SELECTED MINIMAX AND MINISUM
LAYOUT PROBLEMS
By
VINCIE DARRYL THORNTON
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
1977
To Cathy
ACKNOWLEDGMENTS
The author expresses his sincere appreciation to Professors Richard
L. Francis and Timothy J. Lowe, as well as to the other members of his
committee, for their help and guidance in the preparation of this work.
Thanks go to Drucilla N. Bouffard for typing both the draft and
final copies of the manuscript.
The author is also indebted to M. F. Wehling for the drawing of
the figures included in the present work.
Additionally, the author is most appreciative for the assistance
in proofreading the manuscript he received from Robin Andrews and
Joyce Welch.
TABLE OF CONTENTS
Page
ACKNOWLEDGMENTS . . . . . . . . .. . . . . iii
ABSTRACT . . . . . . . . . . . . . . .. vi
CHAPTER
I INTRODUCTION . . . . . . . . . . . 1
The Facility Layout and Design Problem . . .. 1
Related Work . . . . . . . ... .. 3
Motivation for the Research . . . . . .. 10
Overview . . . . . . . . . . 11
II LAYOUT PROBLEMS ON THE LINE INVOLVING DIVISIBLE ACTIVITIES 13
Introduction . . . . . . . .. . . 13
Problem Formulations . . . . . . . .. 13
Related Work . . . . . . . . . 16
Motivation for Research . . . . . . .. 19
Solution Procedures, Example Problems, and
Conclusions . . . . . . . .. .. . 23
Solving PMM . . . . . . . . .. 25
Solving PMS . . . . . . . . .. 29
Analyses for PMM and PMS ... . . . . . 32
Dominance Results . . . . . . .. 32
Analysis for P4.M ... . . . . . 45
Analysis for PMS .. . . . . . . .49
Summary of Results .. .... .. . . . 52
III LAYOUT PROBLEMS ON THE LINE INVOLVING INDIVISIBLE
ACTIVITIES . . . . . . . . . . . . 53
Introduction .. . . . . . . . . 53
Problem Formulations. .... . . . .. . 54
Motivation for Research and Related Work . . . 56
Solution Procedures, Example Problems, and
Conclusions ..... . . . . .. . . 57
Solving PMM . . . . . . . . .. 59
Solving PMS . . . . . . . . . 63
Centrally Dominating Objective Functions . . . 64
Definitions and Notation . . . . . . 64
Alternating Order Interchange Procedure (AOIP) 66
Properties of a CDOF . . . . . . . 67
Page
Analysis for PMM . . . . . . . . . 70
CDOF Analysis . . . . . . . . . .. 75
Summary of the Chapter . . . . . . . 79
IV MINIMAX AND MINISUM LAYOUT PROBLEMS IN THE PLANE .. 80
Introduction .. . . . . . .. ... . 80
Problem Formulations . . . . . . . .. 81
Related Work .. . . . . . . . . 89
Motivation for Research . . . . . . .. 93
Solution Procedures, Example Problems, and
Conclusions . . . . . . . . . . 95
Solving PMM . . . . . . . . .. 98
Solving PMS . . . . . . . ... 102
Analyses for PMM and PMS . . . . . . .. 105
Dominance Results . . . . . . .. 105
Analysis for PMM . . . . . . . .. 122
Analysis for PMS . . . . . . . .. 126
Summary of Results . . . . . . . .. 132
V FUTURE RESERCH . . . . . . . . . .. 133
Introduction . . . . . . . . . . 133
Additional Research on Linear Layout Problems . .. 133
Indivisible Activities with Varying Lengths .133
Linear Layout Problems with Pairwise Weights .134
Centrally Dominating Objective Functions . . 135
Additional Research on Planar Layout Problems . . 135
Rectangular Layouts Relative to PMM2 and PMS .135
Modifications and Extensions of P1I and PMS 141
BIBLIOGRAPHY . . . . . . . . . . . . . 142
SUPPLEMENTARY REFERENCES . . . . . . . . . .. 148
BIOGRAPHICAL SKETCH . . . . . . . . . . . .. 150
Abstract of Dissertation Presented to the Graduate Council
of the University of Florida in Partial Fulfillment of the Requirements
for the Degree of Doctor of Philosophy
SELECTED MINIMAX AND MINISUM
LAYOUT PROBLEMS
By
Vincie Darryl Thornton
March 1977
Chairman: Richard L. Francis
Cochairman: Timothy J. Lowe
Major Department: Industrial and Systems Engineering
Selected minimax and minisum layout problems involving the determina-
tion of optimal configurations and relative locations for linear (divisible
and indivisible) and planar, nonoverlapping, interacting facilities (under
the assumption that the distance between points in the plane is recti-
linear) are formulated. It is assumed that the length (for the linear
case) or area (for the planar case) required by each activity is some
known positive constant. In all cases, the pertinent objective function
is to be minimized and'is a function of distances between the component
activities. The problems considered are analyzed; and, solution proce-
dures, example problems, and conclusions for each are provided.
CHAPTER I
INTRODUCTION
The Facility Layout and Design Problem
Facility layout and design problems have traditionally involved,
among many other things, the determination of the physical dimensions of
a facility (e.g., an industrial plant or warehouse) and the relative loca-
tions.of the various components (such as departments or storage bays) making
up the facility so as to minimize some function of the distances between
components. Since such problems occur in many instances having nothing
to do with "facilities" per se, the term "layout problem" is henceforth
used to mean any problem having the fundamental objective of minimizing a
given function of distances by appropriately specifying shapes and relative
locations.
Many variations of layout problems exist and differ in such ways as
the distance function used, how the distances are defined, how much is
prespecified about the shape or dimensions of the overall design, whether
the problem is linear or planar in nature, and whether the problem involves
minimizing a maximum of distances (minimax or MI) or a sum of distances
(minisum or MS). Layout problems can also be categorized as to whether
they are discrete or continuous. The former type occurs when each facility
or activity is considered to correspond to the union of a number of grid
squares (in the planar case) and only the center points of the squares are
of interest in the measurement of distances, whereas for the continuous
layout problem, the region (linear or planar) to be occupied by a facility
is considered in its entirety. Another distinction lies in whether the
activities are noninteracting or interacting. In the former instance, the
assignment of a facility to a site or to a region on the line or in the
plane affects the other facilities only by preventing them from occupying
the same site or region. We say that two activities interact if, for
example, the "cost" of assigning one facility to a given locality is a
function of the placement of the other.
For layout problems on the line, there is normally only one way in
which distance is defined; however, this is not the case for the planar
situation, where either the Euclidean or the rectilinear distance, among
others, may be of interest. In many instances, travel between or within
facilities occurs along a rectilinear network of aisles or streets so that
the appropriate measure is rectilinear distance.
The objective of this research is to determine MM and MS configurations
and relative locations for linear and planar, nonoverlapping, interacting
facilities under the assumption that the distance between points in the plane
is rectilinear. We will also assume that the length (for the linear case)
or area (for the planar case) required by each activity is some known posi-
tive constant; and constraints on the shape of the overall facility or the
facility components may or may not be given. In all cases, the pertinent
objective function is to be minimized and is a function of distances
between the component activities.
Related Work
A class of problems closely related to layout problems are normally
termed "location problems" and generally involve finding locations for
new facilities relative to known existing facilities such that some func-
tion of the distances between new and existing and/or new facilities is
minimized. The basic distinction usually made between location problems
and layout problems is that the activities to be assigned may be idealized
as points for location problems, while this is not possible for layout
problems.
Since location problems may occur as subproblems in layout problems
and since the geometric structure and solution procedure for the location
problems often provide insight for the layout problems, location theory
literature is directly related to the present research. This literature
is quite extensive, and no attempt will be made here to provide a detailed
discussion of the general location theory literature. The works by Cabot,
Francis, and Stary [6], Elshafel and Haley [14], Francis [16], Francis and
Goldstein [22], Francis and White [2h], Geoffrion [25], Hanan and Kurtzberg
[27], Lea [45], Revelle, Marks, and Liebman [58], Scott [59], and Whybark
and Khumawala [65] each contain an extensive bibliography and discussion
dealing with location literature.
As we shall see subsequently, another field related to the present
research is that of machine scheduling (specifically, the one-machine job
sequencing class of problems). Since the specific works of interest here
from the scheduling literature will be pointed out later, and as only a
particular area of scheduling theory is directly pertinent to layout problems,
we will make no in-depth discussion of the general scheduling literature.
Such a treatment of scheduling theory is provided in the works of Conway,
Maxwell, and Miller [91, Kan [32], and Martin-Vega [50].
We will now review literature directly related to the present research.
Some of the discussion to follow has been adapted from the works of Francis
[18] and Cabot, Francis, and Stary [6], each of which contains extensive
reviews of the literature on layout problems.
A problem occurring in various forms in the location literature is
the assignment problem (A.P.), which can be stated as follows: Given n
activities to be assigned to n locations, find an assignment of activities
to locations in such a way as to minimize the total cost incurred. It
appears that Koopmans and Eeckman [38] were the first to point out the
relevancy of the A.P. to location problems. Later, Moore [51] discussed
plant layout applications of the A.P.
A problem more involved than the A.P. is the generalized assignment
problem (G.A.P.), which can be stated as follows: Given n "divisible"
activities, each requiring a given total number of sites (which may be
different for each activity), and m sites, find an assignment so that the
total incurred cost is minimized. Mallette and Francis [49] show that
some discrete warehouse layout problems with noninteracting activities can
be formulated and efficiently solved as a G.A.P. Francis and White [24]
point out that when the objective function of the A.P. is changed to the
maximum cost of a facility assignment, the resulting problem is a MM or
bottleneck A.P. A discrete, noninteracting layout problem, corresponding
to a special case of the bottleneck transportation problem, is also presented
in [24] and referred to by the authors as a IM G.A.P.
Several continuous MS layout problems which have received attention
in the literature involve noninteracting facilities and are analogous to
the G.A.P. Suppose a warehouse is to be constructed in a region L of the
plane, which represents a warehouse floor or a plot of land. The warehouse
is to store n items where item i will take up a set of points in L represented
by a compact set S. of known positive area A., and any two sets S. and S.,
i # j, cannot overlap. An arrangement of the Si; i s N E {1, ..., n},
satisfying these conditions and denoted by {S, ..., Sn } is called a layout.
The collection of all such admissible layouts is denoted by Hn(L:A). It
is also assumed that for each item i, a function fi(X) is known such that
the total cost of moving item i in and out of storage is given by
f f.(X) dX. The problem of interest for this situation is
Si
n
minimize F(S, S )S i = fi(X)dX
1 n S 1
1
s.t. {Sl, ..., Sn} E Hn(L:A)
Francis [21] has developed sufficient conditions for solving a version of
this problem where there exists a function f(X) and functions gi(y) such
that fi(X) = gi[f(X)]. In addition, he has shown that a substantial class
of layout problems can be so formulated. Francis [191 has also solved a
related warehouse layout problem where additional rectangularity restric-
tions are imposed upon the sets SI, ..., S In a subsequent paper, Corley
1"' n
and Roberts [10] developed necessary and sufficient conditions for a general
solution to the above problem; a G.A.P. is obtained when.their problem is
discretized. They also give urban districting applications to the problem,
while Larson and Stevenson [40] consider a related problem. Both Corley
and Roberts, and Francis construct minimum solutions by utilizing the
previous results of Francis [17] for constructing contour lines of the
relevant functions. Recently, Lowe and Hurter [47], [48] have extended the
results of Corley and Roberts.
Very little research has been done on continuous M layout problems
analogous to the MM G.A.P. and involving noninteracting sets. With refer-
ence to the above discussion, define
Gi(S.) = max [fi(X) : X e Si],
G(Si, ..., S n) max [G.i(S.) : i E N .
The problem of interest is to find a layout in H n(L:A) to minimize
G(S1, ..., Sn). This problem occurs in warehouse layout and in stadium
design contexts. Francis [21] gives examples in both contexts, and he
provides sufficient conditions for a I1 layout corresponding to the problem
when it may be assumed that there exists a single function f(X) and functions
gi(y) such that fi(X) = gi[f(X)], i E N, that g (y) > gi+l (y) for all y,
1 < i < n-l, and that each function gi(y) is nondecreasing.
A step beyond the G.A.P. is the quadratic assignment problem (Q.A.P.),
the formulation of which is due to Koopmans and Beckman [38]. The Q.A.P.
is known to be at least as difficult as the traveling salesman problem
[57). When facilities interact and are discrete, the corresponding layout
problems will lead to a Q.A.P. or to special cases of the Q.A.P. Consider
n new activities to be assigned to n sites, let Xij equal one if activity i
is assigned to site j and zero otherwise, let cij be a fixed cost of assigning
-~J
activity i to site j, and let whj dik = wjh dki be the cost due to transport
of items between activities h and j when activity h is assigned to site i
and activity j to site k. The Q.A.P. may then be stated as follows:
S cn n n n n n
in cX. + X d X
i= j=1 2 h=l i=l j=1 k=l hi hj ik jk
n n
s.t. X. 1, j E N; X. 1, i E N; and,
i=l j=1
X.. = 0, 1; i, j E N
A recent review of the Q.A.P., emphasizing its use in solving problems of
placing logic packages in wiring networks, has been presented by Hanan and
Kurtzberg [27], who list forty-two references. Pierce and Crowston [56]
give some tree-search algorithms for solving the problem, and also list
twenty-seven references not given by Hanan and Kurtzberg. Since no com-
putationally satisfactory procedures exist for finding exact optimal
__
solutions to the Q.A.P. when the number of facilities exceeds fifteen [27],
the Q.A.P. is essentially unsolved and heuristic methods are used almost
exclusively. However, Karp and Held [35] have shown that an analogous but
simpler problem, which they call the module placement problem, can be solved
exactly using dynamic programming when all of the c.. equal zero and the
sites are all located on a single line; however, the effort involved in
their solution procedure is of order n2n. Adolphson and Hu [I] studied
a special case of the module placement problem, while Simmons [60], [61]
considered a similar problem. Other related papers include those of
Hillier [28], Hillier and Connors [29], Lawler [43], [44h, and Nugent,
Vollman, and Ruml [53]. Also, since the plant layout problem can be
formulated as a Q.A.P., or special case thereof, the papers by Armour and
Buffa [2] and Buffa, Armour, and Vollman [5] on CRAFT (a computerized
layout algorithm based on an improvement heuristic) are of interest.
As best as can be determined, no study has been made of what might
be called the MM Q.A.P. This problem would have the same constraints as
the usual Q.A.P., but with objective function to be minimized defined for
any assignment X = (X.i) as follows:
max [max (i : Xi = 1, Vij), max (whjdik : iXjk = 1, Vh,i,j,k)].
Various versions of a MM multifacility location problem, which permit the
consideration of all locations in the plane rather than limiting facility
locations to a finite number of points, have appeared in the works of
Dearing [11], Dearing and Francis [12], and Wesolowsky [64]. Also,
Francis [16] has efficiently solved a related problem, which he refers to
as the Minimax Facility Configuration Problem. His problem is, loosely
speaking, one of finding a facility layout such that the identical com-
ponents thereof will be as close to each other as possible. Although the
components are interacting in this problem, the interaction is of a simple
nature in the sense that they all have the same relative effect on the
assignment of one another so that the problem is a substantially simpler
version of the MM Q.A.P. stated above.
Given the difficulty of solving the Q.A.P., it comes as no surprise
to find that continuous analogs of the Q.A.P. and the MM Q.A.P. have received
relatively slight attention in the literature. Portions of the book by
Larson [39, pp. 78-85, 255-258] and the paper by Newman [52] give some
consideration to continuous layout problems with interacting facilities,
where some measure of average travel distance is of interest.
We will consider several of the works cited above in more detail as
they relate to some of the layout problems presented in subsequent chapters.
Additionally, literature even more closely related to the present research
than many of the above-mentioned works will also be discussed in later
chapters after the pertinent problem of interest has been introduced. We
feel that the relationship between said existing literature and the present
research can be better appreciated by this method of presentation. Such
work not discussed above, but to be considered below, includes that of
Chan and Francis [7], [8], Papineau, Francis, and Bartholdi [55], Francis
and Papineau [23], [54], and Lowe [46].
Motivation for the Research
Layout problems, including plant layout problems, office layout
problems, laying out files on a computer tape, arranging electric modules
on a circuit board, and determining the relative location of the instruments
on an aircraft cockpit panel, to name but a few applications, continue to be
of fundamental interest to people of many diversified fields of endeavor.
In spite of the interest in and importance of these problems, relatively
little fundamental research is being conducted on layout problems. Much
of the current research on layout problems involves computerizing tradi-
tional, normally heuristic, approaches to these problems. Kaimen [311
discusses some one hundred and twenty such computer programs, while
Vollmann and Buffa [63] provide a good discussion of some of the difficul-
ties and inadequacies of such traditional approaches to solving layout
problems.
The (aforementioned) traditional approaches have not resulted in a
cumulative body of sound analytical knowledge; however, there is consider-
able evidence that the development of such a body of knowledge is possible
for layout problems just as has been the case for location problems.
Francis [20] provides the following as possible reasons for the paucity
of the analytical study of layout problems as compared to that of location
problems: layout problems are often smaller in scope than location problems,
it may be difficult to measure the value of a given layout, it may be diffi-
cult to agree upon the class of facility shapes to choose from, solutions
to layout problems are often easy to criticize after the fact, and one of
the basic underlying analytical problems is known to be at least as diffi-
cult as the traveling salesman problem. Because of the importance of layout
problems, it would seem that they deserve more analytical attention even
when the difficulties involved are considered.
A fundamental thesis adopted in the present research is that by
studying and thoroughly solving analytical, but perhaps idealized, models
of the real layout problems, the solutions thus obtained will provide
design guidelines for and insight into the real layout problems and will
help to furnish a basis for further analytical research on layout problems
incorporating more and more of the characteristics of real problems. Hence,
the motivation to study MM and MS layout problems is twofold: first to add
to the development of a cumulative body of knowledge for layout problems,
and second, to provide worthwhile and useful guidelines or benchmarks for
the layout process. We hope that the present research will satisfy both
of these objectives.
Overview
Chapters II, III, and IV each contain a study of MM and MS layout
problems formulated in terms of interacting facilities or activities.
Chapters II and III treat MM and MS problems on the line (involving "divisi-
ble" and "indivisible" activities, respectively), while Chapter IV deals
with such problems in the plane. In each of these chapters, we first formu-
late the problems under consideration, then we present related work, and
give a motivation for the research. In each of the former chapters, we
then provide solution procedures, example problems, and conclusions for
12
the problems on the line, followed by problem analyses and a summary of
the chapter. In Chapter IV we furnish similar material for a certain class
of problems in the plane. The solution procedures for the problems of
Chapter II will provide insight for the work of Chapter IV. We will trans-
form the problems of Chapter IV into equivalent problems with the aid of
a rotation first applied to MM location problems by Francis [15]. In
Chapter V, we give ideas for future research, particularly relative to
extensions and modifications of the problems of Chapters II, III, and IV.
CHAPTER II
LAYOUT PROBLEMS ON THE LINE INVOLVING DIVISIBLE ACTIVITIES
Introduction
This chapter deals with two problems involving the location of n
classes of "divisible" activities on the line, where n is a positive
integer and n 2 2. Two different objective functions are considered.
One problem is of the MM form and entails the assignment of the n
activities to n nonempty, nonoverlapping, and compact sets, SI,...,Sn,
so that each set S. has a given positive measure, or length, Li, i E N
{1,2,...,n}, and such that the maximum of n associated functions of the
maximum distance between each corresponding set and any other set is
minimized. The second problem of interest is of the MS form and differs
from the first only in that the sum, rather than the maximum, of the
functions is to be minimized.
We formulate each problem, present related work, and give motivation
for the research. Then, we provide solution procedures, example problems,
and conclusions for each, followed by analyses of both problems and a
summary of the chapter.
Problem Formulations
In order to state the problems of interest, define a layout on the
line to be a collection S = {S1,S2,...,S }, where Si is any compact subset
of the real line of length Li (a known positive constant) for i s N
and S. and S. are nonoverlapping in the sense that they share no interior
1 J
points, for i, j e N, i # j. Each set S. is assumed to be divisible, in
the sense that it is represented by the union of the elements of a finite
collection of closed nonoverlapping (and not necessarily adjacent) intervals
S ,S i,...,S. where p.i 1 is to be determined V i. For each activity
1 2pi 1
i, the set S. specifies the intervals taken up by the activity. We define
1
L E Z {L. : i e N}. Since the objective functions we shall consider are
all nondecreasing in distance, without loss of generality (w.o.Z.o.g.),
we restrict activities to lie in the interval [O,L]. Hence, a layout S
can be thought of as a partitioning of the interval [O,L], consistent with
the above definitions, so that adjacent subsets share common endpoints.
Further, to introduce a definition needed later, we say that a lay-
out is symmetric if there exists an ordering of activity indices, say [1],
[2],...,[n], such that S l] is a single closed interval with midpoint at
L/2 and length Lll], and S] is the union of two nonadjacent closed inter-
vals, each of length L i/2, with one interval abutting the leftmost
point, and the other interval abutting the rightmost point, of S[ ,
2 5 i s n. For such a symmetric layout, it is quite natural to say that
activity [2] nests about activity [1], activity [3] about [2], and so forth.
In order to define an objective function for every layout, S =
{S1,...,S n, for the MM and MS problems of interest, the following defini-
tions and notation are established. Let d(Si,S.) denote the maximum
distance between any point in S. and any point in S., that is, for i,
j N, i # j,
d(S.,S.) = max {jx.-xjl : x. e S., x. S} (]
1i J 11 j 2 j j
Let D.(S) denote the maximum distance from any point in S. to any point
in S. where i # j, that is,
D.(S) = max {d(Si,Sj : j e N, j # i} (2)
Given any layout S, we use Di(S), defined by (2), as the measure
of distance of interest for activity i, i E N. Di(S) is a conservative
over-estimate of the average distance between activity i and any other
activity, and its use implies the problem contexts of interest are ones
where it is more important to be concerned about "worst" or "greatest"
distances than about average distances.
For i e N, fi is a given nondecreasing real-valued function with
domain (O,-), and we define G.(S) by
G.(S) E fi [D.(S)], i e N ,(3)
and define G(S) by
G(S) E max {Gi(S) : i E N} (4)
We refer to the problem of finding a layout to minimize (4) as PMM
(problem minimax). Such a layout will be referred to as a 1M layout.
Also, we define Fi(S) by
F.(S) = fi [Di(S)], i e N
1
where fi is a nondecreasing function, not necessarily the same as in (3),
and define F(S) by
n
F(S) C Fi(S) ,(6)
i=l
We refer to the problem of finding a layout to minimize (6) as PMS (problem
minisum). Such a layout will be referred to as a MS layout. Generally
it will be convenient to use G(S) and F(S) to distinguish between objective
functions for MM and MS problems, respectively.
Related Work
The layout of facilities on the line is also referred to as "linear
layout" in the literature. Newman [52] studies a linear layout problem
in which the average distance between two regions under an equal likeli-
hood assumption is minimized. Newman obtains a lower bound on the average
distance, and constructs a sequence of linear layouts which attains this
average distance only in the limit. Papineau and Francis [54] derive a
closed form solution to the MM version of Newman's problem for an arbitrary
number of regions. Their problem corresponds to PMM with fi[Di(S)] =
Di(S), i s N, and the solution thereto is found as follows: evenly divide
the largest subset, place one piece of this set at each end of the interval
[0,L], and place the remaining subsets in any fashion in the unallocated
portion of the interval. Lowe has solved a more general case of this
problem where fi[Di(S)] = wiDi(S) with wi a nonnegative weight; the
I
answers he obtains are similar to those of Papineau and Francis, except
that each object class is "split" and there are fewer alternative optima.
The results by Lowe suggest the study of PMM.
Chan and Francis [8] study the problem of locating n facilities on
the line so that facilities i and j are at least a certain distance apart
and either a weighted maximum or a weighted sum of the distances between
the centers of the facilities is minimized. They consider several forms
of the weighting function and the interdistance constraining function
and obtain closed form optimal layouts in each case.
Related to PMS is the module placement problem (MPP) formulated by
Karp and Held [35] as a dynamic programming problem. The problem consists
of assigning modules to locations on a line so as to minimize a weighted
sum of distances between all pairs of modules. Lawler [44] points out
that the dynamic programming approach solves the problem in order n2n
running time. Elmaghraby [13] shows that a closely related sequencing
problem can be formulated and solved in the spirit of Karp and Held's
approach. Adolphson and Hu [1] consider a special case of the MPP where
adjacent locations on the line are a unit distance apart; and, they present
an algorithm requiring 0(n log n) operations for solving the problem when
the associated graph is a rooted tree. A sequencing problem studied by
Horn [30] is identical to the layout problem studied by Adolphson and Hu.
Pratt [57] solves a special case of the MPP where all the modules are of
the same size, and he points out that if we could efficiently solve the
general MPP (one with arbitrary distances), even for the special case of
nonnegative weights, we could then solve the traveling salesman problem
efficiently.
Another class of problems, related to the MPP and to the MS problems
addressed by Chan and Francis [8], involve the problem of positioning
indivisible records in a linear storage medium in such a way that the
expected access time is minimized. This problem arises quite often in
the context of storage applications for computers. Works on such problems
include the papers by Bergmans [4], Grossman and Silverman [26], and Yue
and Wong [67]. Loosely speaking, the solution is (e.g., see [8], [36],
and/or [66]) to assign the most frequently accessed record to the center
of the storage medium and then repetitively to place the next most fre-
quently accessed record alternating between the position immediately to
the left of those already placed and the position immediately to the
right. Pratt [57] terms this ordering procedure "alternating order."
Simmons [61] presents a M\ one-dimensional space allocation problem
somewhat related to PMS. The application context of interest is one of
arranging rooms along one side of a corridor. As indicated in a later
note by Simmons [60], Francis has pointed out that the previous space
allocation problem is analogous to a multitask, one-machine job sequencing
problem. It can be observed that there are close relationships between
linear layout and sequencing problems in scheduling contexts, as demon-
strated by the work of Elmaghraby and Simmons. Also, the layout problems
herein considered include as subproblems sequencing problems studied by
Kan [32] and Lawler [41], [42].
The closest related work to PMS appears to be that of Lowe [46],
where the special case for which fi[Di(S)] = wiDi(S), wi a nonnegative
weight, i e N, is solved. The results by Lowe intimate PMS.
Motivation for Research
One principal motivation for studying layout problems on the line
is the location of classes of machines, work places, or other activities
along an aisle, conveyor belt, or sidewalk. The function f. could
correspond to the relative importance of or amount of travel associated
with activity i. The overall aim dictated by the specific problem of
interest may be best accomplished by either a MM or a MS objective function,
depending on which is more appropriate.
Lowe [46] gives the following possible example of the MS problem
studied here. Suppose that a manager must arrange n departments (separate
activities) along an aisle, where each department has a space requirement
Lj and any individual in department j may find it necessary to travel to
the location of any individual located in department i, i j. The function
fj may correspond to the relative importance of department j (salaries,
e.g.). The value D (S) corresponds to the maximum travel which would be
encountered by activity j with respect to layout S. The individuals com-
prising activity j desire that the overall layout will be such that the
maximum travel distance from activity j to any other activity is as small
as possible. The manager uses the functions fi to reflect the relative
importance of each activity with respect to the overall layout process.
In this situation, the manager seeks a layout S* such that any other
layout S for which F.(S) < F (S*) implies that F.(S) > F.(S*) for at least
j 2. 1
one i, i # j; S* is then called "efficient" in this sense. Finding an S
such that F(S) is minimized gives an efficient layout. Once the manager
assigns locations for the activities, the individual activities may arrange
their departments within their assigned locations in any manner desirable.
An example similar to the above where G(S) instead of F(S) is the
appropriate objective function could be easily imagined. Additional
examples of linear layout problems of interest occur in contexts discussed
in many of the related works mentioned in the sections dealing with related
work above and in Chapter I.
Additional motivation for studying problems with special structure,
such as the linear layout problems considered here, is the previously
stated fact that more general problems like the MPP are still largely
intractable and the insight gained from the present research could well
lead to the subsequent solution of much more general problems. In fact,
PMS could be considered to be a conservative version of the MPP in the
sense that
I ijd(si, S ) 3 v D (S)
ij J
where wij is the weight corresponding to modules i and j, and w
max w.ij : i E N}.
While the above motivation may be both valid and important, it is
believed that the solutions to such layout problems on the line may well
provide valuable insight into more challenging and generally more conse-
quential layout problems in the plane. This belief has been substantiated
in the past by, for example, the fact that many location problems in the
plane can be solved by considering two one-dimensional subproblems [6],
[11]. Another example is the work by Karp et al. [36] and by Yue and
Wong [67] on two-dimensional storage allocation problems; their solution
procedures resulted only after the analogous one-dimensional problem had
been solved. Yet another example is provided by the fact that the work
by Papineau and Francis [5]4 on a linear layout problem subsequently led
to the solution by Papineau, Francis, and Bartholdi [55] of a similar but
more complex layout problem in the plane, where distances are rectilinear,
and activities must be of given areas, although shapes are not prespecified.
Roughly speaking, minimax solutions are such that activities of largest,
and possibly second largest area, are "outside," and other activities
are "inside." The solutions are thus quite analogous to those obtained
in [54]. Depending upon the data, minimax solutions may be obtained such
that outside activities are not "indivisible," but consist of the union
of two disjoint rectangles instead.
One of the major differences between formulations of the MPP in the
plane and on the line involves the question of shape. For formulations
on the line, it is generally assumed that each activity has the same
"shape," a closed interval; in this sense the activity is "indivisible."
For planar problems perhaps the shape requirement most analogous to the
interval requirement would be a square. However, it is often not clear
that requiring all shapes to be squares, or even rectangles, is always
desirable. Indeed, it is known (see, for example, [55], as well as Chapters
6 and 7 of [24]), that there exist optimum solutions to some planar layout
problems which are symmetric and have activities with non-square shapes
"nested" about one another. It therefore seems of interest to examine
layout problems on the line involving divisible activities.
An examination of the literature discussed to this point indicates
a number of reasons for studying layout problems on the line. Some problems,
such as arranging rooms along a corridor, items in a warehouse along an
aisle, or a parking lot with respect to a building (when both the lot and
building must face a common street) are closely approximated by layout
problems on the line. In addition, for the parking lot problem, it is
quite reasonable to assume the parking lot is divisible, with parts of
the lot on either side of the building.
The primary reason for studying layout problems on the line, however,
is that by so'doing we may gain insight into planar problems by first
attacking analogous, and simpler, problems on the line. If the problems
on the line are difficult to solve, it is certain that analogous planar
problems will be more difficult. However, if the problems on the line
are tractable then the analogous planar problems may also be solvable,
and insight gained in solving problems on the line may be helpful in
solving the planar problems. The existing literature demonstrates that
by studying layout problems on the line with divisible activities, we
may gain useful insight into analogous planar problems which have nested
and symmetric solutions. Also, there are minimax problems for which we
know that specified activities of largest "size" should be "outside,"
whether the problems are planar or on the line. Further, we know that
the notion of alternating order is important both for planar problems
and problems on the line.
For the problems we shall study, we shall see that the notions of
nesting, and symmetry, are important. Also, the question of which activity
should be outside is of interest, as are ordering procedures quite analogous
to alternating order procedures. Additionally, these same notions will be
fundamentally involved in planar analogs (see Chapter IV) of the linear
layout problems considered in this chapter.
Solution Procedures, Example Problems, and Conclusions
In this section we give solution procedures, as well as numerical
examples for the problems of interest. Also we examine the procedures
for insights they provide into the problems. We give algorithms for
solving PMM and special cases of PMS, and establish that the general
version of PMS is equivalent to a one-machine multijob sequencing problem
known to be difficult in a well-defined sense.
Perhaps the major analytical result of this chapter is a dominance
result, first established by Lowe [46] and proven subsequently in a dif-
ferent manner. The result states that given any layout S there exists a
symmetric layout S* such that Di(S*) < Di(S), i E N. Since PMM and PMS
each have objective functions which are nondecreasing in the distance
measures Di(S), i E N, a direct consequence of the dominance result is
that it is enough to consider only symmetric layouts in order to solve
either PMM or PMS.
It is convenient to denote by S(p), p e N, the symmetric layout such
that activity p takes up the region [0, L /2] U [L-L /2, L], and such
P P
that, otherwise, activities nest about one another in increasing index
order ([1], [2], etc.) from the midpoint of the interval [0,L] outwards.
For versions of PMM and PIS we shall find that for some activity index,
say p*, the layout S(p*) will be optimal. Most of the computational
effort entails finding p*, the index of the "outside" activity.
For p e N, define
CLp E {Li : i N}
Also, let
L* E L/2 + min {Li/2 : i E NC
and
L** L min {Li/2 : i N}.
It can be directly verified that, for a symmetric layout,
min {Di(S) : i e N} > L*
and
max {D.(S) : i s N} < L**
As a notational convenience, we write f. f-i+1 to mean fi(x) 2
fi+l(x) V x e [L*, L**]. Also, "fi fi+ is a nondecreasing function"
should be taken to mean that fi(x) fi+1(x) is a nondecreasing function
of x, x E [L*,L**].
Solving PIM
Prior to stating procedures for solving PNMM, it is convenient to
state an algorithm due to Lawler [42] for finding a sequence of jobs to
be performed in a multitask, one machine sequencing problem, with pre-
cedence constraints among the jobs, which will minimize the maximum
completion cost. For each job j,.the completion cost, c (t), is assumed
to be a nondecreasing function of time, t. We will state the algorithm
for the special case where some job, say job f, must be performed first
and where job j, j E M E {l,...,m}, can be performed any time after the
completion of job f, where f i M. We let tf be the processing time of
job f and tj be the processing time of job j, j e M.
Lawler's Algorithm.
0) Define M E {l,...,m}, t = tf + t + ... + t .Let j = m.
f 1 m
1) Let [j] c M be any index such that c[j](t) = min {ck(t) : k e M}.
Job [j] will be worked on last among the jobs in M.
2) Delete [j] from M.
3) If M is non-empty go to 4). If M is empty, go to 5).
4) Replace t by t t[j]. Subtract one from j'and go to 1).
5) Stop. Work first on job f, second on job [1], third on job [2],.
..., last on job [m].
To see how Lawler's algorithm is of use in solving PMM, consider a
symmetric layout where activity p, p e M, nests all other activities and
I
activity [i] nests around activity [i l] for 2 i n-l, where [i],
i = l,...,n-l, is some ordering of the indices {l,...,n} {p} = N {p}.
We then have D[i](S) = [L + L[1] + ... + L[i]]/2 for i = l,...,n-l. Thus,
by letting t = L/2, t = L./2 for i E N {p}, c (t) =- and c. = f
f 1 i
for i e N {p}, we can use Lawler's algorithm to find the minimax ordering
[n-l],...,[l] of the indices N {p}. Letting T(p) be the layout where
activity p nests around all other activities and [i] nests around [i-l],
2 < i < n-1, where [i] is the ordering of N {p} found from the use of
Lawler's algorithm, it then follows that G(T(p)) is no larger than the
objective function value of any symmetric layout with p fixed as the
outside activity. By determining T(p) for all p e N, we can then find a
minimax layout which solves PMM. We now state Algorithm T4-l, which
solves PMM.
Algorithm MM-1.
1) Let p = 1 and go to 2).
2) Use Lawler's algorithm to find the ordering [i], 1 < i < n-l, of
the indices N {p}.
3) Using the ordering found in Step 2), compute G(T(p)).
4) If p.= n, go to 5). Otherwise let p = p + 1 and go to 2).
5) Any layout T(p*) such that G(T(p*)) < G(T(p)), p e N, is a MM layout.
Stop.
When we may number activities so that the functions fl,...,fn are
ordered, that is,
fi 2 fi+1, 1 < i < n-1
then Algorithm MM-1 simplifies to Algorithm MM-2 below. We remark, if
there is a tie for which activity to number n, it is helpful to choose,
from those which tie, an activity having greatest length, since so doing
serves to make Dn(S(n)) and consequently G (S(n)) as small as possible
and does not increase Di(S(n)) nor consequently Gi(S(n)) V i 1 n.
Algorithm MM-2.
1) Construct the layout S(n) and compute G(S(n)). Go to 2).
2) Let k be the smallest activity index such that
G(S(n)) = Gk(S(n)).
If k = 1, go to 6). Otherwise go to 3).
3) Define the set
I' = {i : 1 i k, Li > L CLk}.
If I' = p, go to 6). Otherwise go to 4).
4) Define the set
I" = { i : i I', fi(L Li/2) < fk[(L + CLk)/2]}.
If I" = <, go to 6). Otherwise go to 5).
5) For p e I', compute
G(S(p)) = max {f (L L /2), max [G.(S(n)) : 1 *
*
Let p* be any index in I" such that
G(S(p*)) = min [G(S(p)) : p I"].
S(p*) is a MM layout. Stop.
6) S(n) is a MM layout. Stop.
We remark that when all lengths are the same, say Li = k, i e N,
that the condition defining I', Li > L CLk, becomes > (n k) ,
which is impossible (we note that k < n-1 by Remark 1 below), so that
I' = 4, and S(n) is a MM layout.
We now illustrate Algorithm MM-2. The analysis needed to justify
the two algorithms is presented subsequently.
As a numerical example, we take n = 6 and suppose that fi(x) = wix
for 0 5 x < L, where wi > 0, i s N, so that each function fi(x), is linear
in its argument (a positive weight times the argument). Thus, we can
order the functions f. > f i+ 1 < i < 5, for 0 < x < L. Consider the
data {Li : 1 < i < 6} = (1, 6, 4, 1, 1, 11 and {wi : 1 < i < 6} = {11.5,
11, 10.25, 10, 2, 1}, which has the specified ordering. The necessary
s..t arre provided in proper order as follow.
Step 1): G(S(n)) = max {Gi(S(6)) : 1 < i < 6}
= max {86.25, 115.5, 128.125, 130, 27, 13.5} = 130.
Step 2): G(S(6)) = G (S(6)), k = 4.
Step 3): L CLk = 14 12 = 2, I' = {2, 3}.
Step 4): fk[(L + CLk)/2] = 130, f2(L L2/2) = 121 < 130,
f3(L L3/2) = 123 < 130, I" = {2, 3}.
Step 5): G(S(2)) = max {f2(L L2/2), Gl(S(6))}
= max {121, 86.25} = 121,
G(S(3)) = max {f3(L L3/2), G2(S(6)), G1(S(6))}
= max {123, 115.5, 86.25} = 123,
G(S(p*)) = G(S(2)).
S(2) is a MM layout where 2 nests about 6, i + 1 nests about i, 3 s i 5 5,
and 3 nests about 1.
Solving PMS
It is unfortunately the case that PMS is, except in certain special
cases, difficult. Given the symmetric layout S with activities nested
from inside to outside in the order 1, 2, ..., n, we note that, for
1 i n-1,
Fi(S) = fi[Di(S)] = fi[t0 + (t + ... + ti)] (7)
where t0 = L/2, th = Lh/2, 1 5 h < i. F(S) is the sum of (8) and (9)
below:
n-i
Sfi[to + (tI + .". + ti) (8)
i=1
fn[t0 + t 1 + ... + tn_)] (9)
For activity n fixed on the outside, the problem of finding a symmetric
layout to minimize (8) is seen to be equivalent to a one-machine, n 1
job sequencing problem with (7) being the cost incurred for job i. Kan
et al. [33] have established that a special case of this sequencing problem
is intractable in a well-defined sense in that it is in a list of problems
of equivalent difficulty due to Karp [34]; computational experience to
date indicates that all of the problems in this list are intractable.
There are sequencing algorithms due to Lawler [4i], and to Kan et al. [331,
which may be used to minimize (8). Unless we assume more problem structure,
the repeated use of one of these algorithms, with each possible activity
fixed in turn on the outside, is the best way we know to solve PMS.
If we assume PMS has more structure, however, we can obtain some
additional results. Suppose we define the functions h h2, ..., hn by
h.(x) = fi(x) fi+1(x), 1 i n-l
hn(x) = fn(x)
Given rather restrictive assumptions about the functions hl, ..., h ,
and the lengths L1, ..., Ln, we have the following result.
Property 1: Suppose the functions hl, h2, ..., hn are nondecreasing,
and that L, L ... L L Then for any layout S we have
2 n
F(S(n)) < F(S)
That is, S(n) solves ?;3.
We remark that if f'(x) and fj(x) exist V x E [L*, L**], then the
requirement that fk(x) fj(x) be a nondecreasing function of x,
x E [L*, L**], is satisfied if f'(x) > f'(x) V x E [L*, L**]. (Here a
k j
prime on a function is used to mean the usual first derivative.)
A few comments about Property 1 may be in order. It is noted that
the hypotheses of the property are quite restrictive; however, they could
well be satisfied in many cases (e.g., when fl corresponds to a high
voltage supply in a linear electrical circuit such as would be found in
a modern manufacturing plant; where the supply requires a small site in
the circuit relative to other components, and the associated voltage
drop occurring as a result of current traveling from the supply to the
other components increases drastically as a function of the distance the
current must travel so that f is very large and L is very small).
We remark that Lowe [46] has solved the special case of PMS where
f.(x) = w.x, i > 0, i s N. His solution procedure involves choosing some
activity, say p, for assignment to the outside and then numbering the
remaining activities so that wi/Li wi+i/Li+1. (This numbering scheme
is implied by the result of Smith [62] on a job sequencing problem.)
Then for some p*, p* e N, S(p*) solves this special case of PMS.
To illustrate Property 1, suppose all lengths are the same, that
fi(x) = wix, i e N, and that wl w2 ... > wn > 0. Then hi(x) =
(vi wi+) x > 0 for x 0, 1 i < n-1, and hn(x) = wnx > 0 for x > 0.
Thus the hypotheses of the property are satisfied and we conclude S(n)
solves PMS, a conclusion which agrees with the results of Lowe. This
result is intuitively appealing since it causes the activity with largest
weight to be placed in the middle, activity 2, with next largest weight,
to nest about activity 1; and so forth. Thus the solution obtained is
rather analogous to the alternating order solutions of Bergmans [4] and
Pratt [57]. Alternatively, if the functions fl' f are all identical,
1' n
and L. < L. i 1 i < n-1, then again S(n) solves PMS, giving the intui-
1 -1+1
tively appealing result, analogous to results in [54] and [55], that acti-
vities of larger length should surround activities of smaller length.
Property 1 is a consequence of Property 6. Both properties are
established in the subsequent analysis, which we now present.
Analyses for PMM and PMS
In this section, the analyses necessary to verify the solution
procedures for PMM and PMS given in the previous section will be presented.
We first prove the dominance result stated earlier, which applies to both
problems, and was proven in different fashion by Lowe [h6].
Dominance Results
We now establish that, in order to find an optimal layout for either
PMM or PMS, it suffices to consider only symmetric layouts. Some defini-
tions will facilitate the discussion.
Given any layouts S = {S1,...Sn } and S' = {S',...,S'}, we say that
S' dominates S if Di(S') < Di(S) V i e N. For any layout S = {Sl,...,
Sn), we denote by SI and Si the sets Si n [0, L/2] and Si n [L/2, L]
respectively, and refer to Si and St as the left and right parts of S.
respectively, for i E N.
Given any layout S = {S1,...,Sn ,
(a) We call S a binary layout if, for i E N, the number of empty elements
in the set consisting of the left and right parts of Si is either
zero or one, and each nonempty element is a closed interval.
(b) We call S a double interval layout if it is a binary layout with the
additional property that both Si and St are intervals of positive
length for i E N.
(c) We call S a nested layout if it is a double interval layout with the
additional property that V i, j E N, i # j, with either p = i and q = j,
or p = j and q = i, Sp is to the left of S., and Sp is to the right of
+
Sq.
ci)
(d) We call S'a symmetric layout if it is a nested layout with the
additional feature that, for i s N, S. and S. each have length Li/2.
We shall see that any layout is dominated by a binary layout, and
that for each of the layouts defined in (a), (b), and (c) above, there
exists a dominating layout as defined in (b), (c) and (d), respectively.
Thus it will follow that any layout is dominated by a symmetric layout.
Given any layout S = {S ,...,S, for i E N it is convenient to
I n
define Di(S) and D (S) by
D:(S) = max {d(S., S+) : j E N, j 0 i}
and
+ + -
D (S) = max {d(S S) : j E N, j # i}
with the understanding that Dy(S)(D.(S)) is minus infinity whenever S.
is empty (S. is empty) and/or S (S7) is empty for j e N, j # i. A
i j j
readily established consequence is that, for i E N,
D.(S) = max [D((S), DT(S)] (10)
1 1 1
Property 2: Any layout is dominated by a binary layout.
Proof: Given any layout S, w.o.-.o.g., we may assume for some i that
S. is the union of at least two disjoint intervals. Denote by Sil the
leftmost such interval and by Si2 the next leftmost such interval.
Denote by xl and x2 (x3 and x4) the left and right endpoints of
S il(S ) respectively. Since Sil and Si2 are disjoint, the interval
B E [x2, x ] has positive length. Let x' = x2 + (x x ), define
S!2 = [x2, x'], and B' = [x', x4]. Then construct the layout S' from S
by letting S!' = Si, shifting S. left to the interval S!2, and
I i1 i2 12
shifting activities in the interval B right to the interval B'. All
other intervals in S and S' are identical. (In Figure 1 we show layouts
S and S'.) We make the crucial observation that S' U S' = [x x2] U
[x2, x'] = [xl, x'] and is thus a single interval.
S. B S
Ssil I B I i2 .
0 x x2 x3 x4 L/2 L
S' U S' B'
Sil i2 I I
0 xI x' x L/2 L
1 3 3
Figure 1. Layouts Illustrating S and S' for Property 2
Let I(B) be the set of indices of activities taking up at least
one subinterval, of positive length, of B. As the portions of layouts
S and S' to the left of xl, and to the right of x remain identical,
for j i I(B), j j i, we have D-(S') = D-(S), and D+(S') = D+(S), so
J J J J
that (10) gives
D (S') = D (S), j i I(B), j # i (11)
J j
Since B' is obtained by shifting B to the right, D-(S') < D(S )
k k
for k e I(B). For k e I(B), D(S') = Dk(S), as both distances are
computed between points in the intervals [0, x l and [L/2, L], and S'
and S are identical in these two intervals. Thus
Dk(S') < Dk(S), k e I(B) .(12)
+ +
Now if x = 0 and if S. 4, Di(S) is the distance between the
rightmost point of S and x while D (S') is the distance between the
i 2 i
rightmost point of S!' = S and x, so that x2 < x gives Dt(S') < Dt(S).
1 1 3 3' 2 3 1 1
If x > 0, only activities other than i lie in the interval [O, x ], so
D (S) and DT(S') are identical, and equal the distance between the right-
most point of S1+ S! and 0. Thus, when S # 4 we conclude D (S') <
1 1 1 i
D1(S). When S+ = S + = D (S') = D (S) = Since x is the leftmost
point of both S and S we have D(S') (S), and so we conclude,
point of both S7 and S! we have D.(S') = D7(S), and so we conclude,
1 1 1 1
(13)
Di(S') < Di(S)
From (11), (12), and (13) we conclude S' dominates S.
Making repeated use of the above layout construction procedure in
the interval [0, L/2], and of the completely analogous construction
procedure in the interval [L/2, L], we construct a finite sequence of
layouts, each of which dominates the preceding layout, with the last
layout obtained by the procedure being a binary layout. []
Prior to establishing the dominance of any layout by a double inter-
val layout, we introduce the notion of a reflection, which is needed
subsequently. If x is in the interval [0, L/2] ([L/2, L]) and is a dis-
tance r from L/2, then the reflection of x, denoted by x*, is that point
in [L/2, L] ([0, L/2]) which is a distance r from L/2. For any two points
x, y in [0, L], it is readily verified that d(x, y) = d(x*, y*).
Property 3: Any layout is dominated by a double interval layout.
Proof: Due to Property 2, it is enough to show that any binary layout
is dominated by a double interval layout. Let S be any binary layout.
W.o./.o.g., we may assume that S = for some p, so that S has length
P p
L > 3. Since S is a binary layout, letting x and x denote the left
P L R
and right endpoints of S- gives S- = [xL, xR]. With [x x*] the
P P L R L
reflection of S there is an activity index q for which S has left
P q
and right hand endpoints y and y respectively and [x*, x*] nS =
L R R L q
[W w I c [L/2, L], where x* < w < w < x* and y < w < < y
L R R L R L L L R R
Thus we may choose an interval [z z ] of [w w R for which w <
L R L R L ~
z < zR < w so that
L R R
zR < min (wR, YR' x)
(1)
and
z > max (w y x*) (15)
L L L R
Now construct the layout S' from S as follows: let S' = S. if p # i # q,
i 1
and let S' = S'- U S'+ S' = S'~ U S'+, where
P P P q q q
S = [xL z] U [zR, xR] p S = [zL, zR]
S'- = S- U [z, z], and S' = [yL, zL] U [zR' yR'
q q L L q
Note, since S' differs from S only in that proper subintervals of S and
p
S have been interchanged, that D.(S') = D.(S) for i e N, p # i 0 q. It
remains to show D.(S') < D.(S) for i = p, q.
Since Sp = 4, D (S) = D~(S). Also, as xL is the leftmost point
both of Sp and S'~ we conclude Dp(S') = d(xL, L) = Dp(S). Since zR is
the rightmost point of S'+, D+(S') < d(z 0). Thus D (S') < d(0, zR) <
p p R- p R
d(0, x*) = d(xL, L) = D (S'), (where the strict inequality is due to (14)),
so D (S') = D~(S') = D (S).
p p p
P P P
We next establish D (S') = D (S). If (a), S~ = 4, then D (S) =
q q q q
D+(S) = (0, y ). Now, (b), suppose S~ # and S~ lies to the right of
q 'R q q
S-. Then D+(S) = d(0, yR) and D-(S) < d(xR, L) = d(x 0). (In Figure 2
P q R q R
we show layouts S and S' for (b) above.) As [x*, x*] n [y y ] 4>,
R L L R
x* YR so d(0, x*) < d(0, yR). Thus D-(S) < D+(S), so D (S) = D+(S)
R I q q q q
d(0, yR). For (a) and/or (b), the leftmost point of S'- is z* so D-(S') <
q R q -
d(z*, L), and D(S') = d(0, yR) as z* > 0. Thus, using (1), we have
R q R
DI(S') < d(z*, L) = d(0, zR) < d(0, yR) = D (S') = D (S'). Thus, for (a)
and for (b), we have D (S) = D+(S) = d(0, y ) = D (S').
q q q
S- S
p I I
0 x x L/2 y x* z z y x*
L R L R L R R L
S'- S' S S'- S'+ S' S'
.... I q I s I | P lS
0 xL
zR ZL
L/2
L
ZL ZR Y L
Figure 2. Layouts Illustrating S and S' for (b) in Property 3
We now consider the remaining case where S- lies to the left of S-.
q p
Letting v be the leftmost point of S- and noting that yR is the right-
q q R
most point of S+, it follows from the construction of S' that v is the
q q
leftmost point of S'- and yR is the rightmost point of S'+
CI R q
Since activities in the interval [O, z*] are the same for both S and
S', and since at least one activity other than q lies in the interval
[0, z*], we conclude
R\
I
i I
D+(S') = D+(S)
q q
We now consider two subcases: (a) yR # L, and (b) yR = L.
In (a), Dq(S') = d(v L) = D~(S), and thus (16) implies D (S') =
qQ q q
D (S).
q
In (b), D-(S') = d(v z) and D-(S) = d(v y), and since zR
q q R q q L R
is to the right of yL'
D-(S') > D-(S)
q q
Furthermore, since v > 0, we have
qi
d(0, zR) d(Vq zR) = D(S')
Now with yR = L and some activity other than q lying in the interval
[0, z*], we have
Rv
D+(S') R d(L, z)
Using (17), (18), and (19) we have
D (S') > d(L, z*) = d(O, z )
C1 R R
> D-(S') > D-(S)
- q q
(16)
(17)
(18)
(19)
and from (16) it now follows that
D (S') = D+(S') = D+(S) = D (S)
q q q q
If S' is not a binary layout, Property 2 implies there exists a
binary layout, say S", for which S" dominates S', and such that every
double interval activity of S' is also a double interval activity of S".
Thus the conclusion is established by the use, at most n times, of the
construction procedures of this and the preceding proof. []
Property 4: Any layout is dominated by a nested layout.
Proof: Due to Property 3, it suffices to show that any double interval
layout is dominated by a nested layout. Let S be any double interval
layout.
i) We know, for some activity indices p and q, that S- abuts 0 and S+
q P
abuts L. W.o.L.o.g. we may assume q = p, for if q c p we may construct,
as follows, a new double interval layout S' which dominates S, and is
such that S' abuts 0 and S'+ abuts L. Let a be the length of S-, and
denote by B the closed interval between S- and S-. Construct the layout
q p
S' from S by removing S~ from S, shifting both S~ and B to the right a
p C
distance a to constitute new intervals S'' and B' respectively, and letting
S1- = [0, a]. It is readily verified that D (S) = D (S) = L > max [D (S'),
D (S')], and that Dk(S') = Dk(S) for p # k 0 q.
ii) We may assume, for some p, that S- abuts 0 and S abuts L, so that
p p
activity p nests around the other activities. If'S is nested, the proof
is complete. If S is not nested, there exist distinct i and j such that
S. lies between S. and S while S~ lies between S~ and S- which, since
Sj P j p i
S+ abuts L, implies
p
D-(S) > D-(S) (20)
We choose either to make rearrangements in [0, L/2] if
D (S) > D (S) (21)
1 -
or to make rearrangements in [L/2, L] if DD(S) > D+(S). Due to symmetry
about the point L/2 we may assume, w.o...o.g., that (21) is true. Note
that (20) and (21) imply D+(S) > D-(S), so that
Di(S) = Di(S) (22)
Now, (21) and (22) give
D.(S) = max [D (S), D-(S)] .(23)
1 1 j
Now construct a new layout S' from S as follows. (In Figure 3 we
show layouts S and S'.) Let a be the length of S~, let x. be the left-
most point in ST, and denote by B the closed interval between S7 and ST.
J 1
Remove S7 from S, and shift both S7 and B to the right a distance of a,
resulting in new intervals denoted by S'- and B' respectively. Then let
S- = [x xj +a]. All remaining activities in S' are identical to those
in S.
ISP
I SI BI
I S+ I
,S+ I + I
IS Is +
0 x. a L/2 L
J
I "P "i IJ 1 lj "i I IP L
0 x. L/2 L
Figure 3.
Layouts Illustrating S and S' for Property 4.
Since x. is the leftmost endpoint both of S7 and S!, since S+
J j p
abuts L, and i # p # j, we conclude
D.(S) = d(xj, L) = D.(S')
J J 1
Thus (23) and (24) give
Di(S) = max [DT(S), D.(S')]
S1 1
Since Si = S+, (25) gives
I I'
D.(S) = max [D+(S'), D.(S')]-= D.(S')
1 1 1 1
(24)
(25)
(26)
Now for every k, p # k # i, Sk has either been shifted to the right
or remained unchanged in constructing S', so, also using the fact that
S+ = S'+ abuts L, we have
P P
Dk(S') < Dk(S),
i k p
(27)
Further, for every k, i #
that S- = S'~ abuts 0, we
P P
k # p, S'+ = S+ and so, also using the fact
k k
have
Dk(s') = Dk(S),
i k p
(28)
Now, (27) and (28) give
Dk(S') S Dn(S),
i # k p
(29)
Since S- = S'- and S+ = S'+, we also have
P P p P
D (S') = D (S)
p p
From (26), (29), and (30) we conclude S' dominates S. Also S' is a
double interval layout, and i nests about j. Thus, by repeated use of
the construction procedure of this proof, we obtain a nested layout
which dominates S.
(30)
Property 5: Any layout is dominated by a symmetric layout.
Proof: Due to Property 4, it suffices to show that any nested layout
is dominated by a symmetric layout. Let S be any nested layout with
activities numbered, w.o.L.o.g., so that S. nests around S. for
1 s i n-l. For 1 < i 5 n-l, define CLi H I {Lj :1 i j : i}, and
let u.(v.) be the distance between the left (right) endpoint of S.(St)
and L/2. We observe, for 1 < i < n-1, that ui + vi = CLi, Di(S) =
u. + L/2, D+(S) = v. + L/2, and D.(S) = max [D(S), D (S)] = L/2 +
1 1 1 1 1 1
max (u., v.). If max (u., v.) < CL./2, then u. + v. < CL., which cannot
1 1 1 1 1 1 i 1
be. Thus
D.(S) > (CL. + L)/2, 1 5 i 5 n-I (31)
1 1
Similarly, D-(S) = L/2 + v D(S) = L/2 + u and
n n-1 n n-1
Dn(S) > (CLnl + L)/2 .(32)
To complete the proof, let S* = {S*, ..., S*} be a symmetric layout
1 n
such that S* nests about S*, 1 < i < n-1, and S, S*+" each have length
i+1 1 1 1
Li/2, i e N. A direct computation establishes that
D.(S*) = (CL. + L)/2, 1 < i < n-i (33)
and
Dn(S*) = (CLn1 + L)/2
(34)
It thus follows from (31), (32), (33), and (34) that S* dominates S,
which completes the proof. []
Analysis for PMM
The analysis for the minimax algorithms is rather direct. The
justification of Algorithm MM-1, used when no function dominance exists,
is based entirely upon the fact that some symmetric layout is minimax,
and that, for each possible activity p fixed outside, Lawler's algorithm
finds a best symmetric nesting of the remaining activities.
Given the ordering of the functions, Algorithm MM-1 specializes
to Algorithm MM-2. We first note, given the function ordering, that we
may apply Algorithm MM-1 and obtain T(p) = S(p), p e N. Thus it suffices
in Algorithm M1-2 to consider only the layouts S(1), ..., S(n). We may
further reduce the number of layouts to be considered by identifying
conditions under which layouts will be no better than S(n), and thus may
be disregarded. We now state such conditions, and provide brief proofs.
Remark 1: Let k be the smallest activity index for which
Gk(S(n)) = G(S(n))
(a) k S n-l.
(b) If k I p S n, then
G(S(p)) > G(S(n))
Proof:
a) We know that D (S(n)) = D _(S(n)), so f < fn- implies G (S(n)) <
n n-l n n-1 n
Gn_l(S(n)), implying k < n-1.
b) If k 5 p 5 n, then the location of activity k is identical in the
layouts S(p) and S(n), so that Dk(S(p)) = Dk(S(n)), and thus
G(S(p)) > Gk(S(p)) = Gk(S(n)) = G(S(n))
Corollary: S(n) is a minimax layout if Gl(S(n)) = G(S(n)).
Remark 2: If 1 < p < k < n, where k is as defined in Remark 1, and if
L < L CL
P k
then
G(S(p)) > G(S(n))
Proof: Given the hypotheses, (35) is equivalent to
L L /2 > (L + CL )/2
P ~ k
so, using the fact that f 2 fk we conclude
P k
f [L L /2] f [(L + CLk)/2] fk[(L + CLk)/2]
(35)
Since
G(S(p)) > G (S(p)) = f [L L /2]
p p p
fk[(L + CLk)/2] = Gk(S(n)) = G(S(n))
the conclusion follows.
Remark 3: If 1 s p s n, then
G(S(p)) = max {f (L L /2), max [G.(S(n)) :
Proof: From the definition of S(p), activities 1,
identical locations in S(p) and S(n), so that
Gi(S(p)) = G (S(n)),
1 i p-l]}
..., p-1 share
1 i i < p-1
Also, since activity p is outside for the layout S(p), we have
D (S(p)) > D.(S(p)),
p 1
p+ 1 i < n
and thus with f > f., p + 1 < i < n, we have
p i1
Gp (S(p)) > Gi(S(p))
pT -
and conclude
G (S(p)) > max [Gi(S(p)) : p + 1 < i s n]
Gp(S(p)) = fp[L Lp/2]
G(S(p)) = max {max [G.(S(p)) : 1 5 i 5 p-l],
max [G.(S(p)) : p + 1 < i < n]}
(36) follows on using (37) and (38) in (39).
Corollary: If k is as defined in Remark 1, and there exists p such
that 1 L p < k, and if
f (L L /2) < fk(L Lk/2)
then
G(S(p)) < G(S(n))
Now since
(37)
and
(38)
G (S(p)),
p
(39)
It is now direct to verify that Remark 1 and its Corollary justify
Step 2) of Algorithm MM-2, Remarks 1 and 2 justify Step 3), and Remark 3
and its Corollary justify Steps 4) and 5). Step 1) is an initiation
step, and needs no justification, while Step 6) is a consequence of
Steps 2), 3), and 4).
Analysis for PMS
We now develop the analysis for the MS problem, PMS.
Property 6: Suppose f fk, and fk fj are nondecreasing functions,
and that Lk < L.. If S is any symmetric layout for which S is farther
from zero than is S-, then there exists a symmetric layout S' with S!-
k 3
closer to zero than S', such that
F(S') < F(S)
Proof: Denote by a the distance between the right endpoint of S- and
L, and by 8 the distance between the left endpoint of S and the right
endpoint of S-. Construct a symmetric layout S' from S as follows:
J
shift S7 and St away from L/2 by a distance B L./2 to obtain S'-
and S +; shift Sk and S+ towards L/2 a distance B Lk/2 to obtain
Sk~ and Sk ; shift activities lying between Sk and S-, and between Sk
and S towards L/2 a distance (Lj Lk)/2 0; leave all other activities
unchanged. Effectively, the locations of activities j and k have been
interchanged, with activities between j and k shifted to compensate.
Since each activity i between j and k has been shifted towards L/2,
we have
D(S ) < D(S )
while, for every other activity h, j 0 h # k,
D(S') = D(Sh)
h h
It suffices, due to (40) and (41), to establish that
F(S') + F(S!) < F(Sk) + F(S )
We now consider two cases,
(a) S- does not abut zero. We observe that
k
D(S.) = a + L./2,
J J
D(S') = a + 8, and
j
D(S ) = a + B
k
D(S') = a + L /2
k k
Since a + L./2 <
J
a + 6 and f -
k
f is nondecreasing,
j
fk(a + L /2) f (a + L./2) < fk(a + 8) f (a + s),
j- J j K
so that
f (a + L./2) + f.(a + 0) < f (a + S) + f.(a + L./2).
S J J kj J
(4o)
(41)
(42)
(43)
(44)
(45)
Since Lk < Lj and fk is nondecreasing, (45) implies
fk(a + L /2) + f (a + 8) < fk(a + B) + f (a + L /2)
Now, (42) follows from (43), (44), and (46).
(b) S- abuts zero. In this case we have
Ik
D(S ) = a + L /2,
3 3
D(S ) = a + 3 Lk/2
D(S') =a + a L /2, and D(S')
J 3 k
= a + Lk/2
Since a + L./2 < a + Z Lk/2, we have
J k
fk(a + L./2) f.(a + Lj/2) <
f (a + k L /2) f (a + a Lk/2)
k k j k
As L < L and f and f are nondecreasing, we have, using (49),
k- j j k
f (a + B L /2) + fk(a + Lk/2)
f ff(a + L/2) + fk( + L/2)
f.(a + L./2) + f (a + a L /2)
3 J k k
Using (47), (48), and (50), we obtain the result.
(46)
(47)
(48)
(49)
(50)
With the total cost for any layout S defined by
F(S) = {f.[D.(S)] : i N}
1 1
the repeated use of Property 6 readily establishes the following property.
Property 7: Suppose the functions f. f. is nondecreasing, for
1 1+1
1 S i 5 n-l, the function f is nondecreasing, and that L. < L+ i e N.
For any layout S, we have
F(S(n)) s F(S)
That is, S(n) is an optimum layout for PMS.
We remark, if the hypotheses of Property 7 are not all satisfied,
but some are satisfied, then one may employ those hypotheses which are
satisfied in conjunction with Property 6 to reduce the class of layouts
to be considered in seeking a layout to solve PMS.
Summary of Results
In this chapter, PMM and PMS for divisible activities were formulated
and analyzed. Efficient solution procedures for both problems were derived
and presented. The solution procedure given-for PMM, when applied, solves
the problem in all cases where the f. are nondecreasing functions (continu-
1
ous or otherwise). The solution algorithm for PMS provides a solution
when additional assumptions are made about the f. and/or L. relative to
1 1
each other; in other cases, suitable algorithms from the scheduling litera-
ture must be applied in order to solve the problem.
CHAPTER III
LAYOUT PROBLEMS ON THE LINE INVOLVING INDIVISIBLE ACTIVITIES
Introduction
In this chapter we deal with two problems involving the layout of n
classes of activities on the line, where n is a positive integer and
n > 3. As in the previous chapter, we consider two different objective
functions. One problem is of the IM form and entails the assignment of
the n activities to n nonoverlapping intervals, S ..., S so that all
intervals have the same given positive length (which, w.o.L.o.g., is
assumed to be unity) and such that the maximum of n associated functions
of the maximum distance between each corresponding set and any other set
is minimized. The second problem of interest is of the MS form and
differs from the first only in that the sum, rather than the maximum,
of the functions is to be minimized. The only differences between the
problems in this chapter and the two of the previous chapter are that now
(1) each activity i is indivisible in the sense that it is represented by
a single interval of the real line and (2) the required length of each
activity is equal to unity.
We first formulate each problem and then give a motivation for the
research. Next we give solution procedures, example problems, and con-
clusions for each, followed by analyses of the problems.
Problem Formulations
Define a layout on the line to be a collection S = {S1, S2, ..., S n,
where S is an interval of unit length for i E N {1, 2, ..., n) and S
i i
and Sj are nonoverlapping. In addition, we denote by L the sum of the
lengths required by all of the activities so that L = n. W.o.L.o.g., we
assume that the left hand endpoint of the layout is the point 0 on the
real line. As will become clear, the only layouts which need be considered
are those whose right hand endpoint is at the point L. Hence, a layout
S can be thought of as a partitioning of the- interval [0,L], consistent
with the above definitions, so that adjacent subsets share common end-
points. Equivalently, a layout may be considered a sequencing of n unit
intervals in the interval [O,L].
In order to define an objective function for every layout, S =
{Sl, S2, ..., S }, for the MM and the MS problems, we establish the
following definitions and notation. Let d(S., S.) denote the maximum
distance between any point in S. and any point in S., that is, for i,
2z j
j E N, i 2 j,
d(S., S.) E max {x. x. : x. E S, xj S} (1)
Let D.(S) denote the maximum distance from any point in Si to any point
in Sj, j # i, that is,
Di(S) E max {d(Si, S ) : j e N, j # i}
For i E N, fi is a given nondecreasing real-valued function with domain
(0,-), and
G.(S) E fi[Di(S)], i N (3)
(We remark that no continuity assumption is made about the fi). Also,
we define F.(S) by
1
F.(S) E fi[Di(S)], i E N, (4)
where fi is a nondecreasing function, not necessarily the same as in (3).
Using the above definitions, we can now state the 1M and MS problems.
Associated with the former problem is the objective function G(S) defined
by
G(S) E max [G.(S) : i N] (5)
For the MS problem, we define the objective function F(S) by
n
F(S) = Fi(S) (6)
i=l
For the MM (MS) problem, our interest is in minimizing Expression
5 (6) over all layouts S, that is, we wish to find a MM (VB) layout. These
two layout problems on the line will be referred to, respectively, as PMM
and PMS (Problem MM and MS).
Motivation for Research and Related Work
The motivation for studying the types of layout problems considered
here, as well as a discussion of related work, appears in the previous
chapter. In that chapter, we allowed the activities to be divisible,
so that the solutions obtained are valid only when it is possible to
assign each of the activities except one to two disjoint intervals. We
motivated the divisibility assumption there by reason of gaining insight
into layout problems in the plane. However, there are cases where the
problem on the line is of more interest and where it is desired (even
if not essential) to have each activity assigned to a connected subset.
Such a case would exist, for example, when distances within activities
are of interest as well as distances between different activities.
Physical constraints, such as storage records on magnetic computer tapes,
can require that the activities be indivisible. Our assumption that the
required lengths of all activities are equal is satisfied, for example,
for the case of assigning offices of the same size along a corridor of
a building. The primary motivation for this common length assumption is
that it allows solution of a problem which may well provide insight into,
and allow the future solution of, the more general and less tractable
problem (which is combinatoric in nature) where the lengths need not be
the same.
The portions of the discussion of the literature presented in the
previous chapters which most directly relate to the two problems with
indivisible activities considered here are those concerning: the quad-
ratic assignment problem (see Hanan and Kurtzberg [27], Lawler [43], [44],
and Pierce and Crowston [56]; the work by Karp and Held [35] on the module
placement problem (MPP); the special cases of the MPP considered by
Bergmans [4], Pratt [57], and Adolphson and Hu [1]; the sequencing prob-
lems studied by Elmaghraby [13], Horn [30], and Lawler [41], [42]; the
problems with interdistance constraints and costs studied by Chan and
Francis [8]; and, the layout problem on the-line solved by Lowe [46].
Solution Procedures, Example Problems, and Conclusions
Since the problems we consider here (unlike those of Chapter II) do
not depend upon knowing which activities are assigned to the outside, they
can be formulated as assignment problems (xij = 1 if activity i is in
position j, etc.). Most of the analysis to come involves identifying
objective function structure such that solutions obtained via ordering
procedures such as the one in Section 6, Chapter 6, or in (homework)
Problem 9.26 of Chapter 9 of [24] give optimal answers. We note that
assignment algorithms can be used to solve PMM and PMS even when no
orderings of the functions f. (like we later assume) exist.
Before providing the solution procedures for PMM and PMS, we intro-
duce needed notation and definitions. If we are given an assignment of
intervals and a permutation of the activity numbers, say [1], [2], ...,
[n], such that activities are assigned from left to right in [0,L] in
the order
([n], [n-2], ..., [3], [1], [2], ..., [n-3], [n-l])
if n is odd, or in the order
([n-1], [n-3], ..., [3 [1] ], [41 ], ..., [n-2], [n])
if n is even, then we call the assignment an alternating label assignment
(ALA). We call the interval [j] in an ALA position J. We note that an
ALA is merely a labeling convention which specifies which unit interval a
given activity is assigned, so that any layout can have ALA if the activi-
ties are numbered consistent with the above definition. The property
below follows from (2).
Property 1: If a layout S has ALA, then for n odd,
D[i_](S) = D[i](S) = n+ i odd, i 3, i c N,
2
where DI(S) = n+
2
or for n even,
Di_(S) = D[i](S) = n+i, i even, i e N
For a layout which has ALA, if n and i are both odd (even) we say
that intervals i and i-1, 2 < i E n, are mirror images of each other.
Given a layout S which has ALA, we refer to any layout S' as a reflected
alternating label assignment (RALA) for S if S' can be constructed from
S by (1) interchanging the relative assignment of two activities [i] and
[i-l], n and i odd (even), to respective mirror images or (2) making more
59
than one such interchange. The following property follows from Definitions
(3) (5) and Property 1.
Property 2: If layout S has ALA and S' is any RALA for S, then relative
to PMM and PMS, S' and S are equivalent, that is, G(S') = G(S) and
F(S') = F(S).
Let
n+1
n- if n is odd
2
L*
n+2
2 if n is even
It follows from (2) that for a layout in the interval [O,L],
min {D.(S) : i e N} = L* and
1
max {D.(S) : i e N} = L
1
As a notational convenience, we write f > f. to mean f.(x) > f.(x)
V x e [L*,L]. Also, "fi f. is a nondecreasing function" should be
taken to mean that fi(x) f.(x) is a nondecreasing function of x, V x e
[L*,L].
Solving PMM
The solution procedures given below for PIMM entail (to some extent)
assigning the activity with the "largest" f. to the center and then
1
repetitively placing the activity with the "next largest" f. alternating
1
between the position immediately to the right and immediately to the left
of the position of the activity just placed. More specifically, one
finds an activity which, when placed at one end of [0,L], gives the lowest
cost and assigns it to the end; then determines the activity which, when
placed at the other end of [0,L], yields the_ next smallest cost and assigns
it there. Next, the length of [0,L] is reduced by 2 to give [1, L-l],
the algorithm finds the activity (from those remaining) which, when placed
at one end of [1, L-l] gives the smallest cost and assigns it there; then
determines the activity (from those remaining) which, when placed at the
other end of [1, L-l], yields the next smallest cost and assigns it there;
and, repeats the procedure, each time decreasing the length of the interval
by two, until all activities are assigned.
We now state Algorithm 1M-1, which solves PMM. We note that any
ties which occur in using the algorithm may be arbitrarily broken.
Algorithm MM-1.
1) Arbitrarily label the activities with the indices 1, ..., n. Let
k=n, y=n, and C = {i : i N}.
2) Determine F[k] = min {f.(y) : i c C} and denote the number of an
activity i e C achieving the minimum with-thne index [k]. Delete [k] from C.
3) Determine F[k_] = min {f.(y) : i e C} and denote the number of an
activity i c C achieving the minimum with the index [k-l]. Delete [k-l]
from C.
4) Let k = k-2. If k>2, replace y by y-1 and go to Step 2; otherwise,
go to Step 5 if n is odd or to Step 6 if n is even.
5) Denote the one remaining activity index in C by [1] and go to Step 7
with F[] 1]
6) Denote the two remaining activity indices in C by [11 and [2], and
go to Step 7 with F f (2) andF = f (n+2
[1] [1] 2 [21 [2] 2
7) The resulting ALA specifies a layout S with objective function value
G(S) = max {F i](S) : i c N}. Layout S (and any RALA for S) is a MM layout.
We now illustrate Algorithm MM-1 graphically. Consider the case of
n=6 activities with associated functions fi(y) as shown in Figure 4.
f5 5 = [1]
f 6 = [2]
f2 2 = [3]
----- 4 = [4]
Sf 1 = [51
I I
Sf3 3 = [6]
I I
0I I
0 1 2 3 4 5 6
Figure 4. Graphical Example of Algorithm MM-1.
We see with C = l, 2, ..., 6}:
f3(6) = min {f.(6):ieC} so that activity 3 is numbered [6] and removed from C,
f (6) = min {f (6):iEC} so that activity 1 is numbered [5] and removed from C,
f4(5) = min {f (5):icC} so that activity 4 is numbered [4] and removed from C,
and
f 2(5) = min {f (5):ieC} so that activity 2 is numbered [3] and removed from C,
which leaves only activities 5 and 6, which are numbered [l] and [2],
respectively, so that the solution is given by assigning the activities
from left to right in [0,6] in the order (1,2,5,6,4,3).
We note that Algorithm MM-1 is like the repeated use of Lawler's [57]
algorithm for finding a sequence of jobs to be performed in a multitask,
one machine sequencing problem which will minimize the maximum completion
cost. When we may number the activities such that f > f 1 < i < n-1,
i i+1
Algorithm MM-1 simplifies to Algorithm MM-2 below.
Algorithm MM-2. Number the activities such that f > f 1 < i < n-l.
A MM layout is given by the corresponding ALA.
We now illustrate Algorithm M4-1. Let n=7 and consider the following f.:
Activity i f
Activity i fi(x)
(We note that with these fi the less complex procedure, Algorithm iM-2,
could be used.)
63
The necessary steps are illustrated in the following table.
Table 1. Steps for Example of Algorithm MM-1
C
{1,2,3,4,5,6,7}
{1,2,3,4,5,7}
{1,2,3,4,7}
{1,2,3,4}
{1,3,4}
{1,3}
{l}
F[k]
f6(7) =
f5(7) =
f7(6) =
f2(6) =
f4 (5) =
f3(5) =
fl() =
Activity is numbered with index
6 [7]
5 [6]
7 [5]
2 [4]
4 [3]
3 [2]
1 [1]
A MM layout is given by assigning the activities from left to right in
[0,7] in the order (6,7,4,1,3,2,5) and has objective function value
G(S) = 35.
Solving PMS
When we may assume that the functions f. are such that there exists
1
a numbering of activities for which f f is a nondecreasing function,
i i+l
1 i 5 n-1, Algorithm MS below solves PMS. We note that when each f. is
linear, nondecreasing, and has the same y-intercept, there always exists
such an ordering; and, when the functions are all differentiable, the
numbering is equivalent to a numbering such that f' > f' 1 < i < n-l,
i i+1
and f' > 0 (where the prime signifies the usual derivative).
n
Algorithm S0. Number the activities such that f. fi+ is nonde-
creasing, 1 5 i 5 n-1. A MS layout is given by the corresponding ALA.
As a numerical example of Algorithm MS, consider the data of the
previous example. We see that fl(x) f (x) = x, f3(x) fq(x) = x,
f4(x) f2(x) = x, f2(x) f7(x) = x, f (x) f5(x) = 2x, and f5(x) -
f6(x) = x are nondecreasing. Therefore, if we number the activities
(with the indices [i], 1 < i 7) as we did in the preceding example,
the resultant ALA specifies a MS layout (with objective function value
F(S) = 32+35+30+30+24+14+7 = 172); that is, in terms of the original
activity indices, the left-to-right assignment (6,7,4,1,3,2,5) is a MS
layout. We present in subsequent sections the analysis justifying the
algorithms presented here for solving PMM and PMS.
Centrally Dominating Objective functions
In this section, we introduce the concept of a centrally dominating
objective function (CDOF), relate the concept to PMM and PMS, and provide
certain properties of a CDOF. We give proofs of these properties in a
subsequent section on CDOF analysis. Some definitions and notation will
facilitate the discussion.
Definitions and Notation
Given the layouts S and S', we define
6 i Di(S), i e N
6 E D.(S'), i e N
1 1
We say that activity i is more central than activity j in S(S') or j is
more extreme than i in S(S') if 6. < 6. (6! < 61). We use the notation
Si '<' Sj (Si '>' Sj) to mean i is more central (extreme) than j and
Si '<' Sj (S '>' Sj) to mean i is at least as central (extreme) as j.
We note that if S. '<' S. and S '<' Sk, then S '<' Sk, so that the
relationship '<' is transitive.
Given a layout S which has ALA, if Si is in position i V i E N, we
call S an alternating order (AO) layout and say that S has AO.
Remark 1: If S is an AO layout,
(i) and if n is odd, then
61 < 62 = 63 < 64 = 65 < "' < 6n-3 = 6n-2 < 6n- = 6n
(ii) while, if n is even,
1 62 < 63 = 64 < ... < 6n-3 = 6n-2 < 6n-1 = n
We define the difference functions gjk by
gjk = f fk; j, k e N, j # k
A differentially centrally dominating (DCD) layout S is defined to be an
AO layout such that gi i+l is nondecreasing, 1 < i < n-l.
Remark 2: We assume for PMS that the functions f. are such that a DCD layout
exists, and we number the activities such that is nondecresing for j < k.
exists, and we number the activities such that gjk is nondecreasing for j < k.
Remark 3: Algorithm MS yields a DCD layout.
If an objective function C(') is defined for every layout in [0,L]
and a numbering of the activities defining AO is given, we say that C(')
is a centrally dominating objective function (CDOF) if it has the property
that, whenever a layout S is given such that j < k but S. '>' Sk, it is
j k
true that C(S') < C(S), where S' is the layout obtained from S by inter-
changing the locations of activities j and k. The motivation for the
definition comes from the fact that by making activity j at least as central
as activity k we get a layout S' which dominates S. (Roughly speaking,
it is desirable for activities with smaller numbers to be more central than
activities with larger numbers.)
We will now present a procedure for constructing an AO layout. The
procedure will prove useful in our subsequent analysis. Given a numbering
of the activities defining AO, the procedure is as given below.
Alternating Order Interchange Procedure (AOIP)
1) Begin with a layout S, define S(0) = S, and let i = 1.
2) Denote the number of the activity in position i of S(i-l) by [i].
3) If [i] = i, define S(i) = S(i-l) and go to Step 5; otherwise, go to
Step 4.
4) Construct S(i) from S(i-l) by interchanging the positions of activities
i and [i] and go to Step 5.
5) If i < n, replace i by i+l and go to Step 2; otherwise, go to Step 6.
6) The layout S(n) is an AO layout.
We now illustrate the AOIP. Consider the layout S with n=6 specified
by the left-to-right ordering S = (Sl, S2, S S S S6). The necessary
steps are illustrated in the following table.
Table 2. Steps for Example of the AOIP
i [i] S(i)
0 (Sl, 2, s3 s4 S5, s6
1 3(#1) (s3, S2, Sl S, S, S5, S6)
2 4(#2) (S3, S4, S1, S2, S5, Sg)
3 4(W3) (S,4 S3, S1, 82, 5, S6)
4 5(#4) (s5, s3, S1, s2, s4, S6)
5 5 (S5, 3S S s2, S4, S6)
6 6 (s5, s s', s2, s S6)
Properties of a CDOF
We now give properties of a CDOF. Also, we relate the idea of a CDOF
to the objective functions of PIMM and PMS, as well as to other objective
functions.
Property 3: Given the numbering of activities defining an AO layout, the
AO layout is an optimum layout for any layout problem having a CDOF.
Property 4: Given the numbering of activities such that fi fi+l,
1 < i 5 n-1, PMM has a CDOF.
Remark 4: Properties 3 and 4 justify Algorithm -M-2 for solving PMM.
Property 5: PMS has a CDOF for DCD layouts.
Remark 5: As a consequence of the definition of a DCD layout, Remark 3,
and Properties 3 and 5, we have that Algorithm MS yields an AO layout
for PMS and provides a MS solution.
Another example of a CDOF is
n
F(S) = I f((6i)
i=l
where fl,' "' fn are positive-valued functions with the property that
the function rjk defined as the ratio of fj to fk is nondecreasing V j,
k e N, j < k. To check that F is a CDOF, we let S be a layout such that
j < k but 6 > 6k, construct S' by interchanging the locations of activi-
j K' *J ----- -
ties j and k, and note that F(S') < F(S) is equivalent to fj(6kk)fk(6)
fj(6 )fk (k) or
f6(6k) f (6.)
rjk(6k) ) k) fk(j) rjk (6j)
jk k fkW fk(6j) gk )
which we know to be true since Sk S 6j and rjk is a nondecreasing function.
The three example CDOF's (for PMM, for PMS, and the product F) are
all special cases of one which may be constructed as the composition of
two functions:
a(Y) : E +- E
b(X) = [b(x), ..., b (X)] : En E
CCX) E a[b(X)]n
For PI, a(7) = max [y1, ..., Yn], for PMS a(Y) = yl + + yn, and for
the third example a(Y) = ylY2 "'Yn. In each example bi(xi) = fi(xi) V
i c N. In this notation, the function C is a CDOF if, whenever j < k
but xj a Xk, we have C(X') < C(X), where X' is obtained by interchanging
entries j and k of X.
It should be possible to use the foregoing observations to develop
computable conditions which will tell one when a function is a CDOF. A
clue is that in each case we considered the function a(*) has the property
that any sequence of interchanges of components of the vector Y, once Y
is given, does not change the value of a(Y). Also, a(-) is nondecreasing
in each of its arguments.
Let us suppose a(') is nondecreasing in each of its arguments and
that each bi is also nondecreasing. If we have j < k but xj > xk, then
reducing xj to xk causes some decrease, from C(X) to say C(X), d = C(X) -
C(X). Increasing xk to x. causes some increase, from C(X) to C(X'),
e = C(X') C(X). We then have C(X') < C(X) iff e < d, that is, the
increase incurred from increasing xk must be offset by the decrease due
to decreasing xj if C is a CDOF.
We remark that a sum or maximum of CDOF's is a CDOF and a product of
nonnegative CDOF's is a CDOF. This information may be of value in the same
way that it is useful to know that a sum or maximum of convex functions is
convex. For example, if an objective function is made up of a number of
functions, it might be analytically difficult to establish that it is a
CDOF directly; but, it might be easy to establish that the individual
functions which make up the overall objective function are each a CDOF
(in which case the overall objective function would be a CDOF under mild
assumptions).
Analysis for PSMM
In this section, we present analysis for PMM which will justify the
solution procedures provided earlier. A centrally dominating (CD) layout
S is one such that whenever 6k > 6j, then fk() < f j(6k). That is, if
Sk is more extreme than S., then fk is at least as small as f. at 6k'
as illustrated in Figure 5.
fj (6k ) -i /fj
Figure 5. Illustration of CD
Remark 6: Algorithms 1MM-1 and MM1-2 yield CD layouts.
We now give a property which will be used to prove that every CD
layout is a I4 layout.
Property 6: Suppose S' '>' S so that there exists p, p 0 k, such that
k k
S' = S Suppose S' '>' S It follows that there exists j, k j p,
k p P P
such that S. '>' S and S' '>' S'.
j p k
Proof: Figure 6 aids in the development. Let 12 be the interval in [0,L]
to the right of S', let I be the reflection of S' U I about L/2, let
I = I U I2, and let J = [0,L] (I U S'). Note I # 4. Now, S = S
and S' f I implies
S 0 1
p
P
Since S' '>' S = S' and p # k,
p p k
S' cI
P
Layout S:
L-6
L-6 +1
p
k
k
L/2
Layout S':
S P I Sil
6
k
Figure 6. Layouts S and S' for the Proof of Property 3
Since S, '>' S and SV' / I,
S, 4 I (95
Further, 3S '>' S i=,lies
k k
Sk cJ 1 (10)
By (7) and (9), for every i such that S c I, p # i k. Further, due
to (7) anc (), n't every i such that S. c I is such that S! c I. There-
I
fore, ther-e exis-a j, j p, such thtt S. c I, but S! I. Tow,
J J
S. c I implies
S. ' S' S
j k P
Also, J y o(c), S i 2, j h ipily *3. c J, so
S, '>' S []
The fol.lcoI- n .e .a E.,:d its proof d. Con stra e tht ;every CD .layoui
is a MI l-.Zyut. *The roof of the i:'ia u:;il-ze: the ide2. that if e-
assumie that ther- exist; soim CD. layout -fhich is not MM, yre obtain a
contrai.ct .io. 4 h ba. :i, -ea-7 cf ::he .:oof are as follow. If ,. are
given e. M-I layout, S, and j suprs that there e-:ists CD layout, S',
which is no. MM, thL;n, ther. mist '-.st (a) so-e k: for khic f, (6:) >
s ~ ~ ~ ~ ~ ~ i 2s(a mekfc-- i l
fi(6i), i E N, and (b) some j 4 k so that S '>' S' S'S', which is shown
11 j k j
to imply (since S' is CD) that f (6') < fj(6S), which contradicts (a).
k k < j i
Lemma 1: Every CD layout is a MM layout.
Proof: Let S be a MM layout, let S' be a CD layout, and suppose S' is
not MM. Then there exists k such that
G(S') = fk > fi(i), i E N
(11)
which implies
(12)
fk(6k > 6k
Since f is nondecreasing, (12) implies
k
S' '>' S
k k
which implies there exists p, p k, such that
S' = S
k p
Now, (11) and (14) imply
f (6') > f (6 ) = f (6')
kk pp p k
(13)
(14)
(15)
If 6' < 6' S' CD implies
p k
f (it) > f (.')
contradicting (15); therefore,
S' '>' S'
p k
Due to (13), (14), (16), and p k, the hypotheses of Property 6 are
satisfied; hence there exists j, k f j 0 p, such that
6. > i
J k
6 > 6
Since f. is nondecreasing, (17) implies
fj(6') < f (6)
Since S' is CD, (18) implies
fk(6k) fj(6)
(16)
(17)
(18)
(19)
(20)
Now, (19) and (20) give
fk(6k) f(6)
which contradicts (11). Therefore, S' is MM. []
Remark 7: Lemma 1 and Remark 6 justify Algorithms MM-1 and vM-2 for
solving PMM. (Remark 4 previously justified MM-2 using the concept of
a CDOF.)
This completes the analysis of PMM. We now provide analysis which
relates to the concept of a CDOF.
CDOF Analysis
In this section we provide the analysis (including proofs of previously
stated properties that are restated here for the sake of continuity) which
justifies the discussion of CDOF's presented in an earlier section.
Remark 8: For the AOIP, at the beginning of iteration i, with 6i and
6[i] the distance measures in S(i-l) for activities i and [i] respectively,
either
[i] = i and 6[i] = i
or
[i] > i and 6[i] i i
Proof: At the beginning of iteration i, activities 1, ..., i-i are in
positions 1, ..., i-i respectively, while activity [i] is in position i,
so that [i] > i. If [i] = i then obviously 6 = 6.. If [i] 7 i, then
[i] > i. Also, [i] # i implies activity i is not in position i, and we
know it is not in any of the positions 1, ..., i-i. Thus the smallest 6
1
can be is the distance measure of position i, which is 6[i]- [1
Lemma 2: If C is a CDOF, then the layout S(i) constructed at the end of
iteration i of the AOIP is such that
C[S(i)] < CIS(i-1)] V i E N
Proof: Due to Remark 8, either [i] = i, 6i = 6., and S(i) = S(i-l),
[i] i
in which case the claim is trivially true; or, [i] > i and 6 < 6 .
[i] i
In the latter case we construct S(i) from S(i-l) by interchanging the
positions of activities [i] and i. We know Si '>' S[i] and i < [i] so
that the assumption that C is a CDOF implies the desired result. []
Property 3: Given the numbering of activities defining an AO layout,
the AO layout is an optimum layout for any layout problem having a CDOF.
Proof: Given any layout S, apply the AOIP to it. By virtue of Lemma 2,
C[S(i)] < C[S(i-l)] V i N N
Thus, since S = S(0) and S(n) is the AO layout, we have C[S(n)] < C[S]. []
Property 4: Given the numbering of activities such that fi a fi+1'
1 5 i < n-l, PMM has a CDOF.
Proof: Let S be a layout such that j < k but 6 > 6 k Construct S' by
interchanging the locations of activities j and k, so that 6'. = and
J k
6' = S.. If we define
k
Kjk max [fi(6i) : i s N, j j i k]
we have from (5) that
G(S) = max {Kjk, max [f (6 ), f (6 )]}
jk 3 j k k
G(S') = max {Kjk, max [f (6 ), f (6 )}
Jkj k. k j
Hence, to establish the property, it suffices to show that
max [f.(6 ), f (6.)] ; max [f.(6.), f (D )]
j k k j j k k
Since k 6.j and fk is nondecreasing,
fk(6Wk) fk(j)
Since j < k, f fk so that
3 i
fk(6 ) < f (6j)
(21)
(22)
(23)
Now, (22) and (23) give
max [f (6 ), fk(6k) = fj(6)
Since 6k j 6j and fj is nondecreasing,
fj(Sk) < fj(Sj)
Hence, (23) and (25) give
max [fj.(k), f, (6)] < fj(6.)
3 k K -
The use of (24) and (26) gives (21).
Property 5: PMS has a CDOF for DCD layouts.
Proof: Let S be a layout such that j < k but 6 2 6k, and construct S'
by interchanging the locations of activities j and k, so that 61 = 6
J k
and V = 6j. From (6) we have
F(S) F(S') = fj(6 ) + f (6 ) fj(6) f (6')
= fj(6j) + f (6 ) f (6 ) f (6j)
J 3 k k j k k j
S[f (6j) f (6)J [f.(6 ) f (6 )]
jk kj j k k k(6
= g (6.) g (6 )
.jk j jk k
(24)
(25)
(26)
[]1
As > 6k and gjk is nondecreasing (by Remark 2), gjk(6j) > gjk6k)
and thus F(S) F(S') > 0, which established the property. []
This completes the analysis for CDOF's. We remark that the necessary
analysis for PMS was included.
Summary of the Chapter
In this chapter, we formulated and analyzed PMM and PMS for indi-
visible activities of the same length. We derived and presented efficient
solution procedures for both problems. The solution procedure we gave
for PMM, when applied, solves the problem in all cases where the f.. are
1
nondecreasing functions (continuous or otherwise). The solution algorithm
for PMS provides a solution when the additional assumption is made that
the functions f. are such that there exists a numbering of activities
for which g.i = f. f. is a nondecreasing function, 1 < i < n-l.
ii+1 1 1i+l
CHAPTER IV
MINIMAX AND MINISUM LAYOUT PROBLEMS IN THE PLANE
Introduction
This chapter deals with two problems involving the layout of n
classes of activities in the plane, where n is an integer greater than
one. We consider two different objective functions. One problem is of
the MM form and entails the assignment of the n activities to n nonempty,
nonoverlapping, compact sets, say Sl, ..., Sn, so that each set Si has
a given positive measure, or area, Ai, i E N E {1, 2, ..., n}, so that
the union of the n planar sets form a rectangle in the plane, and such
that the maximum of n associated functions of the maximum rectilinear
distance between each corresponding set and any other set is minimized.
The second problem of interest is of the MS form and differs from the first
only in that the sum, rather than the maximum, of the functions is to be
minimized. These two problems in the plane are analogous to the two
linear layout problems treated in Chapter II.
We first formulate each problem and present an equivalent problem,
discuss related work, and provide a motivation for the research. Then,
we present solution procedures, example problems, and conclusions for
each, followed by analyses of both problems and a summary of the chapter.
Problem Formulations
To be able to state the problems of interest, we define a planar
layout to be a collection S = {Sl, S2, ..., Sn where Si is any compact
planar set of area A. (a known positive constant) for i e N and S. and
S. are nonoverlapping in the sense that the area of S. n S. is zero for
J 1 J
i, j E N, i i J. We define A {A. : i e N}, which is simply the total
1
area requirement of the n activities. We also define US U {Si : i E N},
which is the union of all subsets of the plane assigned to the n activities
under layout S.
We now define several types of planar layouts. A rectangular layout
is a layout S in the plane such that US is a rectangle of area A. In the
present chapter, we consider only planar layouts that are rectangular
layouts; and, unless otherwise noted, any mention of layouts should be
taken henceforth to mean rectangular layouts. An ordered rectangular
layout is a rectangular layout for which there exists an ordering of
1, ..., n, say [11, ..., [n], such that U {S[i] : 1 i j j} is a rec-
tangle of area {A[i] : 1 < i < j} for j c N. Some examples of ordered
rectangular layouts are depicted in Figure 7, where V i E N, S[i] is
either (1) a connected set or (2) Si] = Si U S2 with Ski
k e {1, 21, a connected set.
A doubly rectangular (DR) layout is an ordered rectangular layout
{Sl, ..., Sn} such that Si is the union of a (finite) collection of rec-
tangles, all having the same orientation. We remark that the layouts
shown in Figure 7 are DR. A class of DR layouts called "concentric
square layouts" will now be defined.
Example Ordered Rectangular Layouts
S[31
S1
SI
[2]
S[l]
S2
[2]
I.
SS!
18]
S [4]
s
s[6]
sS7]___
________ s_51________
_s___
Figure 7.
A concentric square (CS) layout is a DR layout made
squares. A CS layout for n = 5 is depicted in Figure 8.
for CS layouts, the labeling 1, ..., n of the sets S. is
up of concentric
We note that,
from the center
outward and such an ordering indeed satisfies the criterion for being
a DR layout.
Figure 8. A CS Layout for n = 5
__
A few remarks about our defining the class of feasible layouts to
include only rectangular layouts seem to be in order at this time.
Because of standard construction practices, most buildings and rooms
have a rectangular shape. Additionally, such things as city blocks,
plots of land, and shopping centers are generally in the form of rec-
tangles. For these reasons, it appears that, rather than being a
restriction of the problem to a less general formulation, our allowance
of only rectangular layouts serves to make the problems treated here less
idealized and more general relative to many actual applications.
In order to define an objective function for every layout, S = {S,1
..., Sn, for the MM and MS problems, we establish the following defini-
tions and notation. For any points X = (xi, Yi) and X = (xj, yj) in
the plane, let
r(X., X.) E x. x. + y. y
1 J 1 J 1 3
denote the rectilinear distance between Xi and Xj. Such a distance is
often used when Xi and Xj are points on a rectilinear network of aisles
(or streets) and the travel distance between X. and X. along the aisle
network is of interest. The convention is adopted that the x y axes
specify the rectilinear aisle network. Let P(S., S.) denote the maximum
rectilinear distance between any point in S. and any point in S., that
is, for i, j e N, i $ j,
P(Si, S ) = max {r(Xi, X Xi Si X S} .
3- *- J^ L 3 <3
Let Ri(S) denote the maximum rectilinear distance from any point in S.
i
to any point in S. where j / i, that is,
J
R.(S) max {P(Si, Sj) : j E N, j # i} (2)
For i E N, fi is a given nondecreasing real-valued function (not necessarily
continuous) with domain (0,c), and
FiCS) E fi[Ri(S)] (3)
Using the above definitions, we now state the MM and MS planar layout
problems of interest. Associated with the former problem is the objective
function F (S) defined by
F (S) = max {Fi(S) : i E N} (4)
For the MS problem, we define the objective function FMS(S) by
n
FMS(S) [ F (S) (5)
i=l
For the MM(MS) problem, our interest is in minimizing Expression
4(5) over all DR layouts S, that is, we wish to find a DR layout S* for
which the objective function value for S* is less than or equal to that
for any other DR layout S; namely, we desire to find a MM(MS) DR layout.
I
These two layout problems in the plane will be referred to, respectively,
as PMM.2 and PMS.2 (problem minimax and minisum in two dimensions).
At this point we will construct problems in terms of the Tchebyshev
distance which are equivalent to, and more amenable to solution than,
PMM.2 and PMS.2. Let
T(X) = T(x, y) F (l/V)(x + y, -x + y) E X' and
T-1(X') T-1(x', y') = (1/C2)(x' y', x' + y') E X
The function T rotates the x y axes 45 degrees counter-clockwise to
provide axes denoted by x' y'. The function T-1 is the inverse function
of T, and rotates the x' y' axes 45 degrees clockwise. The term aligned
is used (without additional provisos) to refer to layouts or subsets
thereof, which are aligned with the x' y' axes; and, we will refer to
aligned CS (ACS) layouts.
For any two points Xi = (xi, yi) and X = (x., y ), we denote the
Tchebyshev distance between X. and Xj by t(X., X.), where
1 3 1 J
t(Xi, Xj) max { xi xj, yi yjl
We will make use of the fact (stated by Beckenback and Bellman [3] and
previously shown to be useful in studying planar location problems by
Dearing and Francis [12], Francis [15], Francis and Papineau [23], and
Papineau et al. [55]) that ---
r(X., X.) = t [T(X), T(Xj)] = v t (X!, X) (6)
1 J 1 1 j
Given any planar layouts S = {S *., Sn} and S' = (S, ..., S,
S' denotes the representation of S using the x' y' axes, while S denotes
the representation of S' using the x y axes. Thus, on replacing points
and sets in (1) through (5) by their representation using the x' y' axes,
and replacing the rectilinear distance by the Tchebyshev distance, we
obtain the following expressions, (7) (11), analogous to (1) (5).
Let Q(S!, S') denote the maximum Tchebyshev distance between any point
in S' and any point in S', that is, for i, j E N, i # j,
i j
Q(S, S'.) max {t(X!, X') : X C S: X' E S'} (7)
i j 1 j i i j j
Let D.(S') denote the maximum Tchebyshev distance from any point in S!
to any point in S' where j # i, that is,
D (S') max {Q(S', S') : j c N, j # i (8)
i 1 J
For i E N, define f. as before and let
1
G.(S') E f.[F2 D.(S')] (9)
1 1 1
Also, define the following two objective functions, G (S') and G M(S'),
which respectively correspond to MM and MS planar layout problems:
G M(S') max {G (S') : i N} (10)
and
n
GMS(S') G i(S') (11)
i=l
For the transformed M/I(MS) problem, our interest is in minimizing
Expression 10(11) over all DR layouts S'. We refer to these two trans-
formed problems as T.PMM and T.PMS, respectively.
Due to Relation (6), we have
P(Si, S.) = v' Q(S!, SI) (12)
from which it follows that
R.(S) = D D.(S') (13)
1 1
which yields
F.(S) = G.(S') (14)
1 1
Relation (14) now gives the following property, in which the notation
F(S) is used as a convenience to mean either F,,(S) or FMs(S), and G(S')
is used to mean the corresponding transformed objective function G MM(S')
or GMS(S').
Property 1: The following two statements are equivalent:
(a) S* is a layout and, for all layouts S,
F(S*) < F(S).
(b) S*' is a layout and, for all layouts S',
G(S*') < G(S').
Due to Property 1, in order to find layouts minimizing F M(S) or
FMs(S), it is necessary and sufficient to find layouts minimizing GMM(S')
or G (S'), respectively, and the latter two problems appear simpler to
analyze than the former two. Once a layout S' solving the transformed
problem (with the Tchebyshev distance) is obtained, then S, the repre-
sentation of S' using the x y axes, solves the original problem (with
the rectilinear distance). We are led immediately to the following remark.
Remark 1: PMM.2 is equivalent to T.PMM
and PMS.2 is equivalent to T.PMS.
As a result of the above discussion, we will deal almost exclusively
in the analysis to follow with the transformed problem corresponding to
the particular planar layout problem of interest at the time, and will
refer to this MM(MS) problem as PMM(PMS) for the remainder of the chapter.
Also, we will delete the use of the primed notation with sets, since all
sets will be understood to be represented using the x' y' axes.
Related Work
Newman [52] has posed, but not solved, PMM for the case n = 2,
fi[Ri(S)] = Ri(S) V i, and with no rectangularity restrictions on the
shape of the layout. Francis and Papineau [233 have solved this problem
for general n > 2. For n = 2, a I1 solution for their problem is an ACS
layout as defined above. For n > 2, an optimal layout may not be of that
form; however, in all cases, the union of all sets in the layout, US,
is either a square or the union of two rectangles, and is such that each
aisle in the rectilinear network makes a 45 degree angle with an edge of
the layout boundary (i.e., the square or each rectangle is aligned).
Papineau, Francis, and Bartholdi [55] have extended the results of Francis
and Papineau to include the consideration of distances within facilities
as well as distances between facilities. The remarks made concerning
the conclusions of [23] are also valid for those of [55]- The problems
considered by Francis and Papineau [23], and Papineau et al. [55]
suggest PMM.
Relative to PMS, Larson [39] considers a situation where a police
car patrols a region S, in the plane of known area A; and, the car is
equally likely to be at any point in S and must travel some rectilinear
distance to respond to an emergency which is also equally likely to
occur at any point in S. It is desired to find a configuration for S
so that the average distance traveled in responding to an emergency is
minimized. Larson does not solve this problem, but he obtains closed
form evaluations of the average distance for the cases where S is a
square, a diamond (square rotated 45 degrees from the usual Cartesian
axes, that is, aligned), and a circle. The values for the average
distances are approximately .667 FA, .660 /A, and .650 /A, respectively.
Karp, McKellar, and Wong [36] have recently solved this problem for the
Zp distance (see [24, p. 293]), which includes as special cases the
Euclidean and Tchebyshev, as well as the rectilinear, distances.
Another class of problems closely related to PMM and PMS is referred
to sometimes as lattice point problems. These problems involve finding a
layout of a given number of identical facilities so that either the maxi-
mum or the total rectilinear distance between facilities is minimized.
The potential locations for the facilities are lattice points in the plane,
which are defined as points whose coordinates are integers. Each lattice
point is the center of a unit square with its sides parallel to the x y
axes. The IM version of the problem has been studied by Francis [16],
who derives a virtually closed form solution and provides a method of con-
structing optimal layouts. Chan and Francis [7] consider the MS lattice
point problem and derive a set of geometric properties and necessary
conditions for an optimal layout; then, based on a symmetry assumption,
they develop an implicit enumeration procedure that will yield all optimal
layouts together with the least total distance between facilities. Related
to these lattice point problems are some packing problems and geometric
problems in number theory, a list of which can be found in [16].
Yue and Wong [66] have studied a lattice point problem which can be
viewed as a specialized version of the quadratic assignment problem (Q.A.P.).
Their problem involves allocating objects to the grid squares of a finite
grid, based on the knowledge of access probabilities only, such that the
expected Tchebyshev distance between consecutive accesses is minimized.
Yue and Wong give applications of their problem in contexts relating to
the storage of archive records and to optically addressable storage of
data; and, they develop a near-optimal heuristic procedure for solving
the problem. Karp, McKellar, and Wong [36] study the same problem except
they consider an infinite grid and consider the Euclidean and rectilinear
distances as well as the Tchebyshev distance. Karp et al. develop a simple
placement heuristic which uses only relative frequencies of access and
consists of placing the-most frequently accessed object in the center of
the grid and then filling "shells" of grid squares which are equidistant
from the center with a set of next most frequently accessed objects. By
analyzing the continuous analog of their Euclidean distance problem, they
were able to provide an argument that indicates their heuristic yields a
solution.which is within an additive constant of the optimal answer. They
point out that the additive constant is probably much too large for a
useful bound to be thus obtained. In addition, they indicate that no such
simple heuristic exists for the rectilinear and Tchebyshev metrics. In
the special case where all access probabilities are equal, heuristics
within an additive constant of the optimal do exist, but their implementa-
tion requires the numerical solution of differential equations. Karp et al.
also point out that, for the general case of unequal probabilities and
when the Tchebyshev or rectilinear distance is of interest, "the stumbling
block is in finding the optimal solution to the continuous problem" analo-
gous to their discrete problem.
The continuous analog of the problem considered by Karp et al. involves
finding two regions R0 and R1 with areas A and Al, respectively, and
R c R1 such that the integral / f d(X X ) is minimized, where
X1, RI XOcRO
X X0 denote points in R R respectively, and d(X X ) is the distance
1 0 1 0 1 0
between points X and X Using the techniques of Bergmans [4], Karp et al.
prove for the Euclidean metric that an optimal solution corresponds to the
situation where R0 and RI are concentric circles. For the special case
of equal access probabilities, for both the rectilinear and Tchebyshev
metrics, they indicate that the same techniques can be used to show symmetry
with respect to the x and y axes as well as to lines at 45 degrees and 135
degrees. We remark that ACS layouts possess these same qualities of
symmetry. (We note that the analysis of Karp et al. is based on an
assumption they never state: an optimum solution exists. An optimum
solution to their problem probably does exist; however, in light of the
experience of Newman [52], this assumption causes some concern.)
Motivation for Research
A prime motivation for studying planar layout problems is the location
of classes of machines, work places, or other activities in a job shop,
manufacturing plant, or office building. The function f. could correspond
1
to the relative importance of or amount of travel associated with activity
i. As usual, the overall goal dictated by the specific problem of interest
may be best accomplished by either a MM or a MS objective function.
One possible example of PMM could occur in the context of urban
design. Suppose that a city planner must arrange n activities (such as
industrial parks, shopping areas, housing areas, etc.) for a new town or
suburb of a city, where each activity has a space requirement A. and persons
at activity j may find it necessary to travel to any other activity i # j.
The function, f., may correspond to the relative importance of activity j
J
(e.g., shopping areas would probably be more of interest than recreational
parks). The value R.(S) corresponds to the maximum travel which would
be encountered by activity j with respect to layout S. It is desired
that the overall layout will be such that the maximum travel distance
from activity j to any other activity is as small as possible. The planner
uses the functions fi to reflect the relative importance of each activity
with respect to the overall layout process. He seeks a layout S* such
that FMM(S*) < FMM(S) for any other layout S. Once the planner assigns
locations for the activities, the individual activities may arrange their
components within their assigned areas in any manner desirable. A possible
example of PIS could be the planar analog of the example suggested by
Lowe [461 for a linear M1S problem (see the section of Chapter II on moti-
vation for research).
Solution procedures for the problems under present consideration may
provide benchmarks and/or serve as design aids for less idealized related
layout problems. Such solutions may serve as useful design guidelines
to be used as measures of merit in.the evaluation of more traditional
heuristic solution procedures.
From an applications point of view, PMS(PMM) is related to the Q.A.P.
(MM Q.A.P.). It is believed that solution procedures for the continuous
problems treated here may provide better insight into the solution of
such discrete problems for which no generally suitable algorithm exists
for solving the problems exactly. Further, it is felt that solutions for
PMM and PMS should subsequently serve as a firm basis upon which to build
a cumulative body of analytical techniques which could well lead to new
approaches to more general and less idealized layout problems.
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