R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
T. T ANIMOTO
Optimization by a method of maximum slope in the complex plane and its application to the transportation problem
Rendiconti del Seminario Matematico della Università di Padova, tome 48 (1972), p. 365-376
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T. TANIMOTO
*)
This paper is made
possible by
the ItalianConsiglio
Nazionale del- le Ricerche.For the sake of
completeness
we include some results from a pre- vious paper onregular graphs,
which has been submitted forpublication elsewhere,
on the basic ideas of the maximization of aspecial simple
linear functional
by
the method of maximumslope
for vectors in thecomplex plane.
The new part consists of considerations of
multiplicities
ofpoints
inthe
complex plane which,
when translated toapplications
tographical problems,
introducesmultiple
arcs. In thefollowing particular applica- tion,
themultiple
arcs are those of abipartite graph.
Maximization
by
the Method ofSteepest Slope.
Consider the
following problem:
Given m distinctpoints
zl , z2,z3 , ..., zm in the open sector D of the
complex plane consisting
of thefirst
quadrant, excluding
the real andimaginary
axes, to find h m ofthe
points
whose vector sum will have a maximumslope (or argument).
if and
only
if. Since all the elements are
*) Indirizzo dell’A.: University of Ma~ssachusetts at Boston, Visiting Profes-
sor, Seminario Matematico, Università di Padova, via Belzoni, 3, 35100 Padova.
positive,
it is easy toverify
thatCOROLLARY.
PROOF. Order the with respect to their
slopes
so that afterrelabelling
thepoints
so that
By
Theorem 1,Considering
ui+ u2 as asingle
term,again by
Theorem1,
The
proof
is immediate from Theorem 1. Thus thetriangular
lawholds so that we shall
call 11 z 11
aquasi-norm
of z, and hereafter weshall refer to it as
simply
the norm of z.LEMMA. A binomial can be of maximum norm if and
only
if itcontains as one term the element of maximum norm.
PROOF. As
before,
we assume thatand let
then, by
Theorem1,
or
Thus
so
that,
ingeneral,
we must haveConversely,
so
that 11 I
isequal
to the maximal-norm binomial above.By repeated applications
of thelemma,
we haveTHEOREM 2. A maximal-normed
h-termed, h m,
sum can befound
by sequentially maximizing
the norms of thepartial
sums ofpoints
in
D, starting
with the element of maximal norm.Thus Theorem 2 describes the solution to the
problem
of how tofind a subset of h m elements from the
given
collection zi , z2 , ..., zmwhose sum will have maximal
slope.
Suppose
as before that thepoints
zi , z2, ..., zm are distinct but that we now allowintegral multiplicities; i.e.,
zi can be counted up totimes,
z2 up to a2times,
etc. We now ask how one obtains a linear combination of some subset of h _ mpoints
which will have a maximumslope
when the sum of all the coefficientsactually
used is agiven
con-stant.
Here we must now determine two different sets of
quantities; viz.,
the subset of h mpoints
in thecomplex plane,
and the proper set ofcorresponding integral
coefficientssatisfying
the constraints that goalong
with them.Note that when the
points
in D areweighted,
the variation of theweights
distributes the norm of the linear combination from one extre-me to the other: e.g., if
and the relative
weights
ci>0 andc2 > o;
whenvaried,
vary the meanvalue between ,the indicated bounds.
Thus,
when one has adescending
chain of norms, in order to maintain a maximum norm of
partial
sumof linear combination of vectors, it is necessary that the dominant terms
are
weighted maximally
and yet maintain a lower bound which is as greatas
possible.
If there are severalpoints
of maximal norm, then the onewith the
greatest magnitude
is the one with the maximal inherentweight.
LEMMA 1.
If,
in a binomial of maximum norm in the Lemma of Theorem 2, the element of maximal norm isrepeated
eltimes,
and the second term in which we assume can berepeated
se-veral
times,
then the maximum number ofrepetitions
of the second term is e2 if the binomial is to maintain the samegreatest
lowerbound 11 U2 11,
as in the Lemma of Theorem 2, where e2=
[À2]; i.e.,
the greatestinteger
contained inX2
and whereThe I’s and R’s are the
imaginary
and real parts of the arguments.PROOF. Let
Then , where I’ denotes a deleted subset
of the index set I.
Let and Ui2 = V2 so that
The greatest lower bound is attained if
or
whence
Thus,
If the denominator for
X2 vanishes,
then andX2
is un-bounded,
so we takei.e.,
LEMMA 2. In order that the vector Ui1 determined
by
where u is any vector, also be the one determined
by
it is sufficient to take where
and
I
is thegreatest
lower boundPROOF.
Since ) )
the extreme À is determined wheni.e.,
when X satisfiesThus
If , then any finite
positive
X will do.THEOREM 3. If
I
Utu2 )) *
...)j I and
can be count-ed up to Qik
times,
then there exist h5:mpositive integers
~ a constant, such that
I
is maximalover all
possible
subsets of h vectors vi of the mui’s.PROOF.
By
Lemma1,
we have that the binomial has dominantnorm;
viz.,
If v~ = u~ , then e2 = a~ , and the sequence would be
(Here,
but let us notdisplay
in thedescending
sequence of norms that element which isalready
containedin the linear combination
partial
sumexpression
on theleft).
By
the Lemma of Theorem 2, in the last twodescending
sequences of norms, the binomial of maximal norm mustnecessarily
contain theexpression
of maximal norm considered as asingle
vector. Thuswith ~tg=~3.
Again
e3 is chosen so that Lemma 2 isapplicable;
viz.,where
X3
is determinedanalogously,
as in the Lemma. Thus we obtain the modifieddescending
sequence of norms with the maximal normed vectorbeing
Ingeneral, proceeding
in this manner,we determine the maximal normed r-termed linear combination
by
taking
the r-1-termed maximal-normedexpression
andfinding
with Uir = vr . Now take
where
where ug is a vector not contained in the linear combination
and [[
ugI I
is itsgreatest
lowerbound; i.e.,
Finally,
the h-termed linear combination of maximum norm with where and vh is determinedby
where uih = vh . To show that this last
expression
is the linear combina- tion with maximal norm and isunique
up toequivalent
norms, assumethat another form exists such that
for another choice of h vectors yi of the
original m
zi’s with differentpositive integral coefficients fi
I withLet
== II Y1 II.
.By
the Lemma of Theorem 2, we findi
that we must have
Y1= vl ;
moreover, formaximality
of the norm ofany linear combination
containing Yi ,
we mustweight
itmaximally:
fi = ei . Also, they
both must contain the element v2 ,since I
> ~ ~
unlessThus,
The
only
way inelVl +e2v2!1
is iff 2 e2 . Applying
Lemma 2 indetermining
the maximal trinomial andusing
the binomials as u(in
Lemma2),
we determine v3 , which is= Y3
again.
The norm of the linear form will exceed the norm of theoriginal; viz.,
if and
only
iff 3 C e3 ,
iff2=Øl.
We see that therepeated application
of Lemma 2 determines the same vectors;
i.e., vi’s;
and that forI
we must havef i ei
with at least one of themThus we see which contradicts our as-
sumption.
We now
apply
theforegoing theory
to thetransportation problem.
Given two finite sets of
points A2 ,
...,Am I
and B=_ { B~ , B2 , ..., B~ }
called the source ororigin points,
and the destina- tion or terminalpoints, respectively,
such that at each sourcepoint At ,
there areexactly
ai items or commodities to betransported
to some, one, or all the destinationpoints;
and at each destinationpoint B; , exactly fij
items or commodities must be received from one, some or all theorigin points.
The cost ofshipping
one item from AL toB;
is ci; . What should be the allocation ofgoods
from all the sourcepoints
tothe destination
points
such that the total cost oftransportation
of allitems A to B is a minimum? Thus the classical formulation for a linear-
programming
solution is to maximize the linear functionalwhere the
Xi¡’S
are theintegral
number of itemstransported from A
ito
B; ,
over allpossible assignments
under the linear contraints of theproblem.
gramming
solution is verylengthy
andtime-consuming.
We now refor-mulate the
problem
in such a way that a direct solution can be found with relative easeby
use of the abovetheory.
Since A1 is stocked with oci items and
Bi
can receive5i items,
letand
Define the
weight
of asingle
connection or arc between Ai andBi by
Summing
theweights
of the individual arcs for anypermissible assign-
ment of
transporting
all of theitems,
we havewhere C is the total cost of the
transportation
and W is the sum ofthe
weights
of the individual connections or arcs of thebipartite graph (counting
themultiplicities
ofarcs).
We note that the conservationprin- ciple
holds in thatso that we have
Thus if we can maximize W over all
possible assignments,
we canminimize
C,
orequivalently,
sinceif we can maximize the ratio c over all
possible assignments,
p we can minimizeC
since K is a constant.D of the
complex plane.
Letso that
Associated with every segment
corresponding
to z1; we have a numberpair ~3;) indicating
theoriginal
maximumpossible degree
of theend
points
Ai andB; ; (the degree
of apoint
is the number of arcsemanating
fromit) viz.,
mi and5i respectively.
With is associat-ed the number
pair a;1) .
Tosimplify
thenotation,
letand
bl).
Then eiv, is the maximum-normedterm with the heaviest
possible weight
ei .Applying
Theorem 3 for the case of thebinomial,
we haveIf
A(v)
andB(v)
denote the endpoint
of the segmentcorrespond- ing
to v in the set A and Brespectively,
thenwhere
a*2
andb*,~
are the residualdegrees
of the ends of the segmentcorresponding
to v2 ;i.e.,
and
À.2
as in Theorem 3.To see with
clarity
how the residualdegrees
of the ends of seg- mentsbehave,
we write out the trinomial case.where
where
and
Thus
by
Theorem 3 weultimately,
after Niterations,
obtain the maximal- normed linearcombination,
where N K is determined
by
the iterative process. Wehave,
ingeneral,
that
where
if
A(vi), i =1,
2, ..., k are distinctpoints
if
A(vi)
for two i’s tand
if
i =1, 2,
..., ~k are all distinctpoints
,if for some two or more i’s are not distinct
points
and where
L’
indicates summationomitting
certaine’s, depending
upon how many and which endpoints
coincide.Thus ei is the number of items
transported
from the sourcepoint corresponding
to thesegment represented by
vi(which
is a well-deter- minedz)
to the terminalpoint
of thesegment corresponding
to vi fori =1,
2, ...,N,
which minimizes the total cost oftransportation,
andthe v;’s
correspond
to thesegments
which define theoptimum scheduling.
The actual minimal cost is
given by
where R is the real part of the
complex argument.
Note that when all of the
segments’
endpoints
are ofdegree
1;i.e.,
then all the we then haveK = N,
andthe
problem degenerates
to thesimple assignment problem
ofassigning
N
people
to Njobs optimally.
Manoscritto pervenuto in redazione il 26 giugno 1972.