Hard Special Case and Other Complexity Results
for SALBP-1
Evgeny R. Gafarov
Ecole Nationale Superieure des Mines, FAYOL-EMSE, CNRS:UMR6158, LIMOS,
F-42023 Saint-Etienne, France,
Institute of Control Sciences of the Russian Academy of Sciences,
Profsoyuznaya st. 65, 117997 Moscow, Russia,
email: [email protected]
Alexandre Dolgui
Ecole Nationale Superieure des Mines, FAYOL-EMSE, CNRS:UMR6158, LIMOS,
F-42023 Saint-Etienne, France
email: [email protected]
Abstract
For the well-known simple assembly line balancing problem
(SALBP-1), we propose a special case for which there is no Branch
and Bound algorithm with polynomial Lower Bound which can solve
instances even for n = 60 operations in appropriate time. Additionally, a SALPB-1 problem with opposite optimization criterion, namely,
maximization of stations used, is considered.
Keywords: Line balance, Number of stations, Complexity results.
Introduction
We consider a single-model paced assembly line (see [1] for definitions
and terminology) which continuously manufactures a homogeneous
product in large quantities (mass production). The simple assembly
line balancing problem (SALBP-1) is to find an optimal line balance
for a given cycle time c, i.e., to find a feasible assignment of the given
operations to stations in such a way that the number of stations used
m is minimized. The SALBP-1 is formulated as follows.
Given a set N = {1, 2, . . . , n} of operations and K stations (machines) 1, 2, . . . , K. For each operation j ∈ N a processing time tj ≥ 0
is defined. The cycle time c ≥ max{tj , j ∈ N } is given. Furthermore, finish-start precedence relations i → j are defined between the
operations according to an acyclic directed graph G. The objective is
to assign each operation j, j = 1, 2, . . . , n, to a station in such a way
that:
- number m ≤ M of stations used is minimized;
1
- for each station k = 1, 2, . . . , m a total load time j∈Nk tj does not
exceed c, where Nk – a set of operations assigned to a station k;
- given precedence relations are fulfilled, i.e. if i → j, i ∈ Nk1 and
j ∈ Nk2 then k1 ≤ k2 .
Surveys on results for NP-hard in the strong sense SALBP-1 are
published periodically (e.g. [1]). Different generalizations and solution
algorithms for the problem are considered, e.g., in [1, 2, 3, 4, 5, 6, 7].
There exists a special electronic library http://www.assembly -linebalancing.de of experimental data to test solution algorithms for the
problem. The best known Branch and Bound algorithm is presented
in [8].
The rest of the paper is organized as follows. In the first section we
present a special case for which all known Branch and Bound (B&B)
algorithms with any polynomial time computed Lower Bound have exponential running time and the best known algorithm [8] is not able to
solve some instances of this special case even for n ≥ 60 in appropriate
time. A simple assembly line balancing problem, where the number of
machines used is maximized on solutions with maximal station loads,
is considered in Sections 2. The maximal number of stations computed
for benchmark instances is presented in Section 3. A modification of
the graph of precedence relations to planar one and the benefits of such
the modification are proposed in Section 4.
1 Running Time of Branch and Bound Algorithms for a Special Case
Partition problem. Given is a set N = {b1 , b2 , . . . , bn } of numbers
b1 ≥ b2 ≥ · · · ≥ bn >0 with bi ∈ Z+ , i = 1, 2, . . . , n, and a number
A ∈ Z+ with A <
j∈N bj . Is there a subset N ⊂ N such that
j∈N bj = A?
The worst case running time of B&B algorithms for the well-known
Knapsack Problem is analyzed, e.g., in [10, 11]. In these papers authors
choose to use only special cases, for which it is fairly easy to find
an optimal solution with a B&B algorithm. However to prove its
optimality, almost all feasible solutions should be considered.
In [11], the following special case of the Knapsack Problem was
presented. Here, we have a set of numbers B = {b1 , b2 , . . . , bn } which
are used in the following parameterized optimization instance:
2
⎧
m+n
⎪
⎪
f (x) =
cj xj → max
⎪
⎪
⎪
j=1
⎪
⎪
m+n
⎪
⎪
⎪
cj xj ≤ ka + 1,
⎨
j=1
⎪
a ≥ 2,
⎪
⎪
⎪
⎪
cj = a, j = 1, 2, . . . , m,
⎪
⎪
⎪
⎪
⎪
⎩ cj = bj · a, j = m + 1, 2, . . . , m + n,
xj ∈ {0, 1}, j = 1, 2, . . . , m + n.
(1)
For the special case all B&B algorithms1 have an exponential number of nodes in a search tree, unlike P = N P [11], i.e. the complexity
n+3/2
time of all B&B algorithms is exponential and equals to ∼ 32 · √2
π(n+1)
[11]. In fact, in (1) there is a sub-instance which is an instance of the
Partition Problem with numbers B = {b1 , b2 , . . . , bn }, i.e. to compute
an "usefull" Upper Bound we have to solve the Partition Problem,
which is NP-hard. As a consequence, if we use a polynomial Upper
Bound the reduction of nodes by rule "a local upper bound ≤ current
record" will be ineffective.
We use a similar idea to construct a special case of SALBP-1 for
which each B&B algorithm with no matter what polynomial time computed Lower Bound has an exponential run time. This makes algorithms ineffective for instances with n ≥ 60 operations.
The following reduction from the Partition Problem to the special
case of SALBP-1 is used. Let us modified an instance of the Partition
problem as follows:
Modified instance of the Partition problem. Given is a set N =
{b1 , b2 , . . . , b2n } of numbers b1 ≥ b2 ≥ · · · ≥ b2n> 0 with bi ∈ Z+ , i =
1, 2, . . . , 2n, and a number A ∈ Z+ with A < j∈N bj . The numbers
bi , i = 1, 2, . . . , 2n are denoted as follows:
⎧
⎨ b2n = 1, n
b2i = 2 · j=i+1 b2j−1 , i = n − 1 . . . , 1,
(2)
⎩
b2i−1 = b2i + bi ,
i = n . . . , 1,
where
b1 , b2 , . . . , bn – numbers from the initial instance. Let
nA =
n
of generality let us assume A = 12 i=1 bi
i=1 b2i + A. Without lost
2n
and as consequence A = 12 i=1 bi . The question is: "Is there a subset
N ⊂ N such that j∈N bj = A"? It is obvious that the modification can be done in polynomial time. If for the initial instance of the
Partition Problem the answer is "YES" (and the same answer has the
i
modified instance) then N contains one and only one number b from
1
where Upper Bounds computed in polynomial time are only used
3
each pair {b2i−1 , b2i }, i = 1, 2, . . . , n. If the number bi is included in
the set N then b2i−1 is included in N , otherwise the number b2i ∈ N .
In the special case of SALBP-1 there are 2n operations. Let
w = min{w|10w ≥ 2A}. Processing times of operations are defined as
follows:
ti = 10w + bi , i = 1, 2, . . . , 2n.
2n
Furthermore, c = 12 i=1 ti . There are no precedence relations between operations.
It is obvious that if and only if for the modified instance of the
Partition Problem the answer is "YES" then the minimal umber of
stations m∗ = 2, otherwise m∗ = 3. As a consequence, if N P = P ,
there is no polynomial time computed Lower Bound with a relative
error equal or less than 32 . That means, for any set of polynomial time
computed Lower Bounds {LB1 , LB2 , . . . , LBX }, there is a modified
instance of the Partition Problem with an answer "NO", for which
LBx = 2, i = 1, 2, . . . , X, although m∗ = 3. For the special case
of SALBP-1, any feasible solution is optimal. However, to prove its
optimality almost all feasible solutions must be considered.
Let us estimate the possible number of feasible solutions. On the
first station there could
2n be processed at least n − 1 operations. Thus,
possible loads of the first station, i.e. the
there are at least n−1
number of feasible solutions which have to be considered is greater
than
2n
n + 1 22n
n + 1 2n
·√ .
≈
=
n
n
n
n−1
nπ
To solve such the instance of SALBP-1 with 2n = 60 a computer must
60
perform more than 210 operations. Let us assume that the fastest
known computer performs 230 operations per second, or less than 247
operations per day. Then a run time of an algorithm will be more than
213
10 > 800 days! That means there are instances of SALBP-1 for which
any B&B algorithm with polynomial time computed Lower Bounds
has an unappropriate running time.
Let us us consider the best known B&B algorithm from the paper
[8]. This algorithms solves all benchmark instances published online
on http://www.assembly -line-balancing.de in less than one second per
instance. Let us analyze some of methods which are used in the algorithm to reduce the search tree.
1. Lower bounds. LB1 = cti , LB2 = |{j ∈ N |tj > c/2}| +
|{j∈N |tj =c/2}|
. LB3 is computed by assigning a weight wj to
2
each operation j:
4
wj =
⎧
1
⎪
⎪
⎨ 2
⎪
⎪
⎩
3
1
2
1
3
if
if
if
if
tj > 2c/3,
tj = 2c/3,
c/3 < tj < 2c/3,
tj = c/3.
(3)
then LB3 = wj . For the special case of SALBP-1 LB1 = 2
and LB2 = LB3 = 0, i.e. all of these Lower Bounds are of no use
for the proposed special case.
2. In the algorithm, to solve subproblems a Bin-Packing B&B algorithm is used. For the considered special case such the algorithm
for the Bin-Packing problem has an exponential running time as
well, i.e. with such the technique the search tree will not be
reduced substantially.
3. The authors use a modification of breadth-first strategy of
branching, which could be unappropriated for this special case
in view of the amount of memory required.
We can conclude the following. Despite the algorithm [8] solves all
benchmark instances in less than 1 second per instance, known B&B
algorithms for SALBP-1 remain exponential and can not solve some
instances with the size n > 60 in an appropriate time. That is why we
consider exact algorithms for the general case of the problem unpromising. Researchers can concentrate on special cases or on essentially new
solution schemes.
2
Maximization of Number of Stations
To propose an essentially new solution scheme for SALBP-1, it is necessary to investigate properties of optimal solutions. We can investigate
not only properties of good solutions to try imitate their character
but properties of poor solutions as well to avoid solutions with their
aspects. In this Section in contrast to standart SALBP-1, where the
number of stations used should be minimized we consider an optimization problem with the opposite objective criteria, in other words the
maximization of the number of stations.
On the one hand, the investigation of a particular problem with
the maximum criterion is an important theoretical task. Algorithms for
such a problem with the maximum criterion can be used to cut bad subproblems in the branching tree of B&B algorithms, to compute Upper
and Lower Bounds for a bi-criterion problems [13]. On the other hand,
such problems have also practical interpretations and applications[12,
14, 15]. Moreover a situation arises when the company is considered as
a customer, and one wants to know the worst variant of a line balance,
which is computed in a ‘black box’ (e.g. in a plant).
5
To make the maximization problem not trivial we assume that all
stations (instead the last one) should be maximal loaded, i.e. for two
stations m1 , m2 , m1 < m2 there is no operation j assigned on the
station m2 which can be assigned on station m1 without violation of
precedence constraints or the feasibility’s condition "total load time
of the station does not exceed the cycle time". Such solutions with
maximal station loads we call active solutions.
Usually in solution algorithms or algorithms of computing of Upper
Bounds, such active solutions are considered. Let m – a number of
stations for a feasible solution with maximal station loads, mmin – the
minimal number of stations and mmax – the maximal number on such
the solutions. It is known [9] that mm
min < 2 and as a consequence
mmax
<
2.
mmin
Denote the maximization problem by max − SALBP − 1.
Firstly, we prove that max-SALBP-1 is an NP-hard problem.
Lemma 1 max-SALBP-1 is NP-hard.
Proof. Use a reduction from the Partition problem. Given an instance
of the Partition
problem with n numbers. Without lost of generality
assume A = 12 ni=1 bi . Let M = 2n2 A. Construct an instance of maxSALBP-1 with 2n + 1 operations, where tj = M + bj , j = 1, 2, . . . , n,
and tj = M, j = n+1, . . . , 2n+1. In addition, let c = (n+1)M +A−1.
Let N1 , N2 , N3 be sets of operations assigned on the first, second
and third stations respectively. The first station should
be maximal
loaded. Thenn ≤ |N1 | ≤ n + 1 and c − M + 1 ≤ j∈N1 tj ≤ c. If
c − M+ 1 > j∈N1 tj then there is an operation i ∈ N2 , ti = M such
that j∈N1 tj + ti < c + 1, i.e. the first station is not maximal loaded.
If and only if for the instance of the Partition Problem the answer is
"YES", then there exists an optimal solution with 3 stations, where N1
contains |N | jobs which corresponds to numbers from N and n − |N |
jobs from the set {n + 1, . . . , 2n + 1}, N2 contains N − |N | operations
which corresponds to numbers from N \ N and n − |N | operations
from the set {n + 1, . . . , 2n + 1} and N3 contains only one operations
from theset {n+1, . . . , 2n+1}. Both stations 1st and 2nd are maximal
loaded i∈N1 ti = i∈N2 ti =
nM + A = c − M + 1. If the answer is
"NO" then mmax = 2 where i∈N1 ti > nM + A = c − M + 1.
Thus, max-SALBP-1 is NP-hard.
As a consequence max-SALBP-1 is not approximated with an approximation ratio ≤ 32 unlike P = N P .
Next, we prove that max-SALBP-1 is NP-hard in the strong sense.
3-Partition problem:
A set N = {b1 , b2 , . . . , bn } of n = 3m positive integers is given, where
n
B
B
i=1 bj = mB and 4 < bj < 2 , j = 1, 2, . . . , n. Does there exist a
6
partition of N into m subsets N 1 , N 2 , . . . , N m such that each subset
consists exactly three numbers and the sum of the numbers in each
subset is equal, i.e.,
bj =
bj = · · · =
bj = B?
bj ∈N 1
bj ∈N 2
bj ∈N m
Lemma 2 max-SALBP-1 is NP-hard in the strong sense.
Proof. Use a reduction from the 3-Partition problem. Given an instance of the 3-Partition problem with 3m numbers. Let M = (mB)2 .
Construct an instance of max-SALBP-1 with 3m + 1 operations, where
tj = M + bj , j = 1, 2, . . . , 3m and t3m+1 = M . In addition, let
c = B + 4M − 1
If for the instance of the 3-Partition problem the answer is "YES",
then there exists an optimal solution with m + 1 stations. Operations
which correspond to numbers from the set N i are assigned on the
machine i, i = 1, 2, . . . , m. The operation 3m + 1 is assigned on the
machine m + 1. If the answer is "NO" then mmax = m, and on a
station, 4 operations are assigned (including the operation 3m + 1).
In [16], it was proven that for SALBP-1 any polynomial time heuristic the worst-case ratio is at least 32 , i.e. for a heuristic algorithm there
3
is a situation for which mm
min ≥ 2 . If we consider SALBP-1 as a generalization of the Bin-Packing Problem (BP ), then this approximation
there can be compared with known results for the BP (e.g., see [16]),
where is a heuristic for which m ≤ 11/9mmin + 1. For BP it is known
that for any polynomial time heuristic the worst-case ratio is at least 32 ,
as well. But it holds only for mmin = 2, m = 3, i.e. the absolute error
equals 1. For mmin > 2 we have a worst case m ≤ 11/9mmin + 1. We
can conjecture, that the same relation holds for SALBP-1, too, but in
[16], authors show that worst-case ratio 32 holds for any absolute error,
i.e., for any polynomial time heuristic, for any mmin ≥ 2, there is an
3
instance for which mm
min = 2 .
Next we prove a similar result for max-SALBP-1.
Lemma 3 For any polynomial time heuristic for max-SALBP-1 , for
max
any given absolute error q ≥ 3, there is an instance for which mm = 32
and mmax − m = q.
Proof. Consider the following reduction from the Partition Problem.
Given an instance of the Partition
n Problem with n numbers. Without
of maxlost of generality assume A = 12 i=1 bi . Construct
an instance
SALBP-1 with a set of operations N = N1 N2 · · · Nq . Assume
c = 2n2 A · n2q . Each set Nl , l = 1, 2, . . . , q, contains 2n + 3 operations,
with processing times tj = Ml + bj , j = 1, 2, . . . , n, and tj = Ml , j =
n + 1, . . . , 2n + 1 and t2n+2 = t2n+3 = Tl = c − ((n + 1)Ml + A − 1),
7
where Mq = 2n2 A and Ml = Ml+1 · (n + 2), l = q − 1, q − 2, . . . , 1. In
addition, assume i → j, ∀i ∈ Nl , ∀j ∈ Nl+1 , l = 1, 2, . . . , q − 1 and
2n + 2 → i, where 2n + 2, ∀i ∈ Nl , l = 1, 2, . . . , q.
Then jobs from the different subsets Nl , l = 1, 2, . . . , q, can not
be assigned to the same station. Jobs 2n + 2 and 2n + 3 from the
same subset are assigned to different stations, too. If for the instance
of the Partition problem the answer is "YES", then there exists an
optimal solution with mmax = 3q stations, where jobs from the subset
Nl , l = 1, 2, . . . , q, are assigned to stations with numbers 3(q − 1) +
1, 3(q − 1) + 2, 3(q − 1) + 3 according to assignment from the proof of
Lemma 1. For the instance mmin = 2q, where jobs from each subset
occupied only 2 stations.
The lemma is proven. 3 Maximal Number of Stations for Benchmark Instances
In this section, an experimental study of maximal number of stations
for benchmark instances is presented. So, the purpose of the study
is to estimate the maximal number of stations for some of benchmark instances presented on http://www.assembly-line-balancing.de.
Although, it is obvious that for SALBP-1 there are instances for which
mmax
3
min = 2 (e.g., see Lemma 1), and there are instances for which
mmax
m
mmin ≈ 2[7], it is interesting to compare the maximal and minimal
numbers of stations on benchmark instances. To maximize the number of stations we constructed a simple B&B algorithm with deep-first
branching strategy and a simple Upper Bound which are based on the
following observation:
Let us compute P redj – a set of all predecessors (not immediate, as
well) for each operation j, j = 1, 2, . . . , n. Then the minimal number for
the station Sj , to whichthe operation j = 1, 2, . . . , n, can be assigned,
ti +tj
j
. Denote M Lk , k = 1, 2, . . . , n,
is computed as Sj = c
= minj {tj |Sj ≤ k} and
is the minimal load of the station k. Let tmin
k
min
tmax
= maxj {tj |S
≤
k}.
Then
M
L
=
max{t
, c − tmax
+ 1} and
j
k
k
k
k
m
U B1 = min{m| tj − l=1 M Ll ≤ 0} is an Upper Bound for the
maximal number of stations.
2· t
Some other trivial Upper Bounds are used, e.g., U B2 = c j −1.
Upper bounds are recomputed for each subproblem. Unfortunately, the
B&B proposed can not solve the majority of benchmark instances in
10 minutes on CPU INTEL CORE 2 DUO 2,4 Hz (only one processer
is used), but the results obtained enable us to estimate the difference
between the minimal and maximal number of stations.
In the following table the experimental results are presented:
i∈P red
8
Instance
n
c
mmin mmax running time (sec)
Arcus1
83 3786
21
24
600,00
Arcus2
111 5755
27
30
600,00
Barthol2 148
84
51
57
600,00
Barthold 148 403
14
16
600,00
Bowman
8
20
5
5
0,00
Buxey
29
27
13
16
600,00
Gunther
35
41
14
16
600,00
Hahn
53 2004
8
9
600,00
Heskiaoff
28
138
8
10
600,00
Jackson
11
7
8
9
0,00
Jaeschke
9
6
8
8
0,00
Kilbridge
45
56
10
12
600,00
Lutz1
32 1414
11
13
600,00
Lutz2
89
11
49
52
600,00
Lutz3
89
75
23
25
600,00
Mansoor
11
48
4
5
0,00
Mertens
7
6
6
6
0,00
Mitchell
21
14
8
10
0,36
Mukherje 94
176
25
27
20,90
Roszieg
25
14
10
11
247,32
Sawyer
30
25
14
17
600,00
Scholl
297 1394
50
54
600,00
Tonge
70
160
23
26
600,00
Warnecke 58
54
31
36
600,00
Wee-mag
75
28
63
63
14,66
Table 1: Minimal and maximal number of stations for benchmark
instances
If the running time less than 600 then the presented mmax is optimal.
After running the algorithm for 60 seconds, we had the same values
mmax for all the instances, except Arcus2 (after 60 sec. mmax = 29)
and Lutz1 (after 60 sec. mmax = 12). These results show that the
maximal founded deviation mmax − mmin does not exceed 20%.
4
Flat Graph of Precedence Relations
In [17] authors propose a transformation of graph G of precedence relations to planar one for the well-known Resource-Constrained Project
Scheduling Problem. The same idea can be used for SALBP-1.
Theorem 1 For any instance of SALBP-1 with n operations and v
precedence relations, there exists an analogous instance with a flat
graph G with n operations and v relations, where n + v ≥ n + v .
9
Figure 1: Transformation of K3,3
Figure 2: Transformation of Kk,k
We obtain an analogous instance from the original one by adding
"dummy" operations (with tj = 0) and deleting all the unnecessary
relations. The proof of the theorem follows from Lemmas 4 and 5.
Lemma 4 If there is a subgraph G ⊂ G that is isomorphic to the
special graph K3,3 , then we can transform it into a flat subgraph by
adding "dummy" jobs (with tj = p = 0) and deleting all the unnecessary relations.
See the transformation rules shown in Figures 1, 2 and 3.
Lemma 5 If there is a subgraph G ⊂ G that is isomorphic to the
special graph K5,2 , then we can transform it into a flat subgraph by
deleting all the unnecessary relations.
See the transformation rule shown in Figure 4.
So, according to the well-known Theorem of Pontryagin and Kuratovsky, the modified graph is planar. According to the well-known
Euler’s Theorem, v ≤ 3n − 6 in such the planar graph.
10
Figure 3: Transformation of K4,4
Figure 4: Transformation of K5,2
The number of precedence relations influences running time and the
theoretical complexity of solution algorithms. The number of precedence relations is estimated by different authors as O(n2 ) (i.e, the "Order strength" on http://www.assembly-line-balancing.de is estimated
according to the number n · (n − 1) of precedence relations). If we consider only instances with planar graphs then the number of relations
is ≤ 3n − 6, i.e. O(n).
So, the fact mentioned in Theorem 1 allows us to reduce the run
time of algorithms (by reduction of unnecessary relations) and estimate
the complexity exacter.
5
Conclusion
In this paper, some complexity results for SALBP-1 were presented.
Namely, we propose a special case of the problem for which there is no
Branch and Bound algorithm with a polynomial time computed Lower
Bound that solve instances of a special case even for n = 60 operations
in appropriate time. Since the best known algorithm [8] solves all
benchmark instances in less than 1 second per instance, we suggest
for the future research not concentrate on exact branch and bound
11
based algorithms which are tested only on benchmark instances, but
concentrate on algorithms for special cases with a good running time
estimated theoretically or on essentially new solution schemes.
Additionally, we propose some results for the maximization version
of problem of SALBP-1 and present a technique to reduce the graph
of precedence relations that provides some advantages.
Acknowledgement
The authors are grateful to Chris Yukna for his help regarding the
English presentation.
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