,ITRAK)
DO 230, J=1, IT
DO 230, Jl=l, K
DO 230, J2=1,
CALL RTRC(6,IC3,IC2,IC1,IC0,ITRAC)
CALL INCR(IR3,IR2,IR1,IRO)
DO 230, I=1,IT
DO 230, I1=1,K
DO 230, I2=O,N
DO 230, I3=1, N+1
IF(IZ.NE.I3)THEN
IF((IZ.EQ.0) .AND. (I3.EQ.N+1)) GOTO 230
CALL INCR(IC3,IC2,ICl,ICO)
IF((J.EQ.I).AND.(J1.EQ.Il).AND.(I3.EQ.J2))THEN

230

231

131

WRITE(45,35)
IC3,IC2,ICl,ICO,IR3,IR2,IR1,IRO,Y
ENDIF

ENDIF

CONTINUE

CALL RTRC(S,IR3,IR2,IR1,IRO,ITRAK)
ANUM = 1.0

DO 231, J=1, IT

DO 231, J1=1, K

CALL RTRC(6,IC3,IC2,IC1,ICO,ITRAC)
CALL INCR(IR3,IR2,IR1,IR0)

DO 231, I=1,IT

DO 231, 11=1,K

DO 231, 12=o,N

DO 231, I3=1, N+1

IF(12.NE.I3)TEEN

IF((I2.EQ.O) .AND. (I3.EQ.N+1)) GOTO 231
CALL INCR(IC3,IC2,IC1,ICO)
IF(I3.NE.N+1)GOTO 231
IF((J.EQ.I).AND.(J1.EQ.11).AND.(I3.EQ.N+1))THEN
WRITE(45,35)
IC3,IC2,IC1,ICO,IR3,IR2,IR1,IR0,ANUM
ENDIF

ENDIF

CONTINUE

CALL RTRC(6,IR3,IR2,IR1,IRO,ITRAK)
DO 232, J=1, IT

DO 232, J1=1, K

DO 232, J2=1, N

CALL RTRC(6,IC3,IC2,IC1,ICO,ITRAC)
CALL INCR(IR3,IR2,IR1,IRO)

D0 232, I=1,IT

DO 232, I1=1,K

DO 232, 12=o,N

DO 232, I3=1, N+1

IF(I2.NE.I3)THEN

IF((I2.EQ.O) .AND. (13.EQ.N+1)) GOTO 232
CALL INCR(IC3,IC2,ICl,ICO)
IE((J.EQ.I).AND.(J1.EQ.Il).AND.(J2.EQ.I3)
.AND.(12.EQ.0))THEN

ANUM = -1.0

WRITE(45,35)
IC3,IC2,IC1,IC0,IR3,IR2,IR1,IRo,ANUM
ENDIF
IF(((J-l).EQ.I).AND.(J1.EQ.I1).AND.
(J2.EQ.I2).AND.(I3.EQ.(N+1)))THEN
ANUM = 1.0

WRITE(45,35)
IC3,IC2,Ic1,ICO,IR3,IR2,IR1,IR0,ANUM
ENDIF

ENDIF

CONTINUE

CALL RTRC(7,IR3,IR2,IR1,IR0,ITRAK)
D0 233, J=1, IT

D0 233, J1=1, K

DO 233, J2=1, N

CALL RTRC(6,IC3,IC2,IC1,ICO,ITRAC)
CALL INCR(IR3,IR2,IR1,IRO)

DO 233, I=1,IT

D0 233, 11=1,K

DO 233, IZ=O,N

DO 233, I3=1, N+1

 

234

265

132

IE(12.EQ.13)GOTO 233

IE((12.EQ.0) .AND. (I3.EQ.N+1)) GOTO 233
CALL INCR(IC3,IC2,IC1,ICO)
IF((J.EQ.I).AND.(J1.EQ.I1).AND.(I3.EQ.J2))THEN
ANUM = -1.0

WRITE(45,35)
IC3,IC2,IC1,IC0,IR3,IR2,IR1,IR0,ANUM
ENDIF
IF((J.EQ.I).AND.(J1.EQ.Il).AND.(IZ.EQ.J2))THEN
ANUM = 1.0

WRITE(45,35)
IC3,IC2,IC1,ICO,IR3,IR2,IR1,IR0,ANUM
ENDIF

CONTINUE

CALL RTRC(8,IR3,IR2,IR1,IR0,ITRAK)

CALL RTRC(6,IC3,IC2,IC1,IC0,ITRAC)

DO 234, J=1, IT

DO 234, J1=1, K

DO 234, J2=0, N

no 234, J3=1, N+1

IF(J2.EQ.J3)GOTO 234
IF((J2.EQ.O).AND.(J3.EQ.N+1))GOTO 234
CALL INCR(IR3,IR2,IR1,IRO)

CALL INCR(IC3,IC2,IC1,ICO)

IFLG = o

ANUM = N+1
WRITE(45,35)
IC3,IC2,IC1,IC0,IR3,IR2,IR1,IR0,ANUM
CONTINUE
ANUM = CLOST
CALL RTRC(l,IR3,IR2,IR1,IRO,ITRAK)
CALL RTRC(4,IC3,IC2,IC1,ICO,ITRAC)
DO 237, I=1, IT
DO 237, Il=1, N
DO 237, 12:1, IIW(Il)

CALL INCR(IC3,IC2,ICl,ICO)

ANUM = CLOST*(1+0.2*I)*(1+0.5*Il)
WRITE(45,35)
IC3,IC2,IC1,ICO,IR3,IR2,IR1,IR0,ANUM
CONTINUE
CALL RTRC(7,IC3,IC2,ICI,IC0,ITRAC)
CALL RTRC(2,IR3,IR2,IR1,IR0,ITRAK)
ANUM = 1.0
Do 265, I=1,IT
DO 265, 11=1,K
CALL INCR(IR3, IR2, IR1, IRO)

CALL INCR(IC3, IC2, IC1, ICO)

WRITE(45,35) IC3,IC2,IC1,ICO,IR3,IR2,IR1,IR0,ANUM
CONTINUE

CALL RTRC(7,IC3,IC2,ICI,IC0,ITRAC)

CALL RTRC(2,IR3,IR2,IR1,IRO,ITRAK)

ANUM = —1.0

DO 266, I=1,K

CALL INCR(IR3, IR2, IR1, IRO)

CONTINUE

DO 267, I=1,IT—1

DO 267, 11=1,K

CALL INCR(IR3, IR2, IR1, IRO)

CALL INCR(IC3, IC2, IC1, ICO)

WRITE(45,35) IC3,Ic2,ICl,Ico,IR3,IR2,IR1,IRo,ANUM
CONTINUE

CALL RTRC(7,IC3,IC2,IC1,ICO,ITRAC)

 

270

278

285

133

CALL RTRC(1,IR3,IR2,IR1,IR0,ITRAK)
DO 270, I=1,IT
DO 270, I1=1,K
ANUM = I/2.0
CALL INCR(IC3, Icz, IC1, ICO)
WRITE(45,35) IC3,IC2,IC1,ICO,IR3,IR2,IR1,IR0,ANUM
CONTINUE
CALL ITBl(IR3,IR2,IR1,IRo,ITRAK,
IC3,IC2,IC1,IC0,ITRAC,IT,K,N)
ANUM = 1.0
CALL RTRC(3,IR3,IR2,IR1,IR0,ITRAK)
CALL RTRC(4,IC3,IC2,IC1,ICO,ITRAC)
DO 278, I=1, IT
DO 278, Il=1, N
DO 278, I2=1, IIW(I1)
CALL INCR(IR3,IR2,IR1,IR0)
CALL INCR(IC3,Icz,IC1,Ic0)
WRITE(45,35) IC3,Ic2,IC1,ICO,IR3,IR2,IR1,IR0,ANUM
CONTINUE
CALL RTRC(9,IR3,IR2,IR1,IR0,ITRAK)
ANUM = 1
DO 285, J=1, IT
DO 285, J1=1, K
DO 285, J2=1, N
CALL INCR(IR3,IR2,IR1,IRO)
CALL RTRC(6,IC3,Icz,IC1,Ico,ITRAC)
D0 285, I=1, IT
DO 285,I1=1, K
DO 285,12=o, N
D0 285,I3=l, N+1
IF(IZ.EQ.I3)GOT0 285
IF((I2.EQ.O).AND.(I3.EQ.N+1))GOT0 285
CALL INCR(IC3,ICZ,ICl,ICO)
IF(I2.EQ.O)GOTO 285
IF((J.EQ.I).AND.(J1.EQ.Il).AND.(J2.EQ.I3))THEN
WRITE(45,35)
IC3,IC2,IC1,ICO,IR3,IR2,IR1,IR0,ANUM
ENDIF
CONTINUE
CALL RTRC(9,IR3,IR2,IR1,IRO,ITRAK)
CALL RTRC(9,IC3,IC2,ICl,ICO,ITRAC)

CALL INCR(IR3,IR2,IR1,IRO)
CALL INCR(IC3,IC2,IC1,ICO)
WRITE(45,35)
IC3,IC2,IC1,ICO,IR3,IR2,IR1,IR0,ANUM
CONTINUE
CALL RTRC(10,IR3,IR2,IR1,IRO,ITRAK)
CALL RTRC(9,IC3,IC2,Ic1,Ico,ITRAC)
ANUM = -FACTOR/1.0
DO 300, J=1, IT
DO 300, J1=1, K
DO 300, J2=1, N
CALL INCR(IC3,IC2,IC1,ICO)
CALL INCR(IR3,IR2,IR1,IR0)
WRITE(45,35)
IC3,IC2,IC1,ICO,IR3,IR2,IR1,IR0,ANUM
CONTINUE
CALL RTRC(10,IR3,IR2,IR1,IR0,ITRAK)

 

5511

600
520

530

134

ANUM = 1.0
DO 305, J=1, IT
DO 305, J1=1, K
DO 305, J2=1, N
CALL INCR(IR3,IR2,IR1,IRO)
CALL RTRC(1,IC3,IC2,ICI,IC0,ITRAC)
DO 305, I=1, IT
DO 305, Il=1, N
DO 305, 12=1, K
DO 305, I3=1, IQK(I1,I2)
CALL INCR(IC3,IC2,IC1,ICO)
IF((J.EQ.I).AND.(J1.EQ.I2).AND.(J2.EQ.11))THEN
WRITE(45,35)
IC3,IC2,IC1,ICO,IR3,IR2,IR1,IR0,ANUM
ENDIF
CONTINUE
CALL RTRC(10,IR3,IR2,IR1,IR0,ITRAK)
ANUM = FACTOR/1.0
DO 310, J=1, IT
Do 310, J1=1, K
DO 310, J2=1, N
CALL INCR(IR3,IR2,IR1,IR0)
CALL RTRC(6,IC3,IC2,IC1,ICO,ITRAC)
DO 310, I=1, IT
D0 310,I1=l, K
DO 310,12=0, N
no 310,I3=1, N+1
IF(IZ.EQ.I3)GOTO 310
IF((I2.EQ.0).AND.(I3.EQ.N+1))GOT0 310
CALL INCR(IC3,IC2,ICl,ICO)
IE((J.EQ.I).AND.(J1.EQ.I1).AND.(J2.EQ.I2)
.AND.(I3.EQ.N+1))THEN
WRITE(45,35)
IC3,IC2,Ic1,Ico,IR3,IR2,IR1,IR0,ANUM
ENDIF
CONTINUE
ANUM = -1oooooo.o
WRITE(45,35) IC3,IC2,IC1,IC0,IR3,IR2,IR1,IR0, ANUM
CLOSE(45)
CALL SORT()
OPEN(88,FILE='ICON.FLE')
READ(88,5511)ICON
FORMAT(5X,Il)
CLOSE(88)
WRITE(44,500)
FORMAT('RHS')
CALL RTRC(2,IR3,IR2,IR1,IR0,ITRAK)
DO 520, I=1, IT
DO 520, I1=1, K
CALL INCR(IR3, IR2, IR1, IRO)
ANUM = DAYL
WRITE(44,600) IR3,IR2,IR1,IR0,ANUM
FORMAT(’ RHS R',4(Il),’ ’,F10.1)
CONTINUE
DO 530, I=1, IT
DO 530, Il=1, N
DO 530, 12:1, IIW(Il)
ANUM = IDMD(I,Il,I2)
CALL INCR(IR3, IR2, IR1, 1R0)
IF(ANUM .EQ. 0.0)GOTO 530
WRITE(44,600) IR3,IR2,IR1,IR0,ANUM
CONTINUE

 

559

135

CALL RTRC(5,IR3,IR2,IR1,IRO,ITRAK)
ANUM = 1.0

DO 540, I=1, IT

DO 540, Il=1, K

CALL INCR(IR3, 1R2, IR1, IRO)
WRITE(44,600) IR3,IR2,IR1,IR0,ANUM
CONTINUE

CALL RTRC(8,IR3,IR2,IR1,IRO,1TRAK)
DO 550, I=1, IT

DO 550, 13:1, N+1
1F(12. EQ. I3)GOTO 550
IF((I2. E9. 0). AND. (13. EQ.N+1))GOTO 550
CALL INCR(IR3, IR2, IR1, 1R0)
IFLG = 1
ANUM = N
IF(ANUM.NE.0.0)
WRITE(44,600) IR3,IR2,IR1,IR0,ANUM
CONTINUE
WRITE(44,601)
FORMAT(’BOUNDS’)
WRITE(*,3553)INCOMP
FORMAT(5X,’INCOMP = ',13)
CALL RTRC(1,IC3,IC2,IC1,Ico,1TRAC)
DO 560, I=1, IT
DO 560, 11=1, N

IF(I. GE. INCOMP)STATUS = o
IF(I. LT. INCOMP-1)STATUS =
IF(I.EQ.INCOMP-1)THEN
DO 559, J=0,N
IF((IV(J, 11, IL 1). E9. 2). OR.
(IV(J, 11, I2, I). EQ. 3))STATUS = 1
CONTINU
ENDIF

D0 560, I3=1, IQK(11,12)

CALL INCR(IC3, IC2, IC1, ICO)
LOLIM = 2
IUPLIM = 3
IF(STATUS.EQ.1)THEN
LOWER = INT(IPROD(I3,I1,12,I)) - LOLIM
IF(LOWER.LT.0)LOWER = o
IUPPER = INT(IPROD(I3,Il,I2,I))+IUPLIM
WRITE(44,704) IC3,IC2,IC1,ICO,IUPPER
WRITE(44,703) IC3,IC2,IC1,ICO,LOWER

2

ENDIF
IF(STATUS.EQ.2)THEN
ERROR = IPROD(13,11,12,I)-INT(IPROD(I3,11,12,1))

IF(ERROR.GT.TOL)THEN
LOWER = INT(IPROD(I3,I1,I2,I)) — LOLIM
IF(LOWER.LT.O)LOWER = o
IUPPER = INT(IPROD(I3,11,12,I))+IUPLIM
WRITE(44,7O4) IC3,IC2,IC1,ICO,IUPPER
WRITE(44,703) IC3,IC2,IC1,IC0,LOWER
ELSE
ANUM = NINT(IPROD(I3,Il,I2,I))
WRITE(44,762) IC3,IC2,IC1,IC0,ANUM
ENDIF
ENDIF
CONTINUE

CALL RTRC(3,IC3,IC2,ICl,ICO,ITRAC)

 

625

640

700
701
702
705
703
704
762
763

75

9090

136

DO 622, I=1,
DO 62L J= 1, 11w (I )
CALL INCR(IC3, 1C2, IC1, 1C0)
ANUM = ININV(I, J)/1.
WRITE(44,762) IC3,IC2,IC1,IC0,ANUM
CONTINUE
CALL RTRC(S,IC3,IC2,ICl,ICO,ITRAC)
D0 625, I=1, K
D0 625, J=1, N
CALL INCR(IC3,IC2,IC1,ICO)
ANUM = ANIT(I,J)
WRITE(44,762) IC3,IC2,IC1,IC0,ANUM
CONTINUE
CALL RTRC(6, ICL IC2, Ic1, Ico, ITRAC)
DO 635, I=1,
DO 635, 11: L IK
DO 635, 12=o, N
DO 635, I3=1, N+1
IF(I2.EQ.I3)GOT0 635
IF((12.EQ.O).AND.(13.EQ.(N+1)))GOTO 635
CALL INCR(IC3,IC2,IC1,ICO)
IF(IV(I2,I3,Il,I).GT.O)THEN
1E(IV(12,13,11,I).EQ.1)
WRITE(44,700)1C3,1C2,1C1,1C0
IF(IV(I2,I3,I1,I).EQ.2)
WRITE(44,705)IC3,IC2,IC1,ICO
IF(IV(I2,13,Il,I).EQ.3)
WRITE(44,702)IC3,IC2,IC1,Ico
ENDIF
CONTINUE
0PEN(88,FILE=’DELTA.FLE’)
CALL RTRC(9,IC3,ICZ,IC1,ICO,ITRAC)
D0 640, J=1, IT
DO 640, J1=1, K
DO 640, J2=1, N
READ(88,5511)IDELTA
CALL INCR(IC3,IC2,IC1,ICO)
IF(IDELTA.EQ.1) WRITE(44,700)IC3,ICZ,IC1,ICO
IF(IDELTA.EQ.2) WRITE(44,705)IC3,IC2,ICl,ICO
IE(IDELTA.EQ.3) WRITE(44,702)IC3,IC2,IC1,IC0
CONTINUE
CLOSE(88)
FORMAT(' BV BOUNDS x',4(11))
FORMAT(' UP BOUNDS x',4(11),12x,'1 o')
FORMAT(' FX BOUNDS x',4(11),12x,'o.o')
FORMAT(’ FX BOUNDS x',4(11),12x,'1 0')

FORMAT(’ LI BOUNDS x',4(11),12x,13)
FORMAT(’ UI BOUNDS x',4(11),12x,13)
FORMAT(’ FX BOUNDS X’,4(Il),1OX,F5.1)

FORMAT(’ LO BOUNDS X',4(I1),10X,F5.1)
WRITE(44,75)
FORMAT(’ENDATA’)
CALL RTRC(lO,IC3,IC2,IC1,ICO,ITRAC)
ICC = IC3*1000+IC2*100+IC1*10+ICO
CALL RTRC(13,IR3,IR2,IR1,IR0,ITRAK)
IRC = 1000*IR3+100*IR2+10*IR1+IRO
CALL SOS(IT,K,N,ITRAC)
CALL ORD(IT,K,N,ITRAC,11w,IQK)
CLOSE(44)
CLOSE(4)

WRITE(*,9090)

FORMAT(5X,'LEAVING IP.FOR’)

7

11

12

l4

l6

17

137

RETURN
LAST CARD 0F IP.FOR
END

SUBROUTINE ROWS(11w,N,K,IT,

IR3,IR2,IR1,IR0)

DIMENSION IIW(15)
WRITE(44,5)

FORMAT(’NAME IP MODEL’,/,’ROWS’)

WRITE(44,?)
FORMAT(’ N ROOOO’)

D0 10, J4=1, IT

DO 10, 11=1, N

DO 10, I2=1,IIW(I1)

CALL INCR(IR3, IR2, IR1, 1R0)
WRITE(44,21) IR3,IR2,IR1,IR0
FORMAT(' L R',4(11))

FORMAT(' E R’,4(Il))

FORMAT(’ G R',4(11))

CONTINUE

CALL WTRK(2, 1R3, IR2, 1R1, IRO)
D0 11, I=1,IT

D0 11, 11=1,K

CALL INCR(IR3, IR2, IR1, IRO)
WRITE(44,21) IR3,IR2,IR1,IR0
CONTINUE

CALL WTRK(3, 1R3, IR2, IR1, 1R0)
DO 12, I=1, IT

DO 12, I1=1, N

DO 12, 12=1, IIW(I1)

CALL INCR(IR3, IR2, IR1, 1R0)
WRITE(44,21) IR3,IR2,IR1,IR0
CONTINUE

CALL WTRK(4, 1R3, IR2, IR1, 1R0)
Do 13, 1=1, IT

DO 13, Il=1, K

DO 13, 12=1, N

CALL INCR(IR3, IR2, IR1, IRO)
WRITE(44,22) IR3,IR2,IR1,IR0
CONTINUE

CALL WTRK(5, 1R3, IR2, IR1, IRO)
DO 14, 1=1, IT

DO 14, 11:1, K

CALL INCR(IR3, IR2, IR1, 1R0)
WRITE(44,21) IR3,IR2,IR1,IR0
CONTINUE

CALL WTRK(6, IR3, IR2, IR1, IRO)
D0 16, I=1, IT

DO 16, 11:1, K

DO 16, 12=1, N

CALL INCR(IR3, IR2, IR1, 1R0)
WRITE(44,21) IR3,IR2,IR1,IR0
CONTINUE

CALL WTRK(7, 1R3, IR2, IR1, 1R0)
DO 17, I=1, IT

DO 17, Il=1, K

DO 17, 12=1, N

CALL INCR(IR3, IR2, IR1, 1R0)
WRITE(44,21) IR3,IR2,IR1,IR0
CONTINUE

CALL WTRK(8, 1R3, IR2, 1R1, IRO)
DO 18, I=1, IT

 

45

18

40

3O
35

50

138

D0 18, 11=1, K
DO 18, 12=o, N
D0 18, 13=1, N+1
IF (12 .NE. 13) THEN
IF ((12 .EQ. 0) .AND. (I3 .EQ. N+1)) GOTO 18
CALL INCR(IR3, IR2, IR1, 1R0)
WRITE(44,20) IR3,IR2,IR1,IRO
ENDIF
CONTINUE
CALL WTRK(9, IR3, 1R2, 1R1, IRO)
DO 40, I=1, IT
DO 40, Il=1, K
DO 40, 12=1, N
CALL INCR(IR3, 1R2, IR1, 1R0)
WRITE(44,21) IR3,IR2,IR1,IRO
CONTINUE
CALL WTRK(1o, 1R3, IR2, 1R1, 1R0)
Do 45, I=1,IT
DO 45,11=1,K
DO 45,12=1,N
CALL INCR(IR3, IR2, IR1, 1R0)
WRITE(44,22) IR3,IR2,IR1,IR0

CONTINUE
CALL WTRK(11, IR3, IR2, IR1, 1R0)
RETURN
LAST CARD OF ROWS
END

SUBROUTINE EXQKNT(IR3,IR2,IR1,IRO,
XIC3,IC2,IC1,ICO,IQK,IT,N,K,ITRAC,
XITRAK,IDOT,OBJX,IIW,CAP )

DIMENSION IQK(15,7),ITRAC(15,4),ITRAK(15,4),
OBJX(200,15,7),IDOT(4,200,7,15),IIW(15),CAP(15,7)
WRITE(44,30)

FORMAT(‘COLUMNS’)
FORMAT(’ ',4(11),' ',4(11),' ’,F10.1)
CALL RTRC(1,IR3,IR2,IR1,IR0,ITRAK)
CALL RTRC(1,IC3,IC2,IC1,IC0,ITRAC)
DO 50, I=1, IT
DO 50, I1=1, N
Do 50, 12=1, K
DO 50, I3=1, IQK(Il,I2)
IF(OBJX(I3,Il,12).NE. 0.0)THEN
CALL WKIT1(IC3,IC2,IC1,ICo,IR3,IR2,IR1,IRo,
1 OBJX(I3,I1,I2))
ENDIF
IF(OBJX(I3,Il,I2).EQ. 0.0)THEN
CALL WKIT2(IC3,IC2,1C1,IC0)
ENDIF
CONTINUE
CALL RTRC(1,IC3,IC2,IC1,IC0,ITRAC)
CALL INCR(IR3,IR2,1R1,IR0)

IR3S = IR3
IRZS = IR2
IRlS = IR1
IROS = IRO

DO 52, I=1, 1T
DO 52, Il=1, N
D0 52, 12=1, K
DO 52, I3=1, IQK(11,12)
IC3s = IC3

ICZS IC2

 

51

52

55

139

1c1s =
ICOS = ICo

D0 51, J0=1,IT

DO 51, J=1, N

DO 51, 31:1, IIW(J)
IF((J.EQ.Il).AND.(JO.EQ.I))THEN

ANUM = IDOT(J1,I3,I2,J)/1.0
IF(IDOT(J1,I3,IZ,J).NE. 0.0) THEN

CALL WKIT1(IC3,ICZ,IC1,ICO,IR3,IR2,IR1,IRO,

ANUM)
IC3 = IC3S

IC2 = IC2S

IC1 = ICIS

Ico = Icos

ENDIF

ENDIF

CALL INCR(IR3,IR2,1R1,IR0)
CONTINUE

CALL WKIT2(IC3,IC2,ICI,1c0)
IR3 = IR3S

IR2 = 1st

IR1 = IRIS

1R0 = IROS

CONTINUE

CALL RTRC(2,IR3,IR2,IR1,IRO,ITRAK)
CALL RTRC(1,IC3,IC2,IC1,ICO,ITRAC)
CALL INCR(IR3,IR2,IR1,IRO)

IR3S = IR3
IRZS = IR2
IRlS = IR1
IROS = IRO

DO 55, Il=1, N

DO 55, 12=1, K

D0 55, I3=1, IQK(11,12)
IC3S = IC3

Iczs = IC2

IC1S = IC1

ICOS = 100

D0 54, J=1, 1T

DO 54, J2=1, K
IF((J.EQ.I).AND.(J2.EQ.I2))THEN
IF(CAP(11,12).NE. 0.0) THEN

CALL WKIT1(IC3,IC2,ICl,ICO,IR3,IR2,IR1,IRO,

CAP(Il,I2))

IC3 = IC3s

IC2 = 1C23

101 = IC1S

ICO = ICOS

ENDIF

ENDIF

CALL 1NCR(1R3,1R2,IR1,IRO)
CONTINUE

CALL WKIT2(Ic3,Ic2,1C1,IC0)
IR3 = IR3S

IR2 = IR2S

IR1 = IRlS

IRO = IROS

CONTINUE

CALL RTRC(4,IR3,1R2,1R1,1R0,ITRAK)
ANUM = —1.0
DO 60, J=1, IT

 

D0 60, J1=1, K

DO 60, J2=1, N

CALL INCR(IR3,IR2,1R1,IRO)

CALL RTRC(1,IC3,IC2,IC1,ICO,ITRAC)

DO 60, I=1, 1T

DO 60, I1=1, N

D0 60, I2=1, K

DO 60, I3=1, IQK(11,12)

IF((J.EQ.I).AND.(J1.EQ.12).AND.(J2.EQ.11))THEN

CALL WKIT1(IC3,IC2,ICl,ICO,IR3,IR2,IR1,IRO,
1 ANUM

ELSE
CALL WKIT2(IC3,IC2,1C1,IC0)
ENDIF

60 CONTINUE
RETURN
C LAST CARD OF EXQKNT
END

SUBROUTINE RCOEFF(IT,N,IIW,IR3,IR2,IR1,IRO,
XIC3,IC2,ICl,ICO )
DIMENSION IIW(15)
IDMD1 = 0
DO 201, I=1,N
IDMD1 = IDMD1 + IIW(I)
201 CONTINUE
ANUM = -1.0
IFLG = 0
DO 220, I=1, IT
DO 220, I1=1, N
DO 220, 12=1, IIW(I1)
CALL INCR(IC3, IC2, IC1, ICO)
IHJl = o
ANUM = -1.0
IFLG = 0
CALL RTRK(I, IR3, IR2, 1R1, 1R0)
DO 202, J =1, IT
DO 202, J1=1, N
DO 202, J2=1, IIW(J1)
CALL INCR(IR3, IR2, IR1, 1R0)
IF((Il.EQ.J1).AND.(I2.EQ.J2).AND.(J.EQ.I))THEN
WRITE(45,35) IC3,IC2,IC1,Ico,IR3,IR2,IR1,IRo, ANUM

35 FORMAT(' ',4(11),' ',4(11),' ’,F10.l)
IFLG = 1
ENDIF

IF(IFLG .EQ. 1) GOTO 203
202 CONTINUE
203 CONTINUE
IHJl = IHJl + 1
ANUM = 1.0
IFLG = 0
CALL RTRK(3, IR3, IR2, IR1, 1R0)
DO 204, IH1=1, (IHJ1-1)*IDMD1
CALL INCR(IR3, IR2, IR1, 1R0)
204 CONTINUE
DO 205, J2=1, IT
DO 205, J3=1, N
DO 205, J4=1, IIW(J3)
CALL INCR(IR3, IR2, IR1, 1R0)
IF((I.EQ.J2).AND.(Il.EQ.J3).AND.(I2.EQ.J4))THEN
WRITE(45,35) IC3,IC2,ICl,ICO,IR3,IR2,IRl,IRO, ANUM
IFLG = 1

 

 

205
206
220

><N

35

226

35

27

141

ENDIF
IF(IFLG .EQ. 1) GOTO 206
CONTINUE
CONTINUE
CONTINUE
RETURN
LAST CARD OF RCOEFF
END

SUBROUTINE INVEN(IR3,IR2,IR1,IR0,IC3,Icz,ICI,Ico,
ITRAC,ITRAK,IIw,IT,N,
Iw1,AMTCST)
DIMENSION IIW(15),ITRAC(15,4),1TRAK(15,4)
INTEGER 1w1(15,4)
FORMAT(’ ',4(11),' ',4(11),' ’,F10.1)
CALL RTRC(1,IR3,IR2,IR1,IR0,ITRAK)
CALL RTRC(3,1C3,IC2,Ic1,Ic0,ITRAC)
DO 136, I=o,
Do 136, 11=1, N
DO 136, 12=1, IIW(Il)
CALL INCR(IC3,Icz, 1C1,Ic0)
IF(I.EQ.IT)THEN
ANUM = -(IWI(Il,I2)/1.0)*(AMTCST)
WRITE(45,35) 103,1C2,1C1,IC0,1R3,IR2,IR1,IRo,ANUM
ENDIF
CONTINUE
ANUM = 1.0
CALL RTRC(3,1R3,IR2,IR1,1R0,ITRAK)
CALL RTRC(3,1C3,IC2,IC1,IC0,ITRAC)
IR = 3
CALL SUB2(IC3,IC2,ICl,ICO,IT,N,ANUM,IR)
ANUM = -1.0
CALL RTRC(3,1C3,IC2,IC1,IC0,ITRAC)
D0 226, I=1,N
DO 226, Il=1,IIW(I)
CALL INCR(IC3,IC2,IC1,ICO)
CONTINUE
CALL SUB2(IC3,IC2,IC1,ICO,IT,N,ANUM,IR)
RETURN

END

SUBROUTINE WKIT1(I3,I2,Il,IO,N3,N2,N1,NO,TEMP)
CALL INCR(I3, 12, 11, IO)

WRITE(45,35) 13,12,11,Io,N3,N2,N1,N0, TEMP
FORMAT(’ ',4(11),' ',4(11),' ’,F10.l)
RETURN

END

SUBROUTINE WKIT2(IC3,IC2,IC1,ICO)
CALL INCR(IC3,IC2,IC1,ICO)

RETURN

END

SUBROUTINE SORT()

DIMENSION IR1(50000),1R2(50000),RA(50000)
OPEN(45,FILE=’TEMP.PRN')

N = 0

Do 10, I=1, 450000

READ(45,27) IA,IB,C
FORMAT(3X,I6,3X,I6,3X,Fll.l)
IF(C.GT.-999999.0)THEN

 

10
11

130
36

142

N = N + 1

IF(N.EQ.50000)PAUSE’CONV....TOO MANY COEFFS.’
IR1(I) 1A

IR2(I) IB

RA(I) C

CONTINUE
CONTINUE
CLOSE(45)
CALL SUSS(N,IR1,IR2,RA)
Do 13o,11=1,N

CALL CONV(IR1(Il),IC3,IC2,IC1,ICO)

CALL CONV(IR2(I1),I83,IBZ,IBl,IBO)

WRITE(44,36) IC3,IC2,IC1,ICO,IB3,IBZ,IBl,IBO,RA(I1)

CONTINUE
FORMAT(’ x',4(11),' R',4(Il),’ ’,F10.1)
RETURN
END
SUBROUTINE CONV(IR,IC3,IC2,1C1,IC0)
IF(IR.GT.9999)THEN
PAUSE’CONV....NUMBER >=10,000'
ENDIF
IC3= IR/1ooo

IC2: (IR-IC3*1000)/100

ICl= (IR-IC3*lOOO—ICZ*100)/1O

ICO= (IR-IC3*1000—IC2*100-IC1*10)

IRT = IC3*1000+IC2*100+IC1*10+ICO
IF(IR.NE.IRT)PAUSE’CONV....IR <> IRT’
RETURN

END

SUBROUTINE SUSS(N,IR1,IR2,RA)
DIMENSION IR1(50000),IR2(50000),RA(50000)
L = N/2+1
IR = N
CONTINUE
IF(L.GT.1)THEN
L=L—1
IRR1=IR1(L)
IRR2=IR2(L)
RRA=RA(L)
ELSE
IRR1=IR1(IR)
IRR2=IR2(IR)
RRA=RA(IR)
IR1(IR)=IR1(1)
IR2(IR)=IR2(1)
RA(IR)= RA(I)
IR=IR-1
IF(IR.EQ.1)THEN
IR1(1)=IRR1
IR2(1)=IRR2
RA(1)= RRA
RETURN
ENDIF
ENDIF
I=L
J=L+L
IF(J.LE.1R)THEN

 

20

10

10

40

143

IF(J. LT. IR)Tn
IF(IR1(J). LTN IR1(J+1))J= J+1
ENDIF
IF(IRR1.LT.IR1(J))THEN
IR1(I)=IR1(J)
IR2(I)=IR2(J)
RA(I)= RA(J)

ENDIF
IR1(I)=IRR1
IR2(I)=IRR2

SUBROUTINE RTRC(IR,IC3,IC2,IC1,ICO,ITRAC)
DIMENSION ITRAC(15,4)

IC3 = ITRAC(1R,1)
IC2 = ITRAC(1R,2)
101 = ITRAC(1R,3)
ICO = ITRAC(1R,4)
RETURN

END

SUBROUTINE PNT(ITRAC)
DIMENSION ITRAC(15,4)
D0 10, I=1,12
WRITE(*,*) (ITRAC(I,J),J=1,4)
CONTINUE
RETURN

LAST CARD OF PNT

D

SUBROUTINE INIT(IT,N,K,IIw,ITRAC,IQK)
DIMENSION IIW(15), ITRAC(15,4),IQK(15,7),IC(4)

DO 10, I=1,4

IC(I) = o

CONTINUE

IR = 1

CALL QUICK(IR,IC,1TRAC)

DO 20, I=1,

DO 20, Il=1, N

Do 20, 12=1, K

DO 20, I3=1, IQK(11,12)

CALL INCR(IC(4),IC(3),IC(2),IC(1))
CONTINUE

IR=2

CALL QUICK(IR,1C,ITRAC)

DO 40, I=1,IT

DO 40, Il=1, N

D0 40, 12=1, IIW(I1)

CALL INCR(IC(4),IC(3),IC(2),IC(1))

CONTINUE

IR = 3

CALL QUICK(IR,IC,ITRAC)

 

60

80

90

100

10

144

Do 60, I=o, IT
Do 60, 11: 1,
DO 60, 12=1, NIIW(II )
CALL INCR(IC(4), IC(3), IC(2), IC(1))
CONTINUE
IR = 4
CALL QUICK(IR,IC,ITRAC)
Do 80, I=1,
DO 80, Il=1, N
DO 80, I2=1, IIW(Il)
CALL INCR(IC(4),IC(3),IC(2),IC(1))
CONTINUE
IR = 5
CALL QUICK(IR,IC,ITRAC)
DO 90, 11=1,K
D0 90, I2=1,N
CALL INCR(IC(4),IC(3),IC(2),IC(1))
CONTINUE

CALL QUICK(IR,IC,ITRAC)

DO 100, I=1,IT

DO 100, I1=1,K

DO 100, I2=O,N

DO 100, 13:1, N+1
IF(I2.NE.I3)THEN

IF((IZ.EQ.O) .AND. (I3.EQ.N+1)) GOTO 100
CALL INCR(IC(4),IC(3),IC(2),IC(1))
ENDIF

CONTINUE

IR= 7

CALL QUICK(IR,IC,ITRAC)

DO 110, I=1, IT

DO 110, I1=1, K

CALL INCR(IC(4),IC(3),IC(2),IC(1))
CONTINUE

IR = 8

CALL QUICK(IR,IC,ITRAC)

DO 120, I=1,

Do 120, Il=1, K

D0 120, 12:0, N+1

CALL INCR(IC(4),IC(3),IC(2),IC(1))
CONTINUE

IR = 9

CALL QUICK(IR,IC,ITRAC)

D0 150, I=1,

CALL INCR(IC(4),IC(3),IC(2),IC(1))
CONTINUE
IR = 10
CALL QUICK(IR,IC,ITRAC)
RETURN
LAST CARD 0F INIT
END

SUBROUTINE QUICK(IR,IC,ITRAC)
DIMENSION ITRAC(15,4),IC(4)

ITRAC(1R,1) = IC(4)
ITRAC(1R,2) = IC(3)
ITRAC(1R,3) = IC(2)
ITRAC(1R,4) = IC(l)
CONTINUE

35
225

35

145

RETURN
END

SUBROUTINE SUBZ(IC3,IC2,ICl,ICO,IT,N,ANUM,IR)
COMMON ITRAK(15,4),IIW(15)

CALL RTRK(IR, IR3, IR2, IR1, IRO)

Do 225, I=1, IT

Do 225, Il=1, N

DO 225, 12=1, IIW(Il)

CALL INCR(IC3, IC2, IC1, ICO)

CALL INCR(IR3, IR2, IR1, IRO)

WRITE(45,35) IC3,IC2,ICl,ICO,IR3,IR2,IRl,IRO, ANUM
FORMAT(' ’,4(Il),’ ’,4(Il),’ ’,F10.1)
CONTINUE

RETURN

END

SUBROUTINE RTRK(I, IR3, IR2, IR1, IRO)
COMMON ITRAK(15,4),IIW(15)
IR3 - ITRAK(I,1)

IR2 = ITRAK(I,2)
IR1 = ITRAK(I,3)
IRO = ITRAK(I,4)
RETURN

END

SUBROUTINE WTRK(I, IR3, IR2, IR1, IRO)
COMMON ITRAK(15,4),IIW(15)

ITRAK(I,1) = IR3
ITRAK(I,2) = IR2
ITRAK(I,3) = IR1
ITRAK(I,4) = IRo
RETURN

END

SUBROUTINE INCR(IR3, IR2, IR1, IRO)
IRo = IRO + 1
IF(IRO .EQ. 10) THEN
IRo = o
IR1 = IR1 + 1
IF(IR1 .EQ. 10) THEN
IR1 = o
IR2 = IR2 + 1
IF(IR2 .EQ. 10) THEN
IR2 o

IR3 = IR3 + 1

SUBROUTINE ITBl(IR3,IR2,IR1,IRO,ITRAK,
IC3,IC2,IC1,ICO,ITRAC,IT,K,N)
DIMENSION ITRAK(15,4),ITRAC(15,4)
FORMAT(' ’,4(Il ,' ’,4(Il),’ ’,F10.l)
CALL RTRC(B,IR3,IR2,IR1,IRO,ITRAK)

DO 276, J=1, IT

DO 276, J1=1, K

DO 276, J2=O, N

DO 276, J3=1, N+1
IF((J2.EQ.O).AND.(J3.EQ.N+1)) GOTO 276
IF(J2.EQ.J3) GOTO 276

CALL RTRC(8,IC3,IC2,ICI,IC0,ITRAC)

 

275
276

C

146

CALL INCR(IR3,IR2,IR1,IR0)

Do 275, I=1, IT

Do 275, I1=1, K

Do 275, 12=o, N+1

CALL INCR(IC3,IC2,IC1,ICO)
IF((J.EQ.I).AND.(J1.EQ.I1).AND.(J2.EQ.I2))THEN
ANUM = 1.0

WRITE(45,35)
IC3,Icz,IC1,1C0,IR3,IR2,IR1,IR0,ANUM

ENDIF
IF((J.EQ.I).AND.(J1.EQ.I1).AND.(J3.EQ.12))THEN
ANUM = - .0

WRITE(45,35)
IC3,IC2,IC1,IC0,IR3,IR2,IR1,IRo,ANUM

ENDIF

CONTINUE

CONTINUE

RETURN

END

CSINCLUDE SOS.FOR
CSINCLUDE ORD.FOR

C

19

20

50

60

SUBROUTINE ORD(IT,M,N,ITRAC,IIw,IQK)

INTEGER ITRAC(15,4),FLAG(15,15,14,7)
DIMENSION IIW(15),IQK(15,7)
DO 5,I=1,IT
DO 5,Il=l,M
DO 5, 12= 1, N
DO 5, I3 1,N
ELAG(12,I3,I,11) = o
CONTINUE

OPEN(78,FILE= ’KRUS.ORD’)
WRITE(*,19)
FORMAT(5X,’ENTERING ORD.FOR’)

WRITE(78,20)

FORMAT(’NAME’

FORMAT(lX,’UP x',411,17x,15)
CALL RTRC(lo,IC3,IC2,IC1,ICO,ITRAC)
INUM = IC3*1000+IC2*100+ICl*10+ICO + 1

CALL RTRC(9,IC3,IC2,IC1,IC0,ITRAC)

Do 50, I=1, IT

DO 50, I1=1, M

DO 50, 12=1, N
INUM = INUM — 1

CALL INCR(IC3,I02,Ic1,IC0)
WRITE(78,21)IC3,IC2,IC1,ICO,INUM

CONTINUE

CALL RTRC(1,IC3,IC2,IC1,IC0,ITRAC)

D0 60, I=1, IT

DO 60, Il=1, N

DO 60, 12=1, M

DO 60, 13:1, IQK(I1,12)
INUM = INUM - 1
CALL INCR(IC3, IC2, IC1, ICO)
WRITE(78,21)IC3,IC2,IC1,ICO,INUM

CONTINUE
CALL RTRC(4,IC3,IC2,IC1,IC0,ITRAC)
DO 100, I=1,IT
INUM = INUM — 1
DO 100,I1=1,N
Do 100,I2=1,IIW(I1)

 

100

835

0(3

147

CALL INCR(IC3, IC2, IC1, ICO)
WRITE(78,21)IC3,Icz,IC1,ICo,INUM
CONTINUE
WRITE(78,835)
FORMAT('ENDATA')
CLOSE(78)
WRITE(*,18)
FORMAT(5X,'LEAVING 0RD.FOR’)

LAST CARD OF ORD.FOR
RETURN
END

SUBROUTINE SOS(IT,M,N,ITRAC)
INTEGER ITRAC(15,4),ELAG(15,15,14,7)

DO 5,I=1,IT

D0 5,II=1,M

DO 5,I2=1,N

Do 5,I3=1,N

FLAG(12,I3,I,11) = 0

CONTINUE
OPEN(78,FILE='KRUS.SOS')

WRITE(*,19)

FORMAT(5X,’ENTERING SOS.FOR’)
WRITE(78,20)
FORMAT('NAME')
FORMAT(lX,'Sl’,2lX,I12)
FORMAT(1X,’S2’,21X,I12)
FORMAT(lX,'83',21X,I12)
FORMAT(4X,’X’,4I1,15X,I12)
CALL RTRC(6,IC3,IC2,IC1,ICO,ITRAC)
DO 100, I=1,IT

IWGHT = lO*N*M*(IT—I+1)
DO 100, Il=1,M

WRITE(78,21)IWGHT
Do 100, I2=O,N
Do 100, 13:1, N+1
IF(I2.EQ.I3)GOTO 100
IF((I2.EQ.0) .AND. (I3.EQ.N+1)) GOTO 100
CALL INCR(IC3, IC2, IC1, ICO)

IF((12.EQ.0).AND.(I3.NE.N+1))

WRITE(78,31) IC3,IC2,IC1,ICO
CONTINUE
CALL RTRC(6,IC3,IC2,IC1,ICO,ITRAC)
DO 200, I=1,IT

IWGHT = lO*N*M*(IT-I+l)
DO 200, Il=1,M

WRITE(78,21)IWGHT
DO 200, 12=o,N
DO 200, I3=1, N+1
IF(I2.EQ.I3)GOTO 200
IF((I2.EQ.O) .AND. (I3.EQ.N+1)) GOTO 200
CALL INCR(IC3, IC2, IC1, ICO

IF((I2.NE.O).AND.(I3.EQ.N+1))

WRITE(78,31) IC3,IC2,IC1,IC0
CONTINUE

WRITE(78,835)

FORMAT('ENDATA')

CLOSE(78)

WRITE(*,18)

FORMAT(5X,’LEAVING SOS.FOR’)

 

 

 

 

REFERENCES

 

149

REFERENCES

Baker, K. R. (1974). Introduction to Segpencing and
Scheduling. New York: John Wiley & Sons.

Baker, K. R., & Peterson, D. W. (1979). An Analytic
Framework for Evaluating Rolling Schedules. Management
Science, 2;, 341-351.

Barnes, J. W., & Brennan, J. J. (1977). An Improved
Algorithm for Scheduling Jobs on Identical Machines.
AIIE Transactions, 2, 25-31.

Barnes, J. W., & Vanston, L. K. (1981). Scheduling Jobs
with Linear Delay Penalties and Sequence Dependent
Setup Costs. Operations Research, 29, 146-160.

Beale, E., & Tomlin, J. (1969). Special Facilities in a
General Mathematical Program System for Non-Convex
Problems Using Ordered Sets of Variables. In J.
Lawrence (Ed.), Proceedings of the 5th International
Conference on Operations Research. Tavistock, London.

Bruvold, N. T., & Evans, J. R. (1985). Flexible Mixed-
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