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International Conference on Industrial Engineering and Systems Management
IESM 2005, May 16 – 19, Marrakech (Morocco)
Minimizing work overload in mixed-model assembly lines
Joaquín Bautista, Jaime Cano and Jordi Pereira
ETSEIB, Universitat Politècnica de Catalunya
Av Diagonal 647, 7th floor, 08028 Barcelona, SPAIN.
{joaquin.bautista, jaime.cano-belman, jorge.pereira} @ upc.es
Abstract
A variant of the mixed-model sequencing problem on assembly lines is the one proposed
by Yano and Rachamadugu. They consider time windows in workstations and the work
overload concept (sum of differences between real completion time in the workstation and
due dates, which depends on the time window value). This paper proposes some
procedures for solving the problem, and compares them with others taken from literature
using a computational experience.
Keywords
Sequencing, Work Overload, JIT, Heuristics.
1. Introduction
Assembly lines are efficient means of making great amounts of products. Initially, using
single-model assembly lines, manufacturing companies were able to produce big quantities
of a product. Nowadays, the mixed-model assembly line makes it possible to manufacture
a product with many and small variations, avoiding important setups. In this way,
companies produce all the variations of the same product simultaneously and match the
production with the changing demands of clients, keeping inventories small. This
variability on the final product involves a great amount of optional extras that can have
different versions as well, therefore the production sequence of the products becomes
important and must be considered. Hence, after an assembly line has been balanced - work
has been spread out over the line stations- considering average processing times and
production mixes, the order in which units have to be introduced into the assembly line
must be considered by taking into account certain criteria.
This paper deals with the criteria of minimizing work overload. The problem considers
jobs with accomplishment times greater or smaller than the cycle time. Since the cycle
time is an average of all job times, there exist jobs with larger work content. If products
with large work content are introduced successively into an assembly line, work overload
can occur and work on products can be left unfinished, due to the limited spare-time
window. In automotive assembly lines, work overload can lead to the work being
unfinished and then having to be completed in the "nursing" area, to the assignment of
more utility workers, or like in the worst case, to the halting of a line so that the work can
be finished in the station, with the cost that it incurs.
Yano and Rachamadugu [9] considered the aim of minimizing the work overload caused
by the sequence and put forward a model of the problem. A case with a single station and
two types of products is studied and a solution procedure is given. Their procedure
constructs a subsequence composed by both kinds of products, which is repeated until it is
no longer possible. Afterwards they extend the procedure for K stations and two products,
computational complexity of which is O(KN2).
International Conference on Industrial Engineering and Systems Management
IESM 2005, May 16 – 19, Marrakech (Morocco)
Continuing with the case of one station and two products, Bolat and Yano [4] gave three
procedures: the first one determines a regenerative sequence, in which the positions of the
worker in the station at the beginning and at the end of the subsequence are the same. This
subsequence is repeated for building the solution until a new subsequence cannot be built.
The second procedure is a greedy algorithm that tries to avoid work overload. The last one
is also a greedy algorithm, which minimizes the idle time. The common objective of the
algorithms is to minimize the work overload.
Later on [5], Bolat and Yano propose as an objective, the minimization of the number of
total special units exceeding the k permitted in sequential parts of size l.
Tsai [8] extends the problem considering the displacement time the worker needs for going
from one finished unit to the next. The author also establishes two objectives: 1) to
minimize total worker displacement and 2) to minimize the unfinished work in the station.
The proposed procedure gives an optimal solution under certain conditions and has
complexity O(logN). When the problem is extended to more than one station, the
aforementioned algorithm is used for determining bounds.
In this paper, we focus on the aforementioned procedures of Yano and Rachamadugu [9],
Bolat and Yano [4][5], Tsai [8], and we extend procedures for the case of one station and
two products, minimizing the work overload. It is organized as follows: In section §2, a
formulation of Yano and Rachamadugu [9] for measuring work overload is mentioned.
Section 3 deals with procedures for one station and two products, one taken from
bibliography and a proposal inspired on Bolat greedy procedures. In §4 we consider more
than two products in one station, and extensions of procedures from section §3 are
proposed. The case with more than two products and multi station is considered in §5.
Section §6 contains computational experiments and results.
2. A work overload formulation
Yano and Rachamadugu [9] proposed a formulation to schedule the work on jobs, in order
to minimize work overload for a given sequence. It is assumed that one or more workers
can be at a station, and the saturation of work in stations is not necessarily complete. It is
also assumed the product moves on a paced belt through the stations. New job gets into the
line at a constant rate (cycle time) and the work on it must be done only when it is inside
the station so that the worker in the station does not disrupt the next one. Thus, given a set
of jobs, its respective processing times and limited station lengths, the order in which jobs
must be introduced into the line has to be determined, minimizing the uncompleted work
(work overload). Since minimizing total work overload is equivalent to maximizing total
completed work, Yano and Rachamadugu establish as an objective function the latter. Let
pik processing time for job i,(i=1,...,N) in station k,(k=1,..,M),
ujk amount of time allocated to job in period j, in station k,
sjk starting time for job in period j, in station k,
Lk length of station k, and
bk number of workers at station k.
In the model below, xij is the binary variable which takes the value of 1 if the job i is
assigned in position j; 0 otherwise. L could also be understood as the number of productive
units in any moment inside the work station, or the maximum time granted for a product in
International Conference on Industrial Engineering and Systems Management
IESM 2005, May 16 – 19, Marrakech (Morocco)
the station - by station, we also mean time window. Despite this being the nomenclature
given by [9], in the next section we have modified some of it.
A mixed integer linear programming is formulated for a set of stations and a set of jobs.
∑
∑jjk
kkubmax (1)
s.t. 1=
∑
jij
xi
∀
(2)
1
=
∑
iij
xj
∀
(3)
ij
iikjk xpu ∑
≤kj,
∀
(4)
1
−
≥jsjk kj,
∀
(5)
kjkjjk uss ,1,1 −−
+
≥ kj,
∀
(6)
kjkjk Ljus
+
−
≤
+1 kj,
∀
(7)
{
}
1,0,0,0
∈
≥≥ ijjj xsu (8)
The objective function maximizes the completed work. Equations (2) and (3) ensure each
job has been allocated and that only one job has been assigned in the jth position,
respectively. Constraint (4) indicates assigned time in position j is at most the processing
time of work assigned in this position. The constraints (5), (6) and (7) control the jobs, so
that they can be started only when the work is inside the station, whether the previous work
has been completed or it has left the station unfinished. Non negative and binary
restrictions are in (8). The size of the problem increases exponentially with the number of
jobs and stations. In [9], authors focused on a special case of the single station where two
kinds of products with alternative processing times.
3. Single station procedures
Consider two kinds of products with different processing times in the station: basic
products (which we call B) with processing time inferior to cycle time (interarrival time of
units into the station); and special products (hereafter called A) with load superior to cycle
time. This can be seen easily in car assembly lines, when a unit may or may not receive an
optional part at a station, such as Global Positioning Systems, Computer, etc. In this paper,
we consider stations to be closed, that is, the work on products can be done only when it is
inside the station boundaries. We assume this through the consideration of various
situations in car assembly lines: a worker can not invade the next station to finish his work
because it may interfere with the job of the next worker; when robots or tools are used,
work can not be done beyond the reach of the automaton; in some cases, line stations make
use of the so-called “pirate cart”, which is a mobile structure that travels with the product
from the lower boundary to the upper, solely within the station limits, containing the tools,
equipment and components the worker needs for doing his job. Therefore, consecutively
long subsequences of products, which demand a greater level of work, are not desirable
because work overload may be produced. Without loss of generality, the cycle time is
assumed to be the unit time. The following notation is used:
ta processing time for a unit A,
International Conference on Industrial Engineering and Systems Management
IESM 2005, May 16 – 19, Marrakech (Morocco)
tb processing time for a unit B,
na number of units A to be sequenced,
nb number of units B to be sequenced, and
L maximum time granted for processing any unit in the station.
The unit is the cycle time, thus, tb<1<ta, ta, tb and L, are expressed in cycles. Our objective
is to find a sequence that allows the completion of the maximum number of jobs, so that
the total lost work (also work overload) at the station is the minimum possible. We will
proceed showing the procedures in literature and then a new proposal.
3.1 The Yano and Rachamadugu procedure
The method of Yano and Rachamadugu [9] is based on the repetition of a stable
subsequence, which is composed by ma units A, and mb units B. The idea is to repeat the
subsequences until there are not enough units to complete more. Then, the remaining units
are allocated in such a way that the work overload produced is minimized. The
subsequence is determined by considering that processing times for products with options
ta are bigger than the cycle time, and therefore there is a limit on the number of units A that
can be consecutively sequenced, without causing work overload. This limit (X) is the
maximum integer satisfying X ≤ (L-1)/(ta -1), and is also the maximum value ma can take.
The procedure tries to regenerate, that is, to bring the worker back to the beginning of the
window (its assumed original position), after a cycle composed by ma units A and mb units
B. The subsequence of ma and mb units, can be determined as follows:
babbaa mmtmtm +=
+
ma <= X, mb integers (9)
If the equality (9) is satisfied, a perfect regeneration is reached, and the problem is
optimally solved using this method. Prove of optimality can be seen in appendix C of Yano
and Rachamadugu [9]. Under the supposition that there are integer values for ma and mb,
maximum utilization is achieved solving the next nonlinear MP model:
(
)
(
)
integers0,
..
max
≥
+≤+
≤
+
+
ba
babbaa
a
babbaa
mm
mmmtmt
Xmts
mmmtmt
For a given value of ma, maximum utilization is reached when the smallest integer value of
mb satisfies (10): ma ta + mb tb ≤ ma + mb
(10)
Under the above conditions, the desired sequence is obtained with this procedure
(henceforth called YR) as follows. Let ma the number of units A in a cycle, mb the number
of units B in a cycle, and C the number of cycles composed by ma units A, and mb units B.
1. Determine the number of consecutive cycles consisting of ma and mb units,
⎣⎦⎣
(
⎦
bbaa mnmnC ,min=
)
, where
⎣
⎦
x is the largest integer less than or equal to x;
2. Determine the number of units A that follows the previous C cycles,
()
aaaa mmCnx ,min
⋅
−= ;
International Conference on Industrial Engineering and Systems Management
IESM 2005, May 16 – 19, Marrakech (Morocco)
3. Determine the number of units of kind B that follows the previous xa units A,
; and
bbb mCnx ⋅−=
4. If necessary, determine the quantity of units A that follows the xb units B,
.
aaaa xmCnx −⋅−=
'
3.2 Proposed procedure
With regarding to the idea of regeneration, we propose a heuristic algorithm (henceforth
called Ud1) that builds pseudo cycles. With this, we firstly try to avoid the priority given to
a specific kind of job. Secondly, we try to follow the regularization idea. In this way, it is
not necessary to repeat a subsequence exactly, and products can be scheduled taking into
account the remaining production (and, if necessary, can be weighted). Nowadays,
assembly companies wish not only to regularize production of products but also to aid in
the reduction of work overload. If priority in scheduling is given to a product with large
work content, products with low work content will be postponed, producing idle time at the
end of the sequence, which also can produce incomplete work. In fact, the avoiding of
idleness helps to reduce overload [9]. On the other hand, if priority is given to products
with low work content, jobs with work content larger than the cycle time will be delayed
and will cause more work overload than is necessary.
The constructive idea of the procedure tries to cause the worker displacements in a cyclic
way (up and down), from the lower to upper limit of the window, avoiding if possible,
ether overload or worker idle time. When both magnitudes are bigger than zero, the smaller
value must be chosen. Magnitudes can be weighted. Let woAt and woBt the work overload
caused by the assignment of a product A, and the idle time caused by the allocation of a
product B (respectively) in period t of the sequence. Thus, given an instance [na , nb , ta , tb,
L], apply the following procedure:
while (na > 0 and nb >0) Do
aux=0
while woAt = 0, sequence(t) ← A, aux=aux+1, na= na-1
while woBt = 0, sequence(t) ← B, aux=aux+1, nb= nb-1
if aux=0
if woAt < woBt
sequence(t) ← A, na= na-1
else
sequence(t) ← B, nb= nb-1
end if
Since the required time for producing all jobs can be known, a lower bound for either
unavoidable work overload (C) or unavoidable idle time (K) can be obtained as follows
with C=max{0,na t
a + nb t
b - [na + nb + L - 1]} and K=max{0,N-(na t
a+nb t
b)},
respectively.
4. Multi product procedure
In previous sections, the approach of the problem assumes that products needing to be
assembled are characterized by the presence or absence of an extra optional job in the
station. Nevertheless, there are products in industry, such as the automobiles, which
present a great variety of versions of the same product. Each one of these options can be
International Conference on Industrial Engineering and Systems Management
IESM 2005, May 16 – 19, Marrakech (Morocco)
distinguishable between others by their work content, so that the approach of the problem
we have deal with above, must consider more than two jobs on products differentiable by
processing times. From this point on, we assume each product, distinguishable by option or
attribute, has a different processing time, and even when the product does not need any
extra option to be completed within a station, it requires a minimum amount of work.
Considering a single station and various products, we propose an extension of the
algorithm of Yano and Rachamadugu described in section 3.1, and furthermore an
extension of Ud1. We name them YR-ex and Ud1-ex, respectively.
4.1 Extension of the Yano and Rachamadugu procedure
Whit regards to the idea of repetitive cycles again, the following algorithm (YR-ex) consists
of finding subsequences composed by mi units of product i(i=1,…,P). While there are
enough unscheduled units, the subsequence is repeated. Otherwise, a new subsequence is
calculated with the remaining products. Let
P the number of total different products i (i=1,…,P),
ni the quantity to be produced of product i,
N the total of products to be assembled,
ti the processing time for a product of class i,
A the set of products with processing time bigger than the cycle time(ti >1),
B the set of products with processing time smaller than the cycle time (ti ≤1), and
xi the maximum consecutive quantity of product i that can be scheduled without
causing overload or idle time. Where
∑
=
=P
ii
nN 1
The subsequence can be found solving iteratively the next MP model:
im
immt
idxmst
mmt
i
P
ii
P
iii
iii
P
ii
P
iii
∀≥
∀≤
∀≤
∑∑
∑∑
==
==
0
},min{.
max
11
11
xi⋅ti ≤ xi-1+L, ∀i∈A, and xi⋅ti ≤ 1-L-xi, ∀i∈B. di represents the remaining production of
product i after determining a subsequence and repeats it conveniently in the sequence we
are building. After a subsequence has been found, it is necessary to schedule all the mi
products that are composing the current subsequence. To arrange products in the
subsequence, we consider alternatively the set A, and then the set B. The order in which
units will be incorporated into the sequence is done according the diminishing value of
index ri=mi|1-ti|. Thus, from the set A, the item with the bigger index (let k) is selected and
the mk products k are consecutively assigned. Then, from the set B, the item with the bigger
index ri is selected (let l), and the ml corresponding products l are assigned; and so on.
4.1 Ud1 procedure extension
An extension of procedure Ud1 is proposed in this subsection. This heuristic algorithm
(which henceforth is called Ud1-ex) also considers one station but differentiates two sets of
products, the sets A and B containing products with processing time bigger and smaller
respectively than cycle time (which is taken as the unit time). The procedure tries to favour
International Conference on Industrial Engineering and Systems Management
IESM 2005, May 16 – 19, Marrakech (Morocco)
the movement of the workers from the lower to the upper limit of its station. Doing this in
two cyclic steps, we attempt to regenerate. It means, in the first step products from set A,
and in the second, from set B are allocated, without going beyond the station limits. For
deciding the kind of product to sequence in the period t, we make use of a dynamic index ri
= ni |1-ti|, where i belongs to the set A or B considered in the current stage t. Since ri
depends on the pending production of product i, it must be updated in each phase. The
product of the set under analysis with the bigger index ri is selected to be assigned in
period t. Thus, assuming the worker starts the job on the first product (t=1) at the beginning
of the station, pos(t)=0, while products are pending of been sequenced, next steps are
repeated:
Step 1 while (wi = 0) Do ∀ i∈A: sequence(t) ← i / ri =max(ri ); update pos(t+1), and di
Step 2 while (wi = 0) Do ∀ i∈B: sequence(t) ← i / ri =max(ri ); update pos(t+1), and di
Where
[
]
0,)(max)( Lttpostw ii −+= , ∀ i∈A, and
[
]
0,1)(min)( −+
=
ii ttpostw ∀ i∈B. A
product will be a candidate only if is has pending production.
5. Multi product and multi station procedure
In this section we approach the problem with multi stations and multi products. The
procedures for one station and multi products mentioned in the above section, are taken to
be used with the multi station procedure of [9], with the intention of getting the lower
bounds of work overload a candidate can produce later in the sequence if it is allocated in
the current period. The sequence is built progressively in such a way that for each
sequencing instant t, and for each product with remaining production, the unit with the best
overload predictor is selected and assigned in the tth position of the sequence. Let
M number of stations (j=1,...,M) in the assembly line,
P total number of different products (j=1,...,P) to be assembled,
N total of units to be assembled,
woi(j) work overload prediction caused in station j from current period on, assuming a
product i is assigned in current position in the sequence,
wi(j,t) work overload caused in station j in period t if a product i is assigned in this
position,
pos(j,t) initial position of the worker in station j in period t, expressed in cycle time units,
c(j) weight associated with station j,
sci(t) total work overload a product i produces if it is assigned in period t of the sequence.
The algorithm proceeds as follows:
For t=1 To T
For i=1 To P
suppose sequence(t) ← i, [ni(t)-1]
determine the bound
[
]
)(),()()( 1jctjwjwotsc M
jiii
∑
=+=
reestablish ni(t)
next i
sequence(t) ← i* / sci*(t)= min{sci(t)}
next t
International Conference on Industrial Engineering and Systems Management
IESM 2005, May 16 – 19, Marrakech (Morocco)
The work overload prediction woi(j) is computed with the two one-station procedures for
comparing the results obtained with each. We use YR-ex procedure for getting a bound,
and Ud1-ex for obtaining an estimation of the work overload, considering the remaining
production in period t, [ni(t), nk(t), with k≠0]. If product i is allocated within period t of the
sequence, the work overload caused in the current period in each station, is obtained as
follows: wi(j,t)=min[max[pos(j,t)+ti(j)-L(j),0],ti(j)] ∀ i ∈ A (11)
wi(j,t)=min[max[1-pos(j,t)-ti(j),0],ti(j)] ∀ i ∈ B (12)
Since a product can have greater or smaller processing time than cycle time, it can cause
work overload or idle time. Then, for each kind of set, A and B, we measure the effect of
the assignment of the product in the position t with (11) and (12). With it also the
unproductive time is taken into account as well. It is assumed for t=1, and ∀j , pos(j,t)=0.
6. Computational experience
For this experiment, we apply the procedure for multi-stations and multi-products
mentioned in §5. Multi-stations procedure use one-station procedures for finding the best
candidate to schedule in position t of the sequence. The one-station algorithms we use in
the experiment are Ud1-ex and YR-ex. Both procedures consider multiple products. In each
period t of the sequence, when YR-ex procedure is used, one obtain a bound of the work
overload, on the other hand, when Ud1-ex is used, we get a predictor. We solve a total of
225 instances of the following characteristics: (a) All problems present 4 different products
and 4 stations; (b) 45 production programs (production mixes) are considered. Production
mixes are grouped in five different blocks; (c) 5 different processing times structures in
stations are considered; and (d) each instance is built combining one production program
of (b), and one processing time structure of (c). For further details about the production
programs and the processing times structures, see appendix. Fist, we used two quality
indicators in the experiment: the average percent deviation (
δ
eb) and the average of
confirmed optimums (Oeb), by processing time structure (e), by mix blocks (b), and for the
total of problems. To obtain the index O, we calculate the work overload bound C as
follows:
[
]
{
}
∑
∑
∑
===
−−−= M
j
P
i
P
ij
j
i
j
i
j
i
j
iLtntnC 111
1,0max
Where P is the total of different products built in on the assembly line (i=1,…P). In table
number 1, results had been obtained using Ud1-ex for determining the predictor of lost
work in station j. When the algorithm YR-ex was used for determining a bound of work
overload, the results in table 2 are obtained.
Ud-1-ex
δ
eb (%) Oeb (%)
Block Block
Str 1 2 3 4 5
δ
e1 2 3 4 5
Oe
1 0,742 0,578 0,349 0,467 0,557 0,54 25,0 0,0 0,0 0,0 0,0 5,00
2 3,686 2,344 1,597 1,874 2,038 2,31 0,0 0,0 0,0 0,0 0,0 0,00
3 1,701 1,352 1,249 1,546 1,846 1,54 50,0 16,7 0,0 0,0 0,0 13,33
4 0,049 0,401 0,612 0,262 0,336 0,33 50,0 16,7 0,0 0,0 0,0 13,33
5 1,028 1,517 1,055 1,012 1,244 1,17 25,0 0,0 0,0 0,0 0,0 5,00
δ
b1,44 1,24 0,97 1,03 1,20 1,178 30,00 6,67 0,0 0,0 0,0 7,333
Table 1. Results of the multi station experiment, with Ud1-ex algorithm
International Conference on Industrial Engineering and Systems Management
IESM 2005, May 16 – 19, Marrakech (Morocco)
Y&R -ex
δ
eb (%) Oeb (%)
Block Block
Str 1 2 3 4 5
δ
e1 2 3 4 5
Oe
1 0,730 0,583 0,338 0,447 0,522 0,52 25,0 0,0 0,0 0,0 0,0 5,00
2 2,591 1,768 1,585 1,754 1,873 1,91 25,0 16,7 0,0 0,0 4,2 9,17
3 1,738 1,292 1,194 1,502 1,577 1,46 50,0 16,7 0,0 0,0 8,3 15,00
4 0,128 0,630 1,225 0,787 0,717 0,70 0,0 0,0 0,0 0,0 0,0 0,00
5 0,712 1,641 1,095 1,158 1,410 1,20 50,0 0,0 0,0 0,0 0,0 10,00
δ
b1,18 1,18 1,09 1,13 1,22 1,160 30,00 6,67 0,0 0,0 2,5 7,833
Table 2. Results of the multi station experiment, with YR-ex algorithm
We also compare the solutions obtained when each procedure is used in the multi-stations
procedure: Ud1-ex and YR-ex. Table 3 shows percentage of times (G) each procedure
obtained a better solution than the other procedure. By processing time structures (e) and
by mix blocks (b), better results are obtained when Ud1-ex is used in the determination of
the best candidate i to schedule in the position t of the sequence.
Winner Algorithm by Block (%) Winner Algorithm by Structure (%)
Block Structure
Proc. 1 2 3 4 5
Gi1 2 3 4 5
Gi
Ud1-ex 65,00 56,67 57,14 70,00 55,00 60,76 44,44 42,22 42,22 86,67 73,33 57,78
YR-ex 70,00 53,33 48,57 40,00 47,50 51,88 66,67 68,89 64,44 15,56 33,33 49,78
Table 3. Winner Algorithm
7. Conclusion
This paper treats with a variant of the problem of sequencing products (mixed models) on a
paced assembly line. We consider the approach in which a product demands a component
(attribute), which has different versions, and requires different processing times in the
application of each. The aim of these procedures is to minimize work overload (lost work)
in all the stations of the assembly line due to the limited time spared in the stations and to
the work loads along a given sequence. Both boundaries of stations are closed, and we
assume that, the displacement time the worker need to go from one product in to the next,
is negligible. We compiled some procedures founded in literature: Yano y Rachamadugu
[9], Bolat y Yano [4] y [5], and we propose the procedure Ud1 and Ud1-ex, based on the
previous procedures. Procedures considering more than two different products and multi
stations are also proposed. The experiment considers 225 problems with for stations and
four different products. Results obtained dealing with the deviation in relation to a bound,
are highly satisfactory, as well for the totality of processing time structures, as the blocks
of programming production.
Acknowledgements
We acknowledge Nissan Motor Ibérica as well as the UPC Nissan Chair for partially
funding this research work as well as providing real data. This research paper has also been
partially funded by DPI2004-03475 grant from the Spanish government.
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International Conference on Industrial Engineering and Systems Management
IESM 2005, May 16 – 19, Marrakech (Morocco)
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Appendix
The next tables, referring to experiment 3, contain 45 production programs grouped in 5
blocks. Block 1 has a production mix with a high demand of one kind of unit; the second
block shows two products with greater demand; block 3 has a balanced mix; one unit with
low demand is given in block 4; and the last block presents a balanced mix.
Production Programs
Block-1 Block-2 Block-3 Block4 Block-5
i
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
1 13 1 1 1 7 7 7 1 1 1 5 5 5 3 3 3 4 5 5 5 1 1 1 1 1 1 1 3 3 3 3 3 3 5 5 5 5 5 5 7 7 7 7 7 7
2 1 13 1 1 7 1 1 7 7 1 5 3 3 5 3 3 4 5 5 1 5 3 3 5 5 7 7 1 1 5 5 7 7 1 1 3 3 7 7 1 1 3 3 5 5
3 1 1 13 1 1 7 1 7 1 7 3 5 3 5 5 4 4 5 1 5 5 5 7 3 7 3 5 5 7 1 7 1 5 3 7 1 7 1 3 3 5 1 5 1 3
4 1 1 1 13 1 1 7 1 7 7 3 3 5 3 5 4 4 1 5 5 5 7 5 7 3 5 3 7 5 7 1 5 1 7 3 7 1 3 1 5 3 5 1 3 1
Table 4. Production Programs for experiment
The table below displays the processing times for products in the different stations. The
first structure presents a similar time to the cycle time; structure two has processing times
distant of cycle time; structure three has unbalanced times between the first two and the
last two stations; structure 4 contains two products that cause overload in all stations;
lastly, overload problems within structure 5 relate to each station with a different product.
Processing times for units in stations
Structure 1 Structure 2 Structure 3 Structure 4 Structure 5
Station Station Station Station Station
i
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
1 0.92 1.03 1.01 0.95 0.91 1.20 0.90 1.00 1.11 1.20 0.85 0.82 1.13 1.19 1.15 1.16 1.15 0.99 1.04 0.96
2 0.97 0.98 1.05 1.04 0.80 1.05 1.13 1.07 1.14 1.13 1.00 0.94 1.14 1.13 1.12 1.18 1.04 1.19 1.00 1.02
3 1.03 1.04 0.99 0.96 1.07 0.88 1.17 0.86 0.83 0.85 1.15 1.19 0.82 0.85 0.84 0.87 0.89 0.98 1.14 0.87
4 1.08 0.95 0.95 1.05 1.14 0.87 1.00 1.14 0.98 0.87 1.10 1.15 0.95 0.87 0.94 0.81 0.95 0.87 0.85 1.18
L(j) 1.08 1.05 1.06 1.06 1.15 1.20 1.20 1.15 1.15 1.20 1.15 1.20 1.15 1.20 1.15 1.20 1.15 1.20 1.15 1.18
c(j) 10 9.5 9.9 9.9 9 9 9.5 10 9.5 10 9 10 10 9.5 9 10 9.5 9.5 9 10
Table 5. Processing times structures for experiment