Resorting Vehicles in an Automotive Manufacturing Environment
Jens Feller
1
, Bernhard Mauersberg
1
, Wolfgang Mergenthaler
1
, Yvonne Feller
2
and Lutz M
¨
uller
3
1
FCE Frankfurt Consulting Engineers GmbH, Altm
¨
unsterstraße 2d, D-65439 Fl
¨
orsheim, Germany
2
Heinz-Nixdorf-Berufskolleg, Dahnstraße 50, 45144 Essen, Germany
3
Porsche Leipzig GmbH, Porschestraße 1,D-04158 Leipzig, Germany
Keywords:
Combinatorial Optimization, Dynamic Optimization, Sequencing, Sorting, Set Theory.
Abstract:
One of the most important concepts in an automotive production process is that of a manufacturing sequence.
Sequencing has a vital influence on a series of important performance indicators such as load balance, setup
time, setup cost, timeliness, flow of material etc. Generating appropriate sequences has thus become a key task
for the automotive production planner. Ensuring the stability of a sequence, once it has been determined, by
sorting its actual version back into the form intended originally, however, with systematic means, is a new issue
gaining recently more importance. This problem represents the main topic of the present paper. Instruments
for physically resorting vehicles are sorting channels and parking spaces. Both instruments are closely related
to one another, as will be shown.
1 INTRODUCTION
Problems and solution methods in the sequencing
domain have been widely studied in the past, both
from the theoretical as well as the applicational point
of view, viz. (Chr
`
etienne et al., 1995; Mergenthaler
et al., 1995; Mergenthaler et al., 1994).
Since combinatorial optimization is at the math-
ematical core of a sequencing problem, an optimal
sequence can be found by using heuristics and local
search techniques, viz. (Reeves, 1996). Once an opti-
mal sequence has been found for a set of vehicles, for
the sake of generality called objects below, it is by no
means guaranteed in practice, that this sequence will
remain stable throughout the production process. A
perennial source of disturbance is the paint shop, for
example, where rework is frequent.
Assume n objects numbered 1, 2, 3, . . .,n arrive at
a certain point in an arrival queue according to a se-
quence, or permutation ρ = {ρ
1
, . . . , ρ
n
}, ρ
1
arriving
first, ρ
n
arriving last and set N = {1, . . . , n}. The ob-
jects need to undergo a certain process starting at this
point in the sequence
˜
ρ = {1, . . . , n}, however, and
must therefore be resorted correspondingly. There
are two instruments to implement the resorting task:
sorting channels and parking spaces. Once an ob-
ject is assigned to a sorting channel it can only be
removed from the bottom of the channel at the end
of the assignment process. Alternatively it can be
parked on a parking space until an assignment to a
channel becomes possible. The quantity of parked ob-
jects should be minimized because of cost considera-
tions. The sorting channels, in this paper, are assumed
to have unlimited capacities, the parking space is lim-
ited.
An example sorting process with 7 objects to re-
sort and a factory layout of three channels and four
parking spaces is given in figure 1. The pipeline of
objects is considered to be processed from right to
left. The arrows show one possible assignment strat-
egy. The process starts by moving object 7 to a park-
ing space. All other objects are directly assigned to
a channel marked by the arrows. Finally object 7 is
moved from the parking space to the first channel.
Based on this assignment the original order {1, . . . , 7}
can be restored by pulling the objects from the bot-
toms of the channels.
Both sorting channels and parking spaces require
investments which naturally the factory planner wants
to minimize. One question of interest in this pa-
per will be, how many sorting channels and parking
spaces should be installed. Once the factory layout
has been fixed, only the channel and parking space
assignment can be controlled by the plant manager. It
is straightforward to ask for the optimal assignment
strategy which minimizes the number of objects to be
parked at any time.
This paper is organized as follows: Chapter 2
184
Feller J., Mauersberg B., Mergenthaler W., Feller Y. and Müller L..
Resorting Vehicles in an Automotive Manufacturing Environment.
DOI: 10.5220/0005087601840191
In Proceedings of the 11th International Conference on Informatics in Control, Automation and Robotics (ICINCO-2014), pages 184-191
ISBN: 978-989-758-040-6
Copyright
c
 2014 SCITEPRESS (Science and Technology Publications, Lda.)

Figure 1: Sorting channels and parking spaces.
shows that the original order can be restored by divid-
ing the arbitrary queue into ascending subsequences
and discusses the tradeoff between sorting channels
and parking spaces. Furthermore it shows that permu-
tations can be subdivided into a set of contiguous sub-
sequences such that statements about maximum park-
ing space can be reduced to statements restricted to
those subsets. An important result says that permu-
tations tend towards contiguous subsets as the size of
the permutations tends to infinity. Based on these re-
sults in Chapter 3 a dynamic optimization model is
formulated in order to calculate an optimal assign-
ment strategy for the factory planner. Chapter 4 de-
scribes an appropriate solution heuristic. Chapter 5
concludes this paper with an example.
2 MAIN RESULTS
2.1 Definitions
Let m be the number of sorting channels and set
M := {1, . . . , m}. Let x
i
∈ M be the channel assigned
to object ρ
i
, i ∈ N. Let P be the set of all permutations
of the objects in N. The set of all possible channel
assignments is defined as X := M
n
. Let S
t
(x, ρ) be
the set of objects parked right after ρ
t
has been drawn
from the pipeline, with x ∈ X , ρ ∈ P,t ∈ N.
2.2 The Importance of Ascending
Subsequences
In this section the importance of ascending subse-
quences in each channel with respect to generating
the original sequence will be discussed.
Lemma 1. Let T
µ
⊆ N, µ ∈ M be m ordered ascending
subsets of N such that
T
µ
= {ρ
1
µ
, . . . , ρ
n
µ
µ
},
n
µ
= card(T
µ
),
ρ
1
µ
≤ ρ
2
µ
·· · ≤ ρ
n
µ
µ
T
µ
∩ T
ν
=
/
0, ν ∈ M − {µ},
[
µ∈M
T
µ
= N and
∀µ ∈ M, l ∈ {1, . . . , n
µ
} ∃ w(l, µ) ∈ N : ρ
l
µ
= ρ
w(l,µ)
Then there is a sequence ζ =
{ζ
1
, . . . , ζ
n
} where ζ
i
∈ M, ∀i ∈ {1, . . . , n} such
that ρ
1
ζ
1
= 1, ρ
k(2,ζ
2
)
ζ
2
= 2, . . . , ρ
k(n,ζ
n
)
ζ
n
= n, k(ν, ζ
ν
) =
∑
ν−1
j=1
δ
ζ
ν
ζ
j
+ 1,ν ∈ N and δ
i, j
=

1, i = j
0, else
is the
Kronecker symbol.
Proof. The proof is by induction. Assume that, after
l ∈ N steps the objects
ρ
1
ζ
1
= 1, ρ
k(2,ζ
2
)
ζ
2
= 2, . . . , ρ
k(l,ζ
l
)
ζ
l
= l (1)
have been drawn from the respective channels
z
1
, . . . , z
l
. This is clearly the case for l = 1. If
ρ
1
ν
6= 1 ∀ν ∈ M then, since there must exist one chan-
nel ν
∗
∈ M containing element 1, i.e. p
y(ν
∗
)
ν
∗
= 1 for
some y(ν
∗
) with 2 ≤ y(ν
∗
) ≤ n
ν
∗
(l) one obtains
ρ
1
ν
∗
> 1 and ρ
y(ν
∗
)
ν
∗
= 1 (2)
violating the assumption of ascending subsequences
in each channel. n
ν
(l), ν ∈ M denotes the number of
objects in channel ν ∈ M after l ∈ N objects have been
assigned.
ResortingVehiclesinanAutomotiveManufacturingEnvironment
185

If, in general, 1 holds for some arbitrary l with
1 ≤ l ≤ n, then there must be some channel ζ
l+1
with
ρ
k(l+1,ζ
l+1
)
ζ
l+1
= l +1 because, otherwise, the assumption
of ascending subsequences would be violated again
Verbally, lemma 1 says that when an arbitrary se-
quence, i.e. a permutation of the numbers 1 through
n, can be partitioned into a series of ascending sub-
sequences, then the original sequence can be restored
by pulling consecutively the bottoms from the subse-
quences. On the other hand, the existence of ascend-
ing subsequences is also necessary for the existence
of the ordered sequence
˜
ρ. If at least one subsequence
is not ordered, then obviously
˜
ρ cannot be restored
by pulling the respective bottoms from the sequences.
This is a basic result for the optimization model where
the number of possible assignments is reduced to as-
signments which must create ascending subsets in the
respective channels.
2.3 Tradeoff Between Channel Size and
Parking Spaces
Recall that the channel assignment x ∈ X is defined
such that
x
ν
= µ ⇐⇒ ∃ l ∈ {1, . . . , n
µ
} : ρ
l
µ
= ρ
ν
, ν ∈ N, µ ∈ M
One can now show that the following holds:
Lemma 2. The set of objects parked at time t under
policy x and for an incoming sequence ρ is equal to
S
t
(x, ρ) =
t
[
i∈{1,...,t},
∏
n
j=t+1,x
j
=x
i
1
{ρ
j
>ρ
i
}
=0
ρ
i
(3)
Proof. After step t all those objects ρ
i
, i ∈ {1, . . . ,t}
must be parked,who have at least one successor within
{ρ
t+1
, . . . , ρ
n
} smaller than ρ
i
, but are assigned to the
same channel. Thus an object must be parked, if
n
∏
j=t+1,x
j
=x
i
1
{ρ
j
>ρ
i
}
= 0 (4)
Equation 4 is identical to the filtering condition in
equation 3
Two special cases may illustrate lemma 2: Special
case 1: Let ρ = {1, . . . , n}. In this case it is obvious
that
S
t
(x, ρ) =
/
0, ∀t = 1, . . . , n (5)
if, for arbitrary µ ∈ Mx
1
= µ, . . . , x
n
= µ. for instance.
There may be other assignments also yielding 5.
Special case 2: Let ρ = {n, . . . , 1}. Then, with m
channels given, exactly d
n
m
e − 1 parking spaces are
needed. This can be seen by noting that a possible
assignment policy would be given by r =
l
n
m
m
and
x
1
= 1, .. . , x
r
= 1
x
r+1
= 2, . . . , x
2r
= 2
. . .
x
(m−1)r+1
= m, . . . , x
n
= m.
In this case channel 1 receives objects ρ
1
through
ρ
r
, requiring exactly r − 1 parking spaces. The same
holds for channels 2 through m − 1. For channel m
exactly (nmod r) − 1 parking spaces are needed, if
(nmod r) > 0, r − 1 else.
Let Z
t
(x, ρ) := card(S
t
(x, ρ)). It then follows im-
mediately from lemma 2 that
Z
t
(x, ρ) =
t
∑
i=1
(1 −
n
∏
j=t+1,x
j
=x
i
1
{ρ
j
>ρ
i
}
)
Two worst case approximations on the number of
parking spaces given the number of channels, or -
the number of channels, given the number of parking
spaces - are readily obained.
Corollary 1. In order to generate m ascending subse-
quences from an arbitrary permutation ρ the number
of required parking spaces A(n, m) satisfies
A(n, m) ≤
l
n
m
m
− 1 (6)
Proof. Assign the first
l
n
m
m
objects to channel 1, the
second
l
n
m
m
objects to channel 2, etc. Then, in the
worst case one deals with special case 2 and the corol-
lary is proven
Corollary 2. Assume there are r parking spaces
available. Then for any permutation ρ there is an
assignment policy such that the number of required
channels B(n, r) satisfies
B(n, r) ≤
l
n
r
m
(7)
Proof. From 6 it follows that r ≤ d
n
B(n,r)
e − 1 must
hold.Therefore
B(n, r)r ≤ B(n, r)d
n
B(n,r)
e − B(n, r). On the other
hand d
n
B(n,r)
e ≤
n
B(n,r)
+ 1.
Hence B(n, r)r ≤ B(n, r)(
n
B(n,r)
+ 1) − B(n, r) = n
and therefore B(n, r) ≤
n
r
. Since B(n, r) is an integer,
7 is correct
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186

In the special case, where there is no parking space
available, there is a simple sufficient criterion for the
necessity of at least m sorting channels.
Corollary 3. A sufficient criterion for the necessity
of at least m sorting channels, if there is no parking
space available, is the occurrence of a contiguous,
descending subsequence of length m.
Proof. Let, for some t with 1 ≤ t ≤ n − m
ρ
t
≥ ρ
t+1
≥ . . . ≥ ρ
t+m
Then, even in the best case, where all the objects
ρ
1
, . . . , ρ
t−1
have been accommodated to their respec-
tive channels due to the fact that
n
∏
j=t,x
j
=x
i
1
{ρ
j
>ρ
i
}
= 1, 1 ≤ i ≤ t − 1
all of the objects ρ
t
, ρ
t+1
, . . . , ρ
t+m−1
can safely
be stored in the channels numbered 1,. . . , m and must
be distributed across them. Therefore ρ
t+m
will only
find channels covered with objects larger than itself.
Hence, a new channel is required.
In the dynamic optimization model in 3 either the
channel quantity or the size of the parking space is
given. The other value is calculated based on Corol-
lary 1 or 2.
2.4 The Impact of Contiguous
Subsequences
Assume that the set N can be partitioned into p sub-
sets N
α
, 1 ≤ α ≤ p such that
N
α
∩ N
β
= , ∀α, β ∈ {1, . . . , p}, α 6= β
p
[
α=1
N
α
= N
N
α
= {ρ
l
(α), . . . , ρ
u
(α)}, α ∈ {1, . . . , p}
where each subset N
α
satisfies the following condi-
tions:
min N
α
=

1 + max N
α−1
, α > 1
1, α = 1
(8)
∀ρ ∈ {minN
α
, min N
α
+1, . . . , maxN
α
} : ρ ∈ N
α
(9)
and, where for each A ⊂ N
α
subset A does not satisfy
(8 - 9) and where finally l(α) and u(α) denote the
lower and the upper limit of the index set N delimiting
subset N
α
.
This way of partitioning N ensures that the subsets
can be arranged such that
1. Any subset contains elements strictly greater than
all the elements in the previous subsets
2. Any subset contains all the elements between its
minimum and its maximum
3. No subset of a contiguous subset is a contiguous
subset
With this definition an important result can be de-
rived.
Lemma 3. For each t ∈ N let α(t) be defined through
ρ
t
∈ N
α(t)
. Then
S
t
(x, ρ) =
[
i∈N
α(t)
,i≤t,
∏
n
j=t+1,x
j
=x
i
1
{ρ
j
>ρ
i
}
=0
ρ
i
Proof. Starting from 3, consider what happens if i ∈
N
1
∪ N
2
∪ · · · ∪ N
α(t)−1
. In this case one knows that,
by definition,
ρ
i
<min
j∈N
α(t)
∪N
α(t)+1
∪···∪N
p
ρ
j
= min
j∈{t+1,...,n}
ρ
j
with the consequence that
n
∏
j=t+1
1
{ρ
j
>ρ
i
}
= 1
and
∏
n
j=t+1,x
j
=x
i
1
{ρ
j
>ρ
i
}
= 1.
Therefore, the filtering condition in the union in-
dex in 3 reduces to the one shown above, proving
lemma 3.
Verbally lemma 3 says that, in order to determine
the set of objects in the parking space at any time,
only those objects qualify, which belong to the same
contiguous subset as the actual object.
In the rest of the section contiguous subsets will
be analyzed and, in particular, the chance to find them
in random permutations. Therefore the following def-
initions are needed:
• Let β(n) be the number of permutations of the
set {1, . . . , n}, which form a contiguous subset
and therefore contains no other proper contiguous
subset.
• Let γ(k, n) be the number of permutations of the
set {1, . . . , n}, which contain at least one contigu-
ous subset of length 1 ≤ k ≤ n.
The following result holds:
Lemma 4.
β(n) = n! −
n−1
∑
k=1
β(k)(n − k)! (10)
ResortingVehiclesinanAutomotiveManufacturingEnvironment
187

Proof. It is obvious that
γ(k, n) = β(k)(n − k)!
because the subset {1, . . . , k} is contiguous, therefore
can be organized in β(k) ways and the remaining
elements can be ordered into (n − k)! ways. Also,
if {1, . . . , n} forms a contiguous subset and there-
fore contains no contiguous subset of smaller size,
from the n! permutations of the set {1, . . . , n} those
must be subtracted, which contain a subset of length
1, . . . , n − 1, i.e.
β(n) = n! −
n−1
∑
k=1
γ(k, n) (11)
Inserting 2.4 into 11 proves the lemma.
10 generates the following sequence, as is
straightforward to verify:
β(1) = 1, β(2) = 2, β(3) = 3, β(4) = 13, β(5) =
71, . . . .
Unfortunately one can show that β(n) tends to n!
for n going towards infinity, meaning that less and
less permutations contain contiguous subsets, thus de-
stroying the hope of being able to apply lemma 3 to
large permutations in a systematic way.
Corollary 4.
lim
n→∞
β(n)
n!
= 1 (12)
Proof. 10 yields, after some rearrangements of terms
n
∑
k=1
β(k)
k!
k!(n − k)!
n!
= 1 (13)
or, explicitly
β(1)
1!
1!(n − 1)!
n!
+
β(2)
2!
2!(n − 2)!
n!
+ · · ·+
β(n − 1)
(n − 1)!
(n − 1)!1!
n!
+
β(n)
n!
1 = 1
Abbreviate
C(n) :=
1!(n − 1)!
n!
+
2!(n − 2)!
n!
+ · · · +
(n − 1)!1!
n!
=
n−1
∑
k=1
1

n
k

If it can be proven that
lim
n→∞
C(n) = 0 (14)
then one obtains
lim
n→∞
n−1
∑
k=1
β(k)
k!
k!(n − k)!
n!
= 0
because β(k) < k! and, therefore
β(k)
k!
< 1 yielding
12 upon using 13. In order to prove 14 one proceeds
along the following lines:
C(n + 1) =
n
∑
k=1
1

n+1
k

=
n−1
∑
k=1
1

n+1
k

+
1
n + 1
=
n−1
∑
k=1
k!(n + 1 − k)!
(n + 1)!
+
1
n + 1
=
n−1
∑
k=1
k!(n − k)!
n!
(n + 1 − k)
(n + 1)
+
1
n + 1
=
n−1
∑
k=1
k!(n − k)!
n!
−
n−1
∑
k=1
k!(n − k)!k
n!(n + 1)
+
1
n + 1
= C(n) −C(n + 1) +
C(n)
n + 1
+
2
n + 1
after some lengthy transformations. Therefore
C(n + 1) =
(n + 2)
2(n + 1)
C(n) +
1
n + 1
(15)
Set
C
∞
= lim
n→∞
C(n) (16)
15 and 16 together produce C
∞
=
1
2
C
∞
which can
only be true for C
∞
= 0 which is equivalent to 14 and
therefore proves the lemma.
Corollary 4 tells us that, as n tends towards infin-
ity, almost every permutation of the set {1, . . . , n} is
a contiguous subset. So, lemma 3, which allows for
permutations with limited parking space, will be less
and less applicable for large n.
3 DYNAMIC OPTIMIZATION
MODEL
The main objective of this chapter is to find an opti-
mal assignment strategy for the factory planner which
minimizes the number of parked objects for a given
factory layout at any time. Based on the results of
chapter 2 the integer programming problem is repre-
sented as a dynamic optimization model.
As in (Neumann and Morlock, 1993) the follow-
ing dynamic optimization model is defined.
ICINCO2014-11thInternationalConferenceonInformaticsinControl,AutomationandRobotics
188

Minimize
∑
T
t=1
c
t
(z
t
, u
t
)
Subject to z
t+1
= f
t
(z
t
, u
t
), t = 1, . . . , T
z
1
= z
a
z
t+1
∈ Z
t+1
u
t
∈ U
t
(z
t
), t = 1, . . . , T
Definitions:
1. In every state t either an object is assigned to a
channel or parked in it. In case of an assignment
the object can be the next object in the queue or
an object which is parked. Let N
0
⊂ N be the
set of objects parked ever in a parking space and
n
0
= |N
0
|. Then T = n + n
0
is the total number of
states and defines the horizon of the optimization
problem.
2. The number of available parking spaces or the
number of available channels should be given.
The other variable can be calculated based on
corollary 1 and 2. Let p
max
be the maximum num-
ber of available parking spaces and m the number
of channels.
3. z
t
= (λ
0
(t), p(t), (λ
1
(t), . . . , λ
p(t)
(t)), (ρ
η
1
1
(t), . . . ,
ρ
η
m
m
(t)) where λ
0
(t) is the next object in the
queue, p(t) the number of occupied parking
spaces, λ
i
(t) ∈ N
0
⊂ N, i = 1, . . . p(t) the objects
on a parking space, ρ
η
µ
(t)
µ
∈ N the top-most ob-
jects of channel µ ∈ M. If no object is assigned
ρ
η
µ
(t)
µ
= 0.
4. z
a
= (λ
0
(1), 0, (0, . . . , 0)) with λ
0
(1) = ρ
1
the first
element in the queue, p(1) = 0 occupied parking
spaces and m empty channels.
5. Λ(t) = ∪
p(t)
i=0
{λ
i
(t)} are candidates waiting for an
assignment at time t.
6. The decision space is described by the union of
two subsets U
i
t
(z
t
), i = 1, 2. The first set U
1
t
(z
t
)
describes the assignment of an object to a channel.
Possible candidates are elements from Λ(t). The
second set U
2
t
(z
t
) describes the alternative that the
next object in the queue is parked.
In detail the subsets are defined as follows:
U
1
t
(z
t
) = {( j
t
, ˜µ
t
) : j
t
∈ Λ(t) ∧ ˜µ
t
∈ M}
U
2
t
(z
t
) = park λ
0
(t)
7. Let κ be the index of the actual element in the
queue starting with κ = 1 in t = 1.
8. Depending on u
t
∈ U
t
the following states z
t+1
=
f
t
(z
t
, u
t
) are obtained as follows:
u
t
= ( j
t
, ˜µ
j
t
) ∈ U
1
t
(z
t
) :
j
t
= 0 :
κ = κ + 1
λ
0
(t + 1) = ρ
κ
λ
i
(t + 1) = λ
i
(t), i = 1, . . . , p(t)
p(t + 1) = p(t)
j
t
> 0 :
λ
i
(t + 1) = λ
i
(t), i = 0, . . . , j
t
− 1
λ
i
(t + 1) = λ
i+1
(t), i = j
t
, . . . , p(t) − 1
p(t + 1) = p(t) − 1
∀ j
t
≥ 0 :
ρ
η
µ
µ
(t + 1) = ρ
η
µ
µ
(t), µ ∈ M \ {˜µ
j
t
}
ρ
η
˜µ
j
t
˜µ
j
t
(t + 1) = λ
j
(t)
u
t
∈ U
2
t
(z
t
) :
κ = κ + 1
λ
0
(t + 1) = ρ
κ
λ
i
(t + 1) = λ
i
(t); i = 1, . . . , p(t)
λ
p(t)+1
(t) = λ
0
(t)
p(t + 1) = p(t) + 1
ρ
η
µ
µ
(t + 1) = ρ
η
µ
µ
(t), ∀µ ∈ M
9. Lemma 1 says that the existence of ordered sub-
sequences is necessary for the existence of the
ordered sequence
˜
ρ. Therefore the set of feasi-
ble assignment candidates in t is
ˆ
Λ(t) = { j
t
∈
Λ(t) ∧ ∃ µ ∈ M : j
t
> ρ
η
µ
(t)
µ
}. In order to make
sure that the optimization problem chooses only
feasible assignments, infinitely high penalty cost
for assigning j
t
∈ Λ(t) \
ˆ
Λ(t) are defined. Also,
exceeding the maximal number of parking spaces
leads to infinitely high cost. Parking an object be-
fore it is assigned leads to setup cost c
p
. The over-
all cost in t with t = 1, . . . , T can be defined as:
c
t
(z
t
, u
t
) = max(p(t + 1) − p(t), 0)c
p
+ δ
t
with δ
t
defined as:
δ
t
=



∞, if j
t
∈ Λ(t) \
ˆ
Λ(t)
∞, if p(t) = p
max
∧ u
t
∈ U
2
t
0, else
4 HEURISTIC
4.1 Basic Ideas
Based on the definitions in 3 in every step the follow-
ing question has to be answered:
1. Should the first object in the channel or one of
the objects in the parking spaces be assigned to a
channel (U
1
) or
2. should the first object in the queue be parked (U
2
).
U
1
includes (p(t) + 1)m different actions which
lead to ((p(t) + 1)m)
n
0
possible states. In total
ResortingVehiclesinanAutomotiveManufacturingEnvironment
189

((p(t) + 1)m + 1)
n
0
states are possible. In this chapter
a heuristic is developed which, in every step, selects
only those states, which are locally likely to produce
a good solution.
Assume that the channels after every assign-
ment are renumbered so that ρ
η
1
1
(t) ≥ ρ
η
2
2
(t) ≥ ·· · ≥
ρ
η
m
m
(t).
Then the state
z
∗
t
= (λ
0
(t), p(t), (λ
1
(t), . . . , λ
p(t)
(t)),
(ρ
η
1
∗
1
(t), . . . , ρ
η
m
∗
m
(t))
dominates
z
t
= (λ
0
(t), p(t), (λ
1
(t), . . . , λ
p(t)
(t)),
(ρ
η
1
1
(t), . . . , ρ
η
m
m
(t))
if
∀ j ∈ M : ρ
η
j
∗
j
(t) ≤ ρ
η
j
j
(t)
∃ j ∈ M : ρ
η
j
∗
j
(t) < ρ
η
j
j
(t)
because z
∗
t
allows for at least one more assign-
ment, such as for instance of the element ρ
η
j
∗
j
(t) + 1.
It is now shown how actions can be chosen which
automatically lead to a state that dominates all other
possible states.
Lemma 5. For every λ
j
(t) ∈
ˆ
Λ(t) a channel µ
∗
j
exists
which is defined by the following equation:
d
µ
j
(t) = λ
j
(t) − ρ
η
µ
(t)
µ
˜
M = {˜µ ∈ M|d
j
(t) ≥ 0}
d
˜µ
1
j
(t) ≤ d
˜µ
2
j
(t) ≤ · · · ≤ d
˜µ
˜m
j
(t), ˜m = |
˜
M|
µ
∗
j
= ˜µ
1
(17)
Proof. If λ
j
is assigned to another possible channel
σ ∈ M with σ 6= µ
∗
j
and λ
j
− σ > 0 in t all top most
elements in t + 1 are the same as in t except the top-
most element of channel µ
i
which changes from ρ
η
σ
σ
to λ
j
and the top most element from µ
∗
j
stays ρ
η
µ
∗
j
µ
∗
j
.
When µ
∗
j
is selected ρ
η
µ
j
µ
j
changes to λ
j
and ρ
η
σ
σ
stays the same.
The states in t + 1 will only be different in the top
most elements ρ
η
σ
σ
and ρ
η
µ
∗
j
µ
∗
j
. All other elements re-
main the same.
Based on (17) the following holds:
λ
j
− ρ
η
σ
σ
≤ λ
j
− ρ
η
µ
∗
j
µ
∗
j
⇔ ρ
η
µ
∗
j
µ
∗
j
≤ ρ
η
σ
σ
which contradicts the required dominating state of z
∗
t
.
In the worst case we still have to assign one of
p(t) + 1 elements in every state.
With
e
λ
j
(t) = λ
j
(t) − ρ
η
µ
(t)
µ
ˆ
Λ(t) = { j
t
∈ Λ(t) ∧ ∃ µ ∈ M : j
t
> ρ
η
µ
(t)
µ
}
e
ˆ
λ
1
j
(t) ≤ e
ˆ
λ
2
j
(t) ≤ · · · ≤ e
ˆ
λ
l
j
(t), l = |
ˆ
Λ|
λ
∗
j
= e
ˆ
λ
1
j
(t) (18)
under the assumption of unlimited parking space
analog to Lemma 5 λ
∗
j
leads automatically to a domi-
nate state.
4.2 Implemented Rules
Based on these ideas the following heuristic is imple-
mented.
In the first iteration step p
max
is calculated using
1. Afterwards ∀z
t
∈ Z
t
is done:
1. Determine
ˆ
Λ(t) = { j
t
∈ Λ(t) ∧ ∃ µ ∈ M : j
t
>
ρ
η
µ
(t)
µ
}.
2. If
ˆ
Λ(t) =
/
0 at least one object causes infinitely
high cost when it is assigned to a channel and no
further analysis of the state is necessary.
3. If
ˆ
Λ(t) 6=
/
0 two actions are possible.
(a) Select λ
∗
j
(t) according to 18 and assign it to the
channel µ
∗
λ
∗
j
(t) determined by 17.
(b) Move the actual object to a parking space. If
p(t) = p
max
no parking space is available. Then
select λ
∗
j
(t) and µ
∗
λ
∗
j
(t) as in 3a. If λ
∗
j
(t) 6= ρ
κ
p(t+1) = p(t)-1 and parking is an option in the
next step.
Future work will analyze in detail if these rules at
least lead to local optimal states in every t = 1, . . . , T .
5 EXAMPLE AND DISCUSSIONS
For illustration the following example
is given. Assume 30 objects numbered
1,2,. . . ,30 arrive at a certain point in an ar-
rival queue according to a sequence ρ =
{5, 12, 20, 26, 30, 19, 29, 18, 14, 8, 13, 21, 2, 11, 25, 15,
22, 9, 17, 4, 3, 24, 28, 1, 6, 16, 10, 27, 23, 7}, 5 arriving
first and 7 arriving last. Three sorting channels
are available. According to lemma 2 the follow-
ing 28 parking candidates can be identified N
0
=
{5, 12, 20, 26, 30, 19, 29, 18, 14, 8, 13, 21, 2, 11, 25, 15,
22, 9, 17, 4, 3, 24, 28, 6, 16, 10, 27, 23}.
ICINCO2014-11thInternationalConferenceonInformaticsinControl,AutomationandRobotics
190

Table 1: Assignment strategies (corollary 1 and chapter 4).
ρ x
c
x
h
5 1 1
12 1 1
20 1 1
26 1 1
30 1 1
19 1 2
29 1 2
18 1 2
14 1 2
8 1 2
ρ x
c
x
h
13 2 2
21 2 2
2 2 2
11 2 2
25 2 2
15 2 2
22 2 2
9 2 3
17 2 3
4 2 3
ρ x
c
x
h
3 3 3
24 3 2
28 3 2
1 3 3
6 3 3
16 3 3
10 3 3
27 3 3
23 3 3
7 3 3
With Corollary 1 a limited number of A(30, 3) ≤
l
30
3
m
− 1 = 9 parking spaces can be determined.
In table 1 x
c
is an assignment strategy developed
based on the ideas in the proof of corollary 1 and x
h
an assignment strategy calculated using the heuristc
4 reducing the quantity of parked objects from 22 to
17. When an object is parked before it is assigned the
number of the assignment strategy is marked bold.
Based on Lemma 1 the ordered sequence
{1,. . . ,30} can be realized by taking the bottoms from
the channels and the original problem is solved.
6 CONCLUSIONS
The heuristic helps the factory planner to find a feasi-
ble solution for his assignment problem which in gen-
eral uses a smaller quantity of parked objects than the
intuitive assignment strategy of 1.
Ongoing research looks for additional improve-
ments. Furthermore it would be interesting to anal-
yse what a flexible layout (channel and parking space
quantity) should look like which supports most of
possible object permutations. Another question could
be how in practice existing channel and parking space
restrictions can be integrated in the model. Instead
of the current heuristic considered above, other so-
lution techniques such as Simulated Annealing or
Genetic Algorithms could be considered. However,
since those techniques usually provide less insight
into the ”mechanics” of a problem under investiga-
tion, the above approach was used.
REFERENCES
Chr
`
etienne, P. et al., editors (1995). Scheduling Theory and
its Applications. Wiley, Chichester.
Korte, B. and Vygen, J. (1995). Combinatorial Optimiza-
tion - Theory and Algorihms. Springer, Berlin.
Mergenthaler, W., Stadler, W., and Zimmer, N. (OCRI, Ot-
tawa, Ontario, Canada, 1994). Dynamic scheduling
and job assignment in an automotive manufacturing
process. The 10th ISPE/IFAC International Confer-
ence on CAD/CAM, Robotics and Factories of the Fu-
ture: CARS & FOF’94, M. Zaremba, pp 478-483.
Mergenthaler, W., Stadler, W., Zimmer, N., and Wilbertz,
H. (1995). Optimizing automotive manufacturing se-
quences using simulated annealing and genetic algo-
rithms. Control Eng. Practice, Vol. 3, pp 569-573.
Neumann, K. and Morlock, M. (1993). Operations Re-
search. Hanser, M
¨
unchen.
Reeves, C. R. (1996). Modern Heuristic Techniques, in:
V. J. Rayward-Smith, I. H. Osman, C. R. Reeves and
G. D. Smith: Modern Heuristic Search Methods. Wi-
ley, Chichester.
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