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Engineering Optimization
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A global optimization method for
packing problems
Jung-Fa Tsai a & Han-Lin Li b a
Department of Business Management , National Taipei University of Technology , No. 1, Sec. 3, Chung Hsiao E. Rd., Taipei, 10608, Taiwan
b
Institute of Information Management, National Chiao Tung University , No. 1001, Ta Hsueh Rd., Hsinchu, 300, Taiwan Published online: 26 Jan 2007.
To cite this article: Jung-Fa Tsai & Han-Lin Li (2006) A global optimization method for packing
problems, Engineering Optimization, 38:6, 687-700, DOI: 10.1080/03052150600603264
To link to this article: http://dx.doi.org/10.1080/03052150600603264
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Vol. 38, No. 6, September 2006, 687–700
A global optimization method for packing problems
JUNG-FA TSAI*† and HAN-LIN LI‡
†Department of Business Management, National Taipei University of Technology, No. 1, Sec. 3, Chung Hsiao E. Rd., Taipei 10608, Taiwan
‡Institute of Information Management, National Chiao Tung University, No. 1001, Ta Hsueh Rd., Hsinchu 300, Taiwan
(Received 10 January 2005; in final form 4 August 2005)
The objective of packing problems is to determine an optimal way of placing a given set of three-dimensional (3D) rectangular cartons within a minimum volume 3D rectangular container. Current packing optimization methods either use too many extra 0–1 variables or find it difficult to obtain a globally optimal solution. This study proposes an efficient method for finding the global optimum of packing problems. First the traditional packing optimization problem is converted into an equivalent program containing many fewer 0–1 variables than contained in current methods. Then the global optimum of the converted program is found by utilizing piecewise linearization techniques. The numerical examples demonstrate that the proposed method is capable of finding the global optimum of a packing problem.
Keywords: Packing; Global optimization; Piecewise linearization
1. Introduction
The objective of packing optimization problems is to seek a minimal volume container which can contain a given set of small three-dimensional (3D) rectangular cartons. All the cartons can be different in size and may be rotated in any orthogonal direction. The problem has many applications in the electronic, manufacturing and distribution industries. Examples include packing several components into a minimal case to form a device, cutting wood or foam rubber into smaller pieces, loading pallets with goods or designing packages. A compact device with a minimal case reduces the manufacturing costs and also increases its competitiveness in the market. Moreover, cutting wood, foam rubber or other materials into smaller pieces with minimal waste leads to lower production costs. An optimal design of a package for packing some specific goods also means high economic relevance in real goods processes. Furthermore, an optimal filling of a container decreases the transportation costs along with ‘side effects’ such as increased traffic activity and negative consequences on environmental resources.
*Corresponding author. Email: jftsai@ntut.edu.tw
Engineering Optimization
ISSN 0305-215X print/ISSN 1029-0273 online © 2006 Taylor & Francis http://www.tandf.co.uk/journals
DOI: 10.1080/03052150600603264
The packing problem appears in many related studies such as knapsack (Fayard and Zissimopoulos 1995), assortment problems (Beasley 1985, Li and Chang 1998), pallet loading (Liu and Hsiao 1997, Terno et al. 2000) and container loading problems (Chen et al. 1995, Scheithauer 1999, Bortfeldt and Gehring 2001, Gehring and Bortfeldt 2002, Pisinger 2002, Andreas et al. 2003). In addition, researchers have dealt with various approaches to the problem. For instance, Dowsland (1991) proposed a heuristic method for solving 3D pack-ing problems, Chen et al. (1995) formulated a mixed integer program for container loadpack-ing problems, and Li and Chang (1998) developed a method for finding the approximately global optimum of the assortment problem. Additionally, Bortfeldt and Gehring (2001), Gehring and Bortfeldt (2002), Pisinger (2002) and Andreas et al. (2003) presented different heuristic algorithms such as the parallel tabu search algorithm, hybrid genetic algorithm and parallel genetic algorithm for solving the container loading problem. Two difficulties of these current methods are now listed. Firstly, methods that utilize bar-relaxation or layer-relaxation tech-niques (Dowsland 1991, Liu and Hsiao 1997) and heuristic algorithms (Bortfeldt and Gehring 2001, Gehring and Bortfeldt 2002, Pisinger 2002, Andreas et al. 2003) may only find locally optimal solutions. Secondly, too many 0–1 variables are used to formulate a packing optimiza-tion problem. For instance, Chen et al.’s model (Chen et al. 1995) contains 3n(n− 1) + 4n 0–1 variables (n is the number of cartons), which might cause a heavy computational burden. To overcome these difficulties, this study proposes another method to solve packing problems. Comparing with current methods, the proposed method has the following advantages:
(i) It can solve real packing optimization problems without bar-relaxation or layer-relaxation techniques.
(ii) It only uses half the 0–1 variables used in Chen et al.’s method (Chen et al. 1995) to formulate a packing problem.
(iii) It is guaranteed to find a global optimum of the packing problem within a tolerable error. The generalization to nonlinear objective functions is achieved by the piecewise linearization techniques.
The rest of this article is organized as follows. In the next section, a detailed description of the packing problem is addressed. Then the reformulation problem is proposed to reduce the number of 0–1 variables. The linearization strategy of the nonlinear objective function is discussed in the fourth section. The fifth section presents the solution algorithm. Numerical examples are examined in the sixth section. Finally, some concluding remarks are included.
2. Problem formulation
Given n rectangular cartons with fixed lengths, widths and heights, a packing optimization problem is to allocate these n cartons within a rectangular container which has minimal volume without the need to fix the y and z values in the nonlinear objective function xyz. Denote x, y and z as the width, length and height of the container (x > 0, y > 0, z > 0), respectively; the packing optimization problem is then stated as follows:
Minimize xyz subject to
1. All of n boxes are non-overlapping.
2. All of n boxes are within the range of x, y and z.
3. x≤ x ≤ ¯x, y ≤ y ≤ ¯y and z ≤ z ≤ ¯z (x, y, z, ¯x, ¯y and ¯z are constants and represent the lower and upper bounds of x, y and z, respectively).
The related terminologies used in the packing model, referring to Chen et al. (1995), are described below:
(Pi, Qi, Ri): Parameters indicating the length, width and height of carton i.
(xi, yi, zi): Continuous variables (for location) indicating the coordinates of the front-left-bottom corner of carton i (xi, yi and zi are integers if the given dimensions of cartons are integers).
(lxi, lyi, lzi): Binary variables indicating whether the length of carton i is parallel to the X-axis, Y -axis or Z-axis. For example, the value of lxiis equal to 1 if the length of carton i is parallel to the X-axis; otherwise, it is equal to 0. It is clear that lxi+ lyi+ lzi= 1.
(wxi, wyi, wzi): Binary variables indicating whether the width of carton i is parallel to the X-axis, Y -axis or Z-axis. For example, the value of wxiis equal to 1 if the width of carton i is parallel to the X-axis; otherwise, it is equal to 0. It is clear that wxi+ wyi+ wzi = 1.
(hxi, hyi, hzi): Binary variables indicating whether the height of carton i is parallel to the X-, Y - or Z-axis. For example, the value of hxi is equal to 1 if the height of carton i is parallel to the X-axis; otherwise, it is equal to 0. It is clear that hxi+ hyi+ hzi = 1.
For a pair of cartons (i, k) where i < k, there is a set of 0–1 variables {Aik, Bik, Cik, Dik, Eik, Fik} defined as:
Aik= 1 if carton i is on the left of carton k, otherwise Aik= 0.
Bik= 1 if carton i is on the right of carton k, otherwise Bik= 0.
Cik = 1 if carton i is behind carton k, otherwise Cik= 0.
Dik= 1 if carton i is in front of carton k, otherwise Dik= 0.
Eik= 1 if carton i is below carton k, otherwise Eik= 0.
Fik= 1 if carton i is above carton k, otherwise Fik= 0.
The front-left-bottom corner of the container is fixed at the origin. The interpretation of these variables is illustrated in figure 1. Figure 1 contains two cartons i and k, where carton i is located with its length parallel to the X-axis and the width parallel to the Z-axis, and carton
kis located with its length parallel to the Z-axis and the width parallel to the X-axis. Then lxi,
wzi,hyi, lzk, wxkand hykare equal to 1. In addition, since the carton i is located on the left-hand side of and in front of carton k, it is clear that Aik= Dik= 1 and Bik= Cik= Eik= Fik= 0.
The packing problem can then be formulated below, referring to Chen et al. (1995):
Model 1:
Minimize xyz (1)
subject to
xi+ Pilxi+ Qiwxi+ Rihxi ≤ xk+ (1 − Aik)M for all i, k, i < k, (2)
xk+ Pklxk+ Qkwxk+ Rkhxk≤ xi+ (1 − Bik)M for all i, k, i < k, (3)
yi+ Pilyi+ Qiwyi+ Rihyi ≤ yk+ (1 − Cik)M for all i, k, i < k, (4)
yk+ Pklyk+ Qkwyk+ Rkhyk≤ yi+ (1 − Dik)M for all i, k, i < k, (5)
Figure 1. Graphical illustration of variables.
zi+ Pilzi+ Qiwzi+ Rihzi≤ zk+ (1 − Eik)M for all i, k, i < k, (6)
zk+ Pklzk+ Qkwzk+ Rkhzk≤ zi+ (1 − Fik)M for all i, k, i < k, (7)
Aik+ Bik+ Cik+ Dik+ Eik+ Fik≥ 1 for all i, k, i < k, (8)
xi+ Pilxi+ Qiwxi+ Rihxi≤ x for all i, k, i < k, (9)
yi+ Pilyi+ Qiwyi+ Rihyi≤ y for all i, k, i < k, (10)
zi+ Pilzi+ Qiwzi+ Rihzi≤ z for all i, k, i < k, (11) where lxi, lyi, lzi, wxi, wyi, wzi, hxi, hyi, hzi, Aik, Bik, Cik, Dik, Eikand Fikare 0–1 variables,
M= max{ ¯x, ¯y, ¯z}, xi, yi, zi ≥ 0, 0 < x ≤ x ≤ ¯x, 0 < y ≤ y ≤ ¯y, 0 < z ≤ z ≤ ¯z, and
x, y, z,¯x, ¯y and ¯z are constants.
The objective of this model is to minimize the volume of the container. Constraints (2)–(8) are non-overlapping conditions used to ensure that none of these n boxes overlaps with each other. Constraints (9)–(11) guarantee that all boxes are within the enveloping container.
The binary variables, lxi, lyi, lzi, wxi, wyi, wzi, hxi, hyiand hzi, are dependent and have the following relationships:
lxi+ lyi+ lzi= 1 ∀i ∈ I, (12)
wxi+ wyi+ wzi= 1 ∀i ∈ I, (13)
hxi+ hyi+ hzi= 1 ∀i ∈ I, (14)
lxi+ wxi+ hxi= 1 ∀i ∈ I, (15)
lyi+ wyi+ hyi= 1 ∀i ∈ I, (16)
lzi+ wzi+ hzi= 1 ∀i ∈ I. (17)
Constraints (12)–(17) describe the allocation restrictions among logic variables. For instance, (12) implies that the length of carton i is parallel to one of the axes. Constraint (15) implies that only one of the length, the width and the height of carton i is parallel to the
X-axis. Using constraints (12)–(17), the following five variables can be eliminated from the model lyi, wxi, wzi, hxiand hyi. Model 1 is then fully converted into Model 2 below:
Model 2:
Minimize xyz (1)
subject to
xi+ Pilxi+ Qi(lzi− wyi+ hzi)+ Ri(1− lxi− lzi+ wyi− hzi)≤ xk+ (1 − Aik)M, (2)
xk+ Pklxk+ Qk(lzk− wyk+ hzk)+ Rk(1− lxk− lzk+ wyk− hzk)≤ xi+ (1 − Bik)M, (3)
yi+ Pi(1− lxi− lzi)+ Qiwyi+ Ri(lxi+ lzi− wyi)≤ yk+ (1 − Cik)M, (4)
yk+ Pk(1− lxk− lzk)+ Qkwyk+ Rk(lxk+ lzk− wyk)≤ yi+ (1 − Dik)M, (5)
zi+ Pilzi+ Qi(1− lzi− hzi)+ Rihzi ≤ zk+ (1 − Eik)M, (6)
zk+ Pklzk+ Qk(1− lzk− hzk)+ Rkhzk≤ zi+ (1 − Fik)M, (7)
Aik+ Bik+ Cik+ Dik+ Eik+ Fik≥ 1, (8)
xi+ Pilxi+ Qi(lzi− wyi+ hzi)+ Ri(1− lxi− lzi+ wyi− hzi)≤ x, (9)
yi+ Pi(1− lxi− lzi)+ Qiwyi+ Ri(lxi+ lzi− wyi)≤ y, (10)
zi+ Pilzi+ Qi(1− lzi− hzi)+ Rihzi ≤ z, (11)
where all variables are the same as defined in Model 1.
Chen et al. (1995) solved Model 2 by treating the nonlinear objective function as xy0z0
where y0and z0are fixed values specified by the user. Model 2 then becomes a linear mixed
0–1 program. The following are two disadvantages in Chen et al.’s model:
(i) Too many 0–1 variables are included in the constraints of Model 2. This number of 0–1 variables substantially increases the computational effect required to solve the problem. (ii) Chen et al.’s method can only find a local optimum of Model 2 without fixing the y and
zvalues in the objective function xyz.
The next section describes a way to reduce redundant 0–1 variables in Model 2. Then an algorithm for finding the globally optimal solution of the packing problem is developed.
3. Problem reformulation for computational improvement
This section reformulates the packing optimization problem to reduce the number of 0–1 vari-ables contained in Model 2. Consider the set of six 0–1 varivari-ables (Aik, Bik, Cik, Dik, Eik, Fik) in constraints (2)–(8), which is used to express the six types of non-overlapping conditions of left–right, behind–front and below–above between a pair of cartons (i, k). In fact, a set of three 0–1 variables is enough to express these six types of relationships. Consider the following proposition:
PROPOSITION1 The conditions of non-overlapping between cartons i and k can be
reformu-lated by introducing three binary variables αik, βikand δikwith definitions given in table 1
with reference to figure 2, where 1≤ αik+ βik+ δik≤ 2.
Table 1. Improved non-overlapping conditions (1≤ αik+ βik+ δik≤ 2).
Condition
number αik βik δik Meaning
1 0 0 1 Carton i is on the right side of carton k 2 0 1 0 Carton i is on the left side of carton k 3 1 0 0 Carton i is behind carton k
4 0 1 1 Carton i is in front of carton k 5 1 0 1 Carton i is below carton k 6 1 1 0 Carton i is above carton k
Figure 2. Graphical illustration of non-overlapping conditions.
From the basis of Proposition 1, constraints (2)–(8) can be reformulated effectively as follows:
PROPOSITION 2 The non-overlapping constraints (2)–(8) are equivalent to the following
inequalities:
xi+ Pilxi+ Qi(lzi− wyi+ hzi)+ Ri(1− lxi− lzi+ wyi− hzi)
≤ xk+ αikM+ βikM+ (1 − δik)M, (2)
xk+ Pklxk+ Qk(lzk− wyk+ hzk)+ Rk(1− lxk− lzk+ wyk− hzk)
≤ xi+ αikM+ (1 − βik)M+ δikM, (3)
yi+ Pi(1− lxi− lzi)+ Qiwyi+ Ri(lxi+ lzi− wyi)
≤ yk+ (1 − αik)M+ βikM+ δikM, (4)
yk+ Pk(1− lxk− lzk)+ Qkwyk+ Rk(lxk+ lzk− wyk)
≤ yi+ αikM+ (1 − βik)M+ (1 − δik)M, (5)
zi+ Pilzi+ Qi(1− lzi− hzi)+ Rihzi≤ zk+ (1 − αik)M+ βikM+ (1 − δik)M, (6)
zk+ Pklzk+ Qk(1− lzk− hzk)+ Rkhzk≤ zi+ (1 − αik)M+ (1 − βik)M+ δikM, (7)
1≤ αik+ βik+ δik≤ 2, (8)
where all variables are the same as defined before.
Proof Constraint (2) means that if and only if Aik= 1 then i is on the left of k; constraint (2)
implies that for αik= βik= 0 and δik= 1, i is also on the left of k. Constraint (2) therefore
is equivalent to constraint (2). Similarly, constraints (3)–(7) are equivalent to constraints
(3)–(7), respectively.
Model 2 can then be transformed into Model 3 below:
Model 3:
Minimize xyz subject to
constraints (2)–(8), (9)–(11)
where αik, βik, δik, lxi, lzi, wyiand hzi are 0–1 variables.
Comparing Model 2 with Model 3 shows that Model 2 contains 3n(n− 1) + 4n 0–1 variables, while Model 3 involves 3/2n(n− 1) + 4n 0–1 variables. Model 3 is therefore computationally less demanding than Model 2. This will be illustrated by some numerical results presented in section 6.
4. Linearization strategy
This article is concerned with packing problems with the nonlinear objective function xyz. Therefore, finding the global optimum of the packing problem is difficult. This section proposes a generalized approach to piecewisely linearize the objective function xyz in Model 3 in order to find its globally optimal solution.
Denote F as a feasible set of Model 3 in which x ≥ y ≥ z, F = {(2)–(8), (9)–(11),
x ≥ y ≥ z}. First, consider the following fact:
An optimization program P1: {Minimize Obj1= xyz, subject to x ≤ x ≤ ¯x, y ≤ y ≤ ¯y,
z≤ z ≤ ¯z, x, y, z ∈ F } is equivalent to the program below.
P2: {Minimize Obj2= ln x + ln y + ln z, subject to x ≤ x ≤ ¯x, y ≤ y ≤ ¯y, z ≤ z ≤ ¯z,
x, y, z∈ F }.
The following propositions discuss the proposed approach of linearizing the logarithmic terms ln x, ln y and ln z.
PROPOSITION 3 A logarithm function ln x, 0 < a1≤ x ≤ am, as shown in figure 3, can piecewise linearly be approximated
ln x= ln ˆx = ln a. 1+ s1(x− a1)+ m−1 j=2 sj− sj−1 2 (|x − aj| + x − aj), (18)
where aj, j = 1, 2, . . . , m, are the break points of ln x, aj < aj+1; and sj are the slopes
of line segments between aj and aj+1, sj = (ln aj+1− ln aj)/(aj+1− aj), for j = 1, 2, . . . ,
m− 1.
Figure 3. Graphical illustration of piecewise linear approximation of ln x.
This proposition can be examined as follows: If x= a1then ln x= ln a1(exact). If x≤ a2then lnˆx = ln a1+ s1(x− a1). If x≤ a3then lnˆx = ln a1+ s1(x− a1)+ s2− s1 2 (|x − a2| + x − a2). Similarly, logarithm functions ln y and ln z, can be approximately linearized as
ln y= ln ˆy = ln b. 1+ t1(y− b1)+ m−1 j=2 tj− tj−1 2 (|y − bj| + y − bj), (19) where tj = (ln bj+1− ln bj)/(bj+1− bj), 0 < b1 ≤ y ≤ bmand b1, b2, . . . , bmare its break points, bj < bj+1, for j= 1, 2, . . . ,m − 1; ln z= ln ˆz = ln c. 1+ r1(z− c1)+ m−1 j=2 rj− rj−1 2 (|z − cj| + z − cj), (20) where rj = (ln cj+1− ln cj)/(cj+1− cj), 0 < c1≤ z ≤ cm and c1, c2, . . . , cm are its break points, cj < cj+1, for j= 1, 2, . . . ,m − 1.
Remark 1 Since ln x, ln y and ln z are concave functions, it is clear that the approximations
bound ln x, ln y and ln z from below; ln x≥ ln ˆx, ln y ≥ ln ˆy and ln z ≥ ln ˆz. The following results are then obtained.
Remark 2 (lower bound) Consider the following program:
P3: {Minimize Obj3 = ln ˆx + ln ˆy + ln ˆz, subject to x ≤ x ≤ ¯x, y ≤ y ≤ ¯y,
z≤ z ≤ ¯z, x, y, z ∈ F }.
Program P3 provides a lower bound on Program P2 due to Remark 1. Consider the following proposition of how to linearize lnˆx:
PROPOSITION4 (linearization) lnˆx in (18) can be re-expressed in a linearized form as follows: lnˆx = ln a1+ s1(x− a1)+
m−1 j=2
(sj − sj−1)(ajuj+ x − aj − gj), (21)
(i) −amuj ≤ x − aj ≤ am(1− uj) for j= 2, 3, . . . , m − 1
(ii) −amuj ≤ gj ≤ amuj for j = 2, 3, . . . , m − 1
(iii) am(uj − 1) + x ≤ gj ≤ am(1− uj)+ x for j = 2, 3, . . . , m − 1
(iv) uj ≥ uj−1 for j= 2, 3, . . . , m − 1 where uj = 0 or 1, and gj = 1 or x, respectively.
Proof If x− aj ≥ 0 then uj = 0 and gj = 0 based on (i) and (ii); which results in
ajuj+ x − aj− gj = (|x − aj| + x − aj)/2. If x− aj <0 then uj = 1 and gj = x based on (i) and (iii); which results in
ajuj+ x − aj − gj = |x − aj| + x − aj.
Therefore, lnˆx in (18) is equivalent to (21). Now consider condition (iv).
Since aj−1< aj, if x < aj (i.e. uj = 1) then x < aj+1and uj+1 = 1. If x > aj+1(i.e. uj+1= 0) then x > aj and uj = 0.
Therefore, it is true that uj ≥ uj−1.
Condition (iv) is used to accelerate the computational speed of solving the problem. Similarly, ln ˆy and ln ˆz can be re-expressed in a piecewise linearized form in the same way.
PROPOSITION5 (range reduction) Let (x∗, y∗, z∗) be the global optimum of Model 3, the range of the objective function xyz is bounded as
n i=1 PiQiRi ≤ x∗y∗z∗≤ x+ y+ z 3 3 for any (x, y, z)∈ F.
Proof Sinceni=1PiQiRi ≤ xyz for any (x, y, z) ∈ F and
3 x∗y∗z∗ ≤3 xyz≤ x + y+ z 3 ,
the proposition is true.
Note that (Pi, Qi, Ri)is the length, width and height of carton i.
5. Solution algorithm
From the above discussion, the proposed solution algorithm is as follows.
Let Sτ, Tτ and Uτ be respectively a set of break points of ln x, ln y and ln z at the τ th iteration. Denote ε as a tolerable error (specified later).
Step 1 Range reduction
Solving a linear 0–1 program: {Minimize x+ y + z, subject to x, y, z ∈ F }. Let the solution be (x, y, z)and the objective value be Obj = x+ y+ z. According to Proposition 5, the range of finding the global optimum of xyz is
n i=1 PiQiRi≤ x∗y∗z∗≤ Obj 3 3 and ln n i=1 PiQiRi ≤ ln x∗+ ln y∗+ ln z∗≤ 3 ln Obj − 3 ln 3. (22) Step 2 Linearization
Let iteration τ = 1, S1= {x, ¯x}, T1= {y, ¯y}, U1 = {z, ¯z}, x > 0, y > 0, z > 0.
Consider the linear program
Min
(x,y,z)Obj (x(2))+ Obj (y(2)) + Obj (z(2)) where
Obj (x(2))+ Obj (y(2)) + Obj (z(2)) = ln x + ln¯x − ln x
¯x − x (x− x) + ln y +ln¯y − ln y
¯y − y (y− y) + ln z +
ln¯z − ln z ¯z − z (z− z) and variable x(2) denotes the approximation variable of x∗in the current iteration, similarly for y(2) and z(2).
Subject to (22), (x, y, z)∈ F , x ≤ x ≤ ¯x, y ≤ y ≤ ¯y, z ≤ z ≤ ¯z.
Let the solution be denoted by the specific values of (x, y, z), namely, (x(2), y(2), z(2)), the approximation to (x∗, y∗, z∗)in the current iteration.
If |(Obj (x(2)) − ln x(2))/ ln x(2)| < ε, |(Obj (y(2)) − ln y(2))/ ln y(2)| < ε and |(Obj (z(2)) − ln z(2))/ ln z(2)| < ε then terminate the process. The optimal solution is
(x(2), y(2), z(2)). Otherwise, go to Step 3.
Step 3
Iteration τ = τ + 1.
Let Sτ = Sτ−1∪ {x(τ)}, Tτ = Tτ−1∪ {y(τ)} and Uτ = Uτ−1∪ {z(τ)}, where ‘∪’ means union of sets.
Denote the number of elements (break points) in Sτ, Tτand Uτas mτ. Consider the following linear mixed 0–1 program:
Min
(x,y,z)Obj (x(τ+ 1)) + Obj (y(τ + 1)) + Obj (z(τ + 1)) where
Obj (x(τ+ 1)) + Obj (y(τ + 1)) + Obj (z(τ + 1))
= ln a1+ s1(x− a1)+ mτ−1 j=2 (sj − sj−1)(ajuj+ x − aj− gj)+ ln b1+ t1(y− b1) + mτ−1 j=2 (tj − tj−1)(bjvj + y − bj − qj)+ ln c1+ r1(z− c1)+ mτ−1 j=2 (rj− rj−1) × (cjoj+ z − cj − pj).
Subject to (22), (x, y, z)∈ F , for all j, the following constraints should be satisfied: − ¯xuj ≤ x − aj ≤ ¯x(1 − uj), − ¯xuj ≤ gj ≤ ¯xuj,
¯x(uj − 1) + x ≤ gj ≤ ¯x(1 − uj)+ x, uj ≥ uj−1,
− ¯yvj ≤ y − bj ≤ ¯y(1 − vj), − ¯yvj ≤ qj ≤ ¯yvj,
¯y(vj − 1) + y ≤ qj ≤ ¯y(1 − vj)+ y, vj ≥ vj−1,
− ¯zoj ≤ z − cj ≤ ¯z(1 − oj), −¯zoj ≤ pj ≤ ¯zoj,
¯z(oj− 1) + z ≤ pj ≤ ¯z(1 − oj)+ z, oj ≥ oj−1,
where uj, vj and oj are 0–1 variables, gj, qj and pj are 1 or, respectively, x, y and z, a1, a2, . . . , amτ ∈ Sτ, a1= x < a2<· · · < amτ = ¯x, b1, b2, . . . , bmτ ∈ Tτ, b1= y <
b2<· · · < bmτ = ¯y, c1, c2, . . . , cmτ ∈ Uτ, c1= z < c2<· · · < cmτ = ¯z, sj, tjand rjare the
slopes in the piecewise linearization of ln x, ln y and ln z, respectively, for j = 2, 3, . . . ,
mτ− 1.
Let the solution be (x(τ + 1), y(τ + 1), z(τ + 1)).
If |(Obj (x(τ + 1)) − ln x(τ + 1))/ ln x(τ + 1)| < ε, |(Obj (y(τ + 1)) − ln y(τ + 1))/ ln y(τ+ 1)| < ε and |(Obj (z(τ + 1)) − ln z(τ + 1))/ ln z(τ + 1)| < ε then terminate the process, and (x(τ+ 1), y(τ + 1), z(τ + 1)) is the optimal solution.
Otherwise, repeat Step 3.
Development (convergence). The above algorithm (run with ε= 0) terminates with the.
incumbent solution (ˆx∗,ˆy∗,ˆz∗)being optimum to Model 3 when τ→ ∞.
Explanation. For iteration τ , let {[lτ
x, uτx]} express the sequence [a1τ, a2τ], [a2τ, aτ3], . . . ,
[aτ mτ−1, a
τ mτ], {[l
τ
y, uτy]} express the sequence [bτ1, b2τ], [b2τ, bτ3], . . . , [bτmτ−1, b
τ
mτ] and {[l
τ z, uτz]} express the sequence[cτ
1, cτ2], [c2τ, cτ3], . . . , [cmττ−1, c τ mτ] where a τ 1 < aτ2 <· · · < amττ, b τ 1 < bτ 2 <· · · < bmττ and c τ 1 < c2τ<· · · < cτmτ. Since sequences{aτ
κ}, {bτκ} and {cτκ} are monotone and bounded, where κ = 1, . . . , mτ and τ → ∞, it is obvious that {[lτ
x, uτx]}, {[lyτ, uτy]} and {[lzτ, uτz]} converge to some intervals [lτ
x, uτx], [lτy, uτy] and [lτz, uτz]. Also, when τ → ∞, by the concavity of ln ˆx, ln ˆy and ln ˆz in (18) and the Mean Value Theorem, ˆx∗ = l∗x = u∗x, ˆy∗= ly∗= u∗y andˆz∗= lz∗= u∗z. Which means lnˆx∗= ln x∗, lnˆy∗= ln y∗and lnˆz∗= ln z∗. By referring to Remark 2, lnˆx∗, lnˆy∗and lnˆz∗ are the lower bounds of Program P2; (ˆx∗,ˆy∗,ˆz∗)is then the optimal solution to Model 3.
6. Numerical examples
To validate the proposed method, two tests are performed. Test 1 is used to demonstrate that the reformulation of packing problems can substantially improve the computational efficiency of Chen et al.’s model. Test 2 comprises packing problems with cubes in which the problem model is simpler than Model 3; carton rotation is not considered. All test problems are solved by LINGO (2002) on a Pentium III 1000 personal computer.
Test 1: The test problems taken from Chen et al. (1995) are solved and the CPU times are compared with those of the proposed method. Table 2 indicates that, compared with
Table 2. Comparison of computation results.
CPU time (hh:mm:ss) Objective value Problem Carton
number number i Pi Qi Ri Chen et al. Proposed method Chen et al. Proposed method
1 1 25 8 6 00:04:43 00:00:18 4368 4368 2 20 10 5 3 16 7 3 4 15 12 6 2 1 25 8 6 03:12:07 00:02:19 5040 5040 2 20 10 5 3 16 7 3 4 15 12 6 5 22 8 3 3 1 25 8 6 46:38:29 00:45:02 5880 5880 2 20 10 5 3 16 7 3 4 15 12 6 5 22 8 3 6 20 10 4
Table 3. Optimal solutions obtained by the proposed method.
Problem number Carton number i xi yi zi (x, y, z)
1 1 0 0 0 (28, 26, 6) 2 8 0 0 3 8 10 2 4 16 11 0 2 1 0 3 0 (30, 28, 6) 2 20 8 0 3 14 0 0 4 8 8 0 5 8 0 3 3 1 10 20 0 (35, 28, 6) 2 25 0 0 3 0 0 3 4 10 8 0 5 5 0 0 6 0 8 0
Chen et al.’s method, the proposed method uses much less CPU time to reach the same objective values. This implies that the reformulation of packing optimization problems improves the computational efficiency. The optimal packing solutions for Problems 1, 2 and 3 of table 2 are listed in table 3.
Test 2: Packing problems with cubes are tested by the proposed method and the results of four problems are presented in table 4. For Problem 4, with eight cubes, the proposed method spends 8 seconds finding the optimal solution. For Problem 5, with 10 cubes, it takes about 2.5 minutes to get the global solution. The other two problems are also solved to obtain the global optima with the tolerable error ε= 0.01.
Taking the test Problem 1 for instance, the detailed results obtained by the developed algorithm are listed in table 5. The globally optimal solution is found in the sixth iteration with six break points. From the sixth iteration to the ninth iteration, one of the break points
(28, 26, 6) is also the obtained optimal solution. Accordingly, the values of ε(x), ε(y) and
ε(z), which represent the errors of ln (x), ln (y) and ln (z), respectively, are 0. The convergence
Table 4. Computational results; all cartons are cubes.
Problem Number Cube side CPU time Objective number of cubes length (x, y, z) (hh:mm:ss) value
4 3 1 (8, 5, 5) 00:00:08 200 3 2 Global optimum 1 3 1 5 5 4 1 (8, 6, 5) 00:02:26 240 3 2 Global optimum 2 3 1 5 6 4 1 (8, 6, 3) 00:04:44 144 3 2 Global optimum 4 3 7 4 1 (9, 8, 5) 00:56:38 360 3 2 Global optimum 3 3 1 4 1 5
Table 5. Detailed results of Problem 1. Number of Objective
break points value (x, y, z) ε(x) ε(y) ε(z) 1 6.4738 (25, 18, 11) 0.034 0.1129 0.3383 2 7.3158 (41, 12, 10) 0.1144 0.2373 0.2851 3 7.6211 (41, 12, 10) 0.1144 0.2373 0.2851 4 7.8335 (41, 12, 10) 0.1144 0.2373 0.2851 5 8.0188 (60, 12, 7) 0.2287 0.2373 0.086 6 8.0925 (28, 26, 6) 0 0 0 7 8.1208 (28, 26, 6) 0 0 0 8 8.1626 (28, 26, 6) 0 0 0 9 8.1832 (28, 26, 6) 0 0 0
Figure 4. Convergence graph of objective value against number of break points.
Figure 5. Sensitivity analysis of ε(·) to number of break points.
graph of objective value against the number of break points is shown in figure 4. Figure 4 reveals that the objective value increases stably and converges to the globally optimal solution while the number of break points increases. The sensitivity analysis of the errors of ln(x), ln(y) and ln(z) against the number of break points is also depicted in figure 5.
7. Conclusions
This article proposes a new method to solve packing problems. First the computational efficiency is improved by reducing the number of 0–1 variables of the problem model and by finding the minimal range of the objective function. Then a piecewise linearization technique is applied to linearize the nonlinear objective function xyz. By solving the linear mixed 0–1 program iteratively the proposed method can find a global optimum within the tolerable error. Numerical examples demonstrate that the proposed method can obtain the global optimum of a packing problem. To further improve the computational efficiency a direction for future development is to implement the proposed method in a distributed computation system. Another direction of development is to use a heuristic method such as tabu search, simu-lated annealing or genetic algorithm to find a sub-optimal solution which is treated as a lower bound of the objective. Based on this solution, the search region of the global optimum can be reduced and the computational time decreased.
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