VARIOUS OPTIMIZERS FOR SINGLE-STAGE PRODUCTION
Wayne
E. Smith1
The following note develops t h e o r e m s which enable one to schedule operations on a single-stage production s y s t e m so a s to minimize
c e r t a i n functions o r o p t i m i z e r s . It is shown that the maximum t a r d i n e s s
of job completion is minimized by o r d e r i n g the jobs according to the date
the job is due to be completed. The s u m of t i m e s until e a c h j o b is c o m pleted may be minimized under various conditions. If all jobs can be
completed by their due d a t e s , a weighted s u m of completion t i m e s can
be minimized, subject to all jobs being completed by their due dates.
This is done by scheduling l a s t the jobs with the l a r g e s t r a t i o of p r o c e s s ing t i m e to weighting f a c t o r among those jobs which will be completed
on time i f scheduled l a s t ; the next to l a s t job is scheduled by applying
the s a m e rule to the remaining jobs, and so forth, until the complete
ordering is obtained.
In the process of production scheduling there are so many factors that must be taken
into consideration that the determination of a best schedule is impossible without a very precise
meaning of the word "best." We shall discuss some fairly simple criteria which might be u s e d
as a means of judging whether o r not a schedule is good. We shall also show how the choice of
an optimizer will affect the characteristics of best schedules, in certain special cases.
We shall be concerned with single-stage production and with optimizers such that there
is never any advantage in delaying the completion of any operation, unless the completion of at
least one other operation is moved forward. It is not difficult to see that a schedule is then
completely determined by the ordering of the objects to be processed; that is, an optimal solution subject to a given ordering is obtained by performing each operation as soon as possible,
consistent with the ordering.
SECTION 1
In most of the scheduling with which we will be concerned, we establish sufficient conditions that an ordering will be a "best" ordering in some sense of the word. We wish now to
establish a sufficient condition that we shall use repeatedly.
We assume that the desirability of any schedule can be measured, and that this measurement can be expressed as a function f over the set of all schedules. Since the schedule is
assumed to be completely determined by the ordering, then f can be considered as a function
of the ordering or as a function over the set of permutations of the n objects to be processed.
We wish to find a permutation n , where n is the order ( r l , r2,. . .,nn), which minimizes
the value of f.
l p r e p a r e d while the author was under contract to the Office of Naval R e s e a r c h , . Logist i c s Branch, and employed on the Management Sciences R e s e a r c h P r o j e c t , University of California, Los Angeles. The author wishes to thank Dr. J . R. Jackson and A. J. Rowe f o r t h e i r
helpful suggestions and comments.
59
W. E. SMITH
60
THEOREM: A sufficient condition that f(?) I f(n) for all permutations IT of the n
objects is that:
(A) there is a real-valued function g of ordered pairs of elements such that if n is
any permutation and n' the permutation obtained from n by the interchange of the
i-th and (i+l)-st elements, then f(n) 5 f(n') if g(ni, ni+l) 5 g(ri+l, ni) and
(B) 5 is such that the i-th object precedes the j-th object if g(i,j) < g(j,i).
PROOF: Given that the permutation li satisfies condition (B) we wish to show that
any n # f can be altered by a series of interchanges of pairs of adjacent elements in such a
manner that the value of the function decreases o r remains the same at every interchange.
Since # n, there is an element ni of n such that ni is immediately preceded by an item which
follows n i in the order . (For example, r i may be the element appearing as 5 where the
j'
j-th position is the first position at which n and do not agree.) Therefore, g(ni, ni-l) 5
g(ni,l, ni) so we may interchange ni and ni-l in the permutation IT, obtaining a new permutation n' with f ( d ) 5 f(n). Also n' has one fewer pairs of elements which disagree in ordering with those of ;. One may proceed inductively, decreasing the number of pairs of elements
which disagree in ordering with those of li and at each step decreasing the value of the function
or leaving the value of the function the same.
The proof becomes more simple if g is a function which relates the elements in a
transitive manner, that is if g(i,j) 5 g(j,i) and g(j,k) s g(k,j) together imply that g(i,k) C g(k,i).
However, this is not necessary, as can be seen from the example in S. M. Johnson's paper [ 11.
It may be possible to simplify the function g so that it is a function of only one of the
elements. In this case rule (B) becomes: ? is such that the i-th object precedes the j-th
object if g(i) < g(j). This does lead to a transitive relationship and is the type of relationship
that is found in most of the following examples.
W e have not stated any method of obtaining the function g. In the following examples
the function g is derived from the function f. Upon interchanging the i-th and (i+l)-st elements of a permutation, the value of f changes by a certain amount. If this change can be
attributed to quantities attached to just the i-th and (i+l)-st items, then one may be able t o
eliminate constant values from both values of the function f, obtaining f(n) = A + B g(ni, ni+l)
and f(n') = A + B g(ri+l, ni), where IT' differs from T only by the interchange of the i-th and
(i+l)-st elements. It may then be possible to simplify this by finding a function h such that
g(i,j) < g(j,i) if h(i) < h(j).
In the specific examples that follow we shall assume we are given the operation times
(which shall include the setup times except in Section VIII) for a finite number of jobs. Also,
the optimizers with which we are concerned are such that the deletion of all idle times cannot
make a schedule worse so we shall assume that no idle time occurs.
n
SECTION II
W e now assume that each operation o r job has a due date and that we wish to minimize
the maximum tardiness. This type of optimizer might be used if the output of the machine is
used in an assembly line production system where the delay on the assembly line is the maximum of the individual delays.
We let Ti be the operation time of the i-th object and Di be the due date of the i-th
object. If the objects are processed in numerical order the tardiness of the i-th object is:
VARIOUS OPTIMIZERS FOR SINGLE-STAGE PRODUCTION
. .
. . . ,n .
6i = T1+ T2 . + Ti - Di. Let A = max
an ordering which minimizes A.
{ai}
THE0REM:z The o r d e r { 1,2,3,.
D15 D2 5 D35
. L Dn.
. . ,n } is an o r d e r which minimizes
..
over i = 1, 2,
61
Our problem is to find
A if
...
, n, and let n ’ be the s a m e o r d e r except with
PROOF: Let n be the o r d e r 1, 2, 3,
i and i+l interchanged. Thus T = T! and D = D’ for j # i or i + 1; Ti= Ti+l, Ti+l = Ti, Di
j
j
j
i
= Df+l and Di+l = Df. Also, 6 = 6! for j # i o r i+l. .Thus A 5 A’ if
1
1
Max (6i, 6i+l) 5 Max ( 6 f , 6f+1)
.
This is by definition equivalent to:
i
Max ( C T j
j=1
- Di,
i+ 1
C Tj
j= 1
- Di+l)
Replacing the Tf by their equal Ti values we have:
i
M=(.C
]= 1
Tj
- Di,
Subtracting from each side
i+l
j=Z1 T j - Di+l)
i+l
j= 1
Tj
5
Max (-Ti
- Di+l,
- Di)
or
-Min (Ti+l
+ Di,
Di+l)
5
-Min (Ti + Di+l, Di)
or
Min (Ti + Di+l, Di) 5 Min (Ti+l + Di, Di+l)
.
Since all Ti are positive, if Di 5 Di+l this condition is met, the Di being the smallest of the
four t e r m s concerned. Therefore, if Di 5 Di+l we have that A 5 A’, and the o r d e r with i
preceding i+l is as good or better than the o r d e r with them interchanged. We have, then, by
the theorem of Section I, that the maximum tardiness can be minimized if the jobs are arranged
according to their due dates. Thus, if it is possible to complete all the jobs by their due dates,
this can be done by ordering them according to due date.
2J. R. Jackson [ 2 ] has generalized this theorem to take into consideration restrictions
on the time at which the processing of each item can s t a r t .
391368 0
- 56 - 5
62
W. E. SMITH
SECTION 111
In scheduling for single-stage production, there is a waiting time, a time that each job
must wait until it can be processed. Thus, if the objects are processed in numerical order,
j-1
the waiting time for the j-th object is C Ti. One optimizer that may be considered is the
i=1
n j-1
sum of all waiting times, c. r, T i . The minimum of this expression occurs at the same
j=1 i=l
n
time as the minimum of the expression C
j
C Ti occurs, since the two expressions differ
j=1 i=l
n
by the constant, Z Ti. We shall use the latter formulation in this paper.
i=l
If the objects are processed in numerical order, the completion time of the j-th object
i
is C
J
j - i=l
Ti. We see that the above problem is the same as the problem of finding an order
n
which minimizes the sum of completion times,
C Ci.
i=l
THEOREM:
n
C Ci is minimized over all orderings of the n objects if the objects
i=1
are arranged in increasing order of values of Ti.
n
n
i
n
PROOF: Z C i = C C T. = c (n+l-j) T According toHardy, Littlewoodand
i-1
i-1 j=1 3 j=1
j'
Polya (Inequalities, p. 261), such a sum is least when the series are monotonic in opposite
5 Tn.
senses. Therefore, the optimizer will have its minimum value if T1 5 T2 5 T3 5
. ..
SECTION IV
We may now generalize the theorem of Section III. Since some jobs may be considered
more important than others, we may weight the jobs accordingly and consider a weighted sum
of completion times. Weighting job i by an amount ai 2 0, the optimizer is then of the form
n.
z aj C j .
j =1
n
THEOREM: The sum
j=1
a. C . is minimized over all orderings of the n objects if
3
J
the objects are arranged so the T./a. form a non-decreasing sequence.
1
1
.
PROOF: Let n be the order 1,2,3,. .,,n and let n' be the same order except with i
and i+l interchanged. Thus T = T! and a. = a! for j i o r i+l;T i = Ti+l , Ti+l = T i ,
a. = ai+l
' ,ai+l = a;.
than 'II' if
+
1
1
Also C j = C! for j # i o r i+l. Now, the order
j
i
3
n
c
i= 1
ajCj
n
5
c a;~;.
j= 1
'II
is as good as o r better
VARIOUS OPTIMIZERS F O R SINGLE-STAGE PRODUCTION
63
By the above equalities, this is true if
a.1 C.1 + ai+l citl
I
a!1 C!1 t aitl
cit1
or
Subtracting ai
(jkl
Tj)
t
aitl
('2=1 T j
j
(1)
t
Titl
aitl T.1
fr?m both sides this becomes
5
ai Titl
.
If ai and aitl a r e not zero, we may divide by their product obtaining
Defining T/O to be infinity, condition (1) and condition (2) a r e equivalent whether o r not any a's
a r e zero.
Therefore if condition (2) is satisfied, the order with i preceding it1 is as good as o r
better than the order with them interchanged. Thus, by the theorem of Section I, the minimum
value of the optimizer is obtained if the objects are arranged in increasing order on Ti/ai.
SECTION V
We now propose to use the same optimizer as in Section 111, the s u m of times until each
job is completed, but add the restriction that each job must be completed by its given due date,
Di
.
THEOREM: If all jobs, 1,2,3,. . . ,n, can be completed by their respective due dates,3
n
then there is an ordering of the jobs with the k-th job last which minimizes c Ci , subject to
i=l
completion of all jobs by their respective due dates if, and only if, the k-th job has the properties:
n
(i) Dk 2 C Ti and
i=1
n
(ii) Tk 2 Ti for all i for which D. > Z Ti
1 - i=l
.
P R O O F Let ?I be the ordering 1,2,3,. . . ,n, and let it be so chosen that with the objects
in this order all jobs can be completed by their due dates. Let In be the set of indices for which
n
Di
2
c Ti. Since the ordering
i =1
II
is such that all jobs a r e completed by their due dates, we have
3We stated in Section I1 that if a s e t of jobs can be completed by their due d a t e s , this can
be done by ordering according to t h e i r due dates.
64
w. E. SMITH
that n is in the set In. Suppose that k is the index of the object with the largest value of Ti
f o r i in the set In. (If this is not unique, we choose one arbitrarily from those which are the
largest.) Now let 1~' be the ordering obtained from IT by interchanging the k-th and n-th
terms. Thus T i = Ti and D. = D! for i P k or n; Tk = TL, Tn = T i , D -D' and Dn = Dk.
1
1
k- n
Nowforj<k,C'
C
j Ti= C
j Ti=C.,andsinceT > T wehaveforksj<n,C!=
1 = i=l
i=l
I
k- n
J
j
= iZ
=1 T i + T n - Tk -- C 1. t T n - T k 5 C j .
C
j T'
i=l i
n
n
A l s o , C L = C T ! = i=l
C T 1. = C n ' Therefore,t&ing
i =1
n
n
the sums of C's, we have that C C; 5 C Ci. Also, all jobs are completed by their respeci =1
i =1
tive due dates under the new ordering, since for i # k or n we have C; 5 Ci 2 Di = D; ;
chs
n
n
2 T i ( D n = D k a n d C h = C Ti'Dk=DL.
i =1
i=1
n
is an ordering which minimizes C Ci, we have found an
i=1
ordering IT' which also minimizes the optimizer. Thus, if the k-th object has properties (i)
and (ii)there is an ordering of the jobs with the k-th job last, which minimizes the optimizer
subject to completion of all jobs by their respective due dates.
If the original ordering IT has a last item which does not have property (ii), then Tn<Tk;
In the particular case that
and in the above steps for k s j
<
71
n, C'
j
<
n
C. so that E C;
i
Ci. Thus IT cannot be an
i=l
i=l
ordering which minimizes the optimizer. The ordering
n
< C
IT
is not even admissible without
property (i).
n
In o r d e r to obtain a complete ordering which minimizes Z Ci, we can determine an
i=l
item to b e scheduled last, one which has properties (i) and (ii). Once having determined the
n- 1
last item, Cn is fixed, and we have the problem of ordering n-1 jobs so as to minimize C Ci,
i=l
subject to the completion of all jobs by their respective due dates. Applying the theorem again,
we can determine the last of these n-1 jobs. Continuing in this manner, we can find a complete
ordering which minimizes the value of the optimizer.
We note that if the objects are arranged in increasing o r d e r on Ti and still meet their
due dates, then one cannot decrease the value of the optimizerby rearrangement, that is, the value
of the optimizer for this o r d e r is a lower bound on its value no matter what the side restrictions
are.
On the other hand, if the objects are arranged in increasing order on Di , and this order
is such that Dj-l
1
C Ti 5 D. for all j (taking Do = 0), then the given o r d e r is the only
i=l
I
arrangement by which all jobs can b e completed on time.
<
SECTION VI
The results of Sections IV and V can be combined to give a method of ordering so as to
n
minimize the weighted sum z ai Ci subject to the restriction that every job must be completed
i=l
VARIOUS OPTIMIZERS FOR SINGLE-STAGE PRODUCTION
65
by its due date, that is, Ci 5 Di. If all jobs can be completed by their due dates, an order which
minimizes the weighted sum may be obtained. This order has as its last job one with the largest
value of Ti/ai from among those with due date as large as the total processing time of all jobs.
Proceeding as before, one can determine the last of the remaining n-1 jobs to be processed and
continue in like manner until the complete ordering is determined.
SECTION VII
Other optimizers that have been considered are:
n
Total tardiness = C max (0, Ci
i =1
- Di)
n
Weighted tardiness =
5
i=l
ai [max (0, Ci
- Di) ] .
Thus far we can give solutions only in special cases. For example, if all the Ti are the
same, the total tardiness is minimized if the objects are arranged in order of due dates. If the
objects are arranged in order of due dates and then have only one tardy item, this order minimizes total tardiness since it has minimized the maximum tardiness. In minimizing total
tardiness, a rule for the ordering of two adjacent operations is as follows: If the first of the
i-th and (it1)-st operations start at time to, and if max {Di, to + Ti}( max{ Ditl, to + Titl},
then the order {i,itl} gives no greater total tardiness than the order { i+l,i}. Since this relation depends on to, one cannot in general expect to reach the optimal solution by only a series
of interchanges of adjacent items.
SECTION VIII
Thus far we have considered only the case where the setup times could be included as
part of the operation times. Under these assumptions, the over-all completion time of n given
jobs is fixed and is just the sum of the T i s , regardless of order. Now let us see what type of
problem we have if we wish to minimize the over-all completion time of all jobs where the setup times are taken into consideration.
Since the sums of the actual operation times is constant, and the over-all completion
time is determined by the sum of operation times and setup times, our problem is equivalent to
minimizing the sum of the setup times. This is a non-symmetric "traveling salesman" type of
problem. T o be more precise, the amount of setup times required for the j-th job, if it follows
the i-th job, is taken to be the "distance" from i to j (not necessarily the same as from j to i).
We then have the problem of finding a cycle o r "tour" starting with 0 (the original setup), passing
through each object once, and only once, and returning to 0, which minimizes the distance traveled. (If one does not c a r e in what form the machine is set up at the end of the operations, then
the distance from j to 0 would all be taken as zero.) Since such problems have been discussed
elsewhere, we shall not consider this problem further [3, 41.
66
W. E. SMITH
SECTION M
The optimizers for which an optimum ordering has been found in this paper are such
that the solution can be obtained from any order (satisfying the given restrictions) by a series
of interchanges of adjacent operations, improving (or not making worse) the value of the optimizer at each step. Thus there are no "local" optima with respect to permutations of adjacent
elements. We have shown that any optimizer which satisfies the conditions of the theorem of
Section I also has this property. Although it should not be too difficult to test to see whether
specific optimizers satisfy these conditions, it would be desirable to characterize more completely those optimizers for which no "local" optima exist. Further research is planned in
this direction.
Another difficult problem to be considered, of course, is that of finding how to determine
an order which minimizes the value of similar functions when several machines or stages of
production are used.
REFERENCES
[l] Johnson, S. M., "Optimal Two- and Three-Stage Production Schedules with Setup Times
Included," Naval Research Logistics Quarterly, Vol. 1, No. 1, 1954, pp. 61-68.
[2] Jackson, J. R., "Scheduling a Production Line to Minimize Maximum Tardiness," Research
Report No. 43,36 pp, Management Sciences Research Project, University of California,
Los Angeles.
[3] Dantzig, G., Fulkerson, R., and Johnson, S., "Solution of a Large Scale Traveling Salesman
Problem," Journal of the Operations Research Society of America, Vol. 2, No. 4, November
1954.
[4] Kuhn, H. W., "The Traveling Salesman Problem," Proceedings of the Sixth Symposium in
Applied Mathematics of the American Mathematical Society, McGraw-Hill Book Co.
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