A Heuristic for Two Dimensional Packing of Rectangular Items in Large Identical Bins

(1)

A Heuristic for Two Dimensional Packing

of Rectangular Items in Large Identical

Bins

Bas Minkes

s1462091

Master’s Thesis Technology and Operations Management

Supervisor RUG: Dr. ir. D. Catanzaro

Co-assessor RUG: Dr. E. Ursavas

Supervisor Philips: W. Mulder

(2)

(3)

3

A Heuristic for Two Dimensional Packing

of Rectangular Items in Large Identical

Bins

Bas Minkes

s1462091

June 30, 2014

Abstract

In this paper we develop a mathematical model and a heuristic approach for two dimensional bin packing. The mathematical model is simplification from a three dimensional model and gives an optimal solution

for bin packing a set of rectangular items in large identical bins. The items in this paper are heavy and need support from either another item or a bin. The developed heuristic in this paper has two stages and handles practical loading constraints. The first stage assigns items to a cart; the second stage assigns carts

to a bin. Within a reasonable amount of time, the heuristic can find a feasible solution which can be optimal. The heuristic has a direct practical contribution in an industrial application and a theoretical

(4)

4

List of Figures

Figure 1: Example configuration of items packed in bins ... 9

Figure 2: Two stage packing model ... 9

Figure 3: Strip packing (left side) and bin packing (right) side ... 11

Figure 4: Bin 1, configuration for the numerical example ... 20

Figure 5: Bin 2, configuration for the numerical example ... 20

Figure 6: Objective function for the first stage of the heuristic approach ... 22

Figure 7: Configuration of carts after the first stage of the heuristic approach ... 25

Figure 8: Bin 1, numerical example of the heuristic ... 26

Figure 9: Bin 2, numerical example of the heuristic ... 26

(6)

6

List of Tables

Table 1: A set of items ... 8

Table 2: Physical restriction for each identical parallel machine... 8

Table 3: Dimensions for every bin ... 8

Table 4: Notation ... 14

Table 5: Corresponding notation used by Chen et al. (1995) and our notation ... 15

Table 6: Notation for the original model from Chen et al. (1995) ... 15

Table 7: A set of items for a numerical example ... 19

Table 8: Physical restrictions for the bins ... 19

Table 9: Dimensions of the cart in the numerical example ... 26

Table 10: Characteristics of the bins after the heuristic ... 26

Table 11: Distribution for different item lengths ... 28

Table 12: Distribution for the heights of the items ... 29

Table 13: Scenario settings for the computational experiments ... 29

Table 14: Experiment settings ... 30

Table 15: Average number of bins used within each experiment ... 32

Table 16: First experiment results for the utilization percentage ... 33

Table 17: Second experiment results for the utilization percentage ... 33

Table 18: Third experiment results for the utilization percentage ... 34

Table 19: Fourth experiment results for the utilization percentage ... 34

Table 20: Fifth experiment results for the utilization percentage ... 35

(7)

7

1. Introduction

A question from a manufacturing company producing quartz glass tubes and rods initiates the topic of this final thesis. The manufacturing company faces a tough competition within the worldwide glass industry. Due to the rise of new technologies like LED, the call for producing at lower costs increases. A

significant share of costs within the manufacturing process is for energy consumption. One part of the process is the annealing of glass in a vacuum furnace, which is done at 1050 °C between 10 and 60 hours. This operation is non pre-emptive and consists of simultaneously annealing several products in a vacuum furnace. The annealing operation is a batch process, where products with equal process times are placed together in a vacuum furnace. After pre-processing we obtain a set of items which have a certain length and height. The width of our items is equal to the width of a bin (i.e. the vacuum furnace). As a result we obtain a two-dimensional problem with the length and height as our two dimensions. The operational planning of the vacuum furnaces is made on a weekly basis and is the focus for this research.

The setting with multiple identical vacuum furnaces is a Parallel Machine Scheduling (PMS) configuration. Additionally, our process is also typified as batch-wise production. To obtain a schedule for our PMS problem, two different types of decisions are necessary:

1) Packing products to a batch.

2) Assigning and sequencing batches to one of the identical parallel machines.

Both decisions are Non-deterministic Polynomial-time problems. No efficient algorithm solves this problem in polynomial time. Martello, Pisinger and Vigo (2000) conclude in their paper that packing rectangular shaped boxes to a bin in two-dimensions is already strongly . In our case, we have a set of two-dimensional items which are packed in a two-dimensional bin. Lenstra, Rinnooy Kan and Brucker (1977) theorized that a make span minimisation when scheduling identical parallel machines is an problem, even for the case with only two identical parallel machines. The complexity of such problems increases exponentially with the number of machines which makes finding an optimal solution unmanageable. The production data from the manufacturing shows that the due dates for our jobs are relatively high and thus we choose to optimize the first decision. Our interest lies in finding the optimal configuration to minimize the number of bins needed to pack all the items (i.e. jobs).

(8)

8

equal since we have identical parallel machines. Our objective in this example is to minimize the number of bins.

Item # Length Height

1 2 2 2 4 2 3 6 2 4 8 2 5 10 2 6 2 3 7 4 3 8 6 3 9 8 3 10 10 3

Table 1: A set of items

Identical Parallel Machine # Length Height

1 10 5

2 10 5

… 10 5

m 10 5

Table 2: Physical restriction for each identical parallel machine

Bin # Length Height

Bin 1 10 5

Bin 2 10 5

… … …

Bin M 10 5

(9)

9

The optimal solution for this configuration is three bins (Figure 1). In this example, the amount of idle space is equal to “0”. Minimizing idle space or the number of bins used is interchangeable (Chen, Lee and Shen, 1995) and results in exactly the same solution.

Figure 1: Example configuration of items packed in bins

In our case we have an additional stage. We place items on a cart and place the carts behind each other in a bin (Figure 2). Several items are placed on top of each other on a cart. The height of a cart is restricted by the height of the bin, which is in our case the machine (i.e. vacuum furnace). Before making the first decision, the length and the height of a cart is unknown and depends on the longest item and the sum of the heights of the items which are placed at the respective cart. The carts are placed behind each other in a bin. These set of carts guarantees that each item is supported by either another item or the bin itself. This loading constraint is for practical purposes since an item usually weights over one hundred kilogram. The physical appearance of the carts itself (i.e. wheels etc.) in a bin in the second stage is ignored.

Figure 2: Two stage packing model

(10)

10

The setup of this configuration is a two-stage model. In the first stage we want to bin pack the items in two-dimensions on a set of carts. The number of carts is unlimited. The second stage involves one-dimensional packing the set of carts into bins. Dyckhoff (1990) reviewed and typified the literature for packing and cutting problems. The author denotes cutting and packing problems with four symbols: . The first symbol denotes the dimensionality of the problem. symbols the kind of

assignment. typifies the assortment of large objects and the assortment of small items. A missing symbol characterizes the possibility that all respective properties are applicable. Using the typology of Dyckhoff (1990) this problem is denoted as . The problem is two-dimensional and two

assignment decisions are made: The items are assigned to a set of carts and the set of carts are assigned to a set of bins. We have identical figures as large objects (bins) and the small items are rectangular figures (items).

In this paper our aim is to find a solution to solve the problem. The solution has a direct practical contribution in our case company. Other industries where two-stage container loading, pallet loading and other forms of bin packing and/or cutting is applicable can benefit from this research as well. To our best knowledge, this problem is not researched in existing literature. The solution in this paper gives new perspective for multi-stage packing and cutting problems. First we suggest a simplified version of a mathematical model from Chen et al., (1995) for finding an optimal solution in a single stage packing problem. The simplified model gives an exact solution for packing a set of items in a set of bins. Secondly, we suggest a two-stage heuristic to solve the problem stated in the previous paragraph. Two linear programming models are used successively to find a solution. With the exact model, we can derive a lower bound for our heuristic. The lower bound gives an indication for the minimal number of bins which should be used in our heuristic and gives in theory an indication whether we found an optimal solution with the heuristic. The research question in this paper is stated as follows:

“How are rectangular items optimally packed in a set of identical bins in two dimensions?”

(11)

11

2. Theoretical Background

Packing problems have a large number of applications such as computer science (assignment of segments on hard disks drives), industry (cutting stock problems) and within transportation and logistics (loading problems) (Fernández, Gil, Baños and Montoya (2013)). We need to consider two dimensions in our case: The length and the height of the items and bins.

Our bin packing problem stems originally from an industrial application, i.e. we have a logistic problem. The items which we want to pack in bins are usually over one hundred kilogram. A vast amount of literature within two-dimensional bin packing originates from computer science and has its application in multi-processor scheduling (Fernández et al., 2013). Within computer science, the influence of the weight of items is usually neglected. However, within our two-dimensional problem we need to consider the weight of the items. For a practical feasible solution, we cannot place items freely in the space of an empty bin. Each item which is packed in bin must be fully supported by either another item or the bin itself. We refer to this additional restriction as the loading constraint.

Recently cutting and packing problems are receiving increased amounts of attention from researchers. Dowsland and Dowsland (1992) made an interesting survey about packing problems. The authors make a distinction between two-dimensional strip packing problems and bin packing problems (Figure 3). A common situation in strip packing is a rectangle of a given width and undetermined height , a set of items and the objective to minimize the required height to fit all the pieces. In bin packing, the width and height of the bins is fixed, and the objective is to fit all the given items in a

minimum number of bins . Our problem refers to bin packing, since the physical dimensions of our bins are fixed by the physical dimension of a machine and our objective is to minimize the number of bins used.

(12)

12

Lodi, Martello and Monaci (2002) build on the paper from Dowsland and Dowsland (1992) by presenting a survey dedicated to two-dimensional bin packing and strip packing problems. The authors review a set of solution approaches for solving two-dimensional bin packing problems. They introduce a set of mathematical models for solving two-dimensional bin packing problems in an exact way. The first mathematical model which they introduce is a column generation approach, borrowed from Gilmore and Gomoryt (1965). Lodi et al. (2002) conclude that a dynamic approach is necessary for handling the generated columns and obtaining a feasible solution. Solving a problem in a dynamic fashion can result in immense computation times and is therefore not our preferred solution approach.

As far as we know, Lodi et al. (2002) present the most recent survey about two-dimensional bin packing and make a distinction within bin packing problems by the respective solution approach. The authors consider mathematical models, approximation algorithms and (meta) heuristics, lower bound techniques and exact enumerative approaches. Nonetheless, the authors do not consider practical constraints such as the weight of items.

The preferred solution approach in our case is a linear programming (LP) model which is a powerful tool to address integer programming and combinatorial optimization (Valério de Carvalho, 2002). However, Valério de Carvalho (2002) concludes in his review paper that it is difficult to find exact solutions to the integer cutting stock problem. Our case is a bin packing problem and we use the same solution approach as in cutting stock problems: Integer linear programming (ILP). The key difference between the LP models reviewed by Valério de Carvalho (2002) and our case is the number of

dimensions. The author is reviewing mathematical models which are solving one-dimensional problems, whereas our bin packing problem has two dimensions. ILP models are still applicable for two dimensions, however finding an exact solution becomes more complex, but can be manageable for small instances.

Chen and Huang (2007) propose a two-level search algorithm for solving a two-dimensional rectangle packing problem. They pack rectangles in two-dimensions. However, they are packing

rectangles for a floor layout and do not consider a loading constraint. Within bin packing problems, items are directly assigned to a bin. As a result, the position of the item within the bin is freely chosen. As a consequence, an approach for two-dimensional bin packing can be infeasible due to a practical constraint. Literature on this subject is scarce.

As an alternative, heuristics using load balancing can be used. De Queiroz and Miyazawa (2013 theorize two heuristic for load balancing which means that items can be placed above each other. They show the practical constraints when packing items in a bin and suggest a level packing heuristic.

However, the items in our case are almost identical in their physical shape and we assume that a simpler heuristic than the heuristics provided by de Queiroz and Miyazawa (2013) can be used.

(13)

13

(14)

14

3. Notation and Problem Formulation

We want to pack a set of items into a set of bins. We consider two dimensions: The length and height. The items have a rectangular shape, and have a fixed orientation (i.e. they cannot be rotated). The number of items, the length and the height of the items are known beforehand. The bins have a rectangular shape, and have an identical length and height which is predetermined. The number of available carts and bins is unlimited. The length and the height of the cart are decision variables. Typically, one item fills the area of a bin between 5% and 10%. The objective is to minimize the number of bins used. We aim to use a notation which agrees with current literature (Table 4).

Total number of items to be packed Total number of available carts Total number of available bins

An arbitrary large number Length of item Height of item Length of cart Height of cart Length of bin Height of bin Table 4: Notation

For the arbitrarily large number we suggest to use equation (1), which is the sum of all the lengths of the items which need to be packed.

∑

(1)

(15)

15

4. Mathematical Model

Chen et al. (1995) present in their research paper an analytical model to capture the mathematical essence of loading non-uniform cartons in containers in three dimensions. The authors formulate a zero-one Mixed Integer Programming (MIP) model, propose some extensions and give some numerical examples in their research paper. They consider multiple containers, multiple carton sizes, carton orientations, and the overlapping of cartons in a container. In our notation, we have a set of items which are packed in a set of bins. For the sake of clarity, Table 5 gives the corresponding notations.

Chen et al. (1995) Our notation

Cartons Items

Containers Bins

Table 5: Corresponding notation used by Chen et al. (1995) and our notation

4.1 Original Mathematical Model

Chen et al. (1995) introduce first the set of decision variables in their model (Table 6). In total the authors suggest 17 binary variables and 3 continuous variables as decision variables.

Binary variable, equal to 1 if item is placed in bin ; otherwise it is equal to 0 Binary variable, equal to 1 if bin is used; otherwise it is equal to 0

Parameters indicating the length, width and height of item Parameters indicating the length, width and height of bin

Continuous variables (for location) indicating the coordinates of the front-left-bottom corner of items of item

Binary variables, indicating whether the length of item is parallel to the X-, Y-, or Z-axis; otherwise it is equal to 0

Binary variables, indicating whether the width of item is parallel to the X-, Y-, or Z-axis; otherwise it is equal to 0

Binary variables, indicating whether the height of item is parallel to the X-, Y-, or Z-axis; otherwise it is equal to 0

Binary variables, indicating whether an item is placed next to another items. is equal to 1 if item is placed left from item ; otherwise it is equal to 0. Similarly for

, they indicate whether they are placed on the right of, behind, in front, of below or above item

Table 6: Notation for the original model from Chen et al. (1995)

(16)

16 ∑ ∑ (2)

Equation (3)-(8) check for overlap between items.

(3)

(4)

(5)

(6)

(7)

(8) Equation (9) guarantees that the model only checks for overlap when a pair of items are placed in the same bin.

(9)

Equation (10) ensures that all the items are placed in only one bin.

∑

(10)

Once an item is placed in a bin, equation (11) ensures that the bin is used.

∑

(11)

(17)

17

(14)

Equation (15) is the binary restriction for all the relevant variables.

(15) Equation (16) ensures that the front-bottom-left corner of an item is greater or equal to zero.

₍₁₆₎

4.2 Simplification of the Mathematical Model

In our case we deal with multiple identical bins, multiple item sizes, a fixed rotation for every item, the overlapping of items in a bin and two dimensions. Our suggestion is a simplification of the original model proposed by Chen et al. (1995) in their paper. We use the same notation as in Section 3 and Section 4.1.

The initial objective is to minimize the number of bins for packing all the items. The objective function changes from equation (2) to equation (17). We would like to minimize the sum of the bins which are used.

∑

(17)

In our problem is the width of items equal to width of the bins(i.e. vacuum furnaces); therefore we delete constraint (7) and (8). Also we have a fixed rotation for every item. After removing the variables

from the original model in Section 4.1 and changing to the equations (3)-(6) transform to constraints (18)-(21).

(18)

18

(20)

(21)

Since we have only two dimensions, we delete and from equation (9), and derive (22).

,

(22)

Equation (10) and (11) do not change, for clarity we provide them here.

∑

(10) ∑

(11)

Equation (14) is deleted and from (12) and (13) we obtain (23) and (24)

( )

(23)

( )

(24)

After deleting some of our variables from equation (15) and (16) the relevant variables and equations are (25) and (26).

(25)

₍₂₆₎

Equation (27) gives a lower bound for the number of bins which is needed to pack all the items. We calculate the sum of the area of every job (∑ ) and divided this by the area of a bin . When ceiling the answer, we derive a lower bound for the number of bins which we need to use to pack all the items

∑

(27)

(19)

19

4.3 Numerical Example

The mathematical model gives an optimal solution for the number of bins which is needed to pack all the items. To get an impression of the mathematical model, we suggest the following numerical example. In this example we want to schedule a set of 25 items with almost similar dimensions (Table 7). Each bin has identical physical restrictions (Table 8). Our objective is to minimize the number of bins which is needed to pack all the jobs.

Item # Length (mm) Height (mm) Item # Length (mm) Height (mm)

1 1340 135 14 1350 210 2 1500 135 15 1520 210 3 1620 135 16 1580 210 4 1480 135 17 1540 135 5 1450 210 18 1540 210 6 1510 135 19 1430 135 7 1510 210 20 1470 210 8 1420 210 21 1290 135 9 1540 210 22 1510 210 10 1400 210 23 1580 210 11 1510 135 24 1550 210 12 1610 210 25 1480 135 13 1560 210

Table 7: A set of items for a numerical example

Physical restrictions

Length (mm) 4500

Height (mm) 945

Table 8: Physical restrictions for the bins

Without giving any bounds as an extra constraint, we are unable to find an optimal solution within three hours of computational time. This is in line with Chen et al. (1995), who conclude in their paper that more efficient solution procedures are needed to solve large scale container loading problems.

Using formula (27), we can easily derive the minimum number of bins which is needed to pack all the items without the use of any MIP models. However, we consider two dimensions when packing the items and we might obtain an infeasible solution from (27). In the models presented in Section 4.1 and 4.2 we do consider the possibility of overlapping items, which is not considered in (27). Our suggestion is using (27) as a formulation for the lower bound in (28).

(20)

20 ∑

(28)

We add the extra constraint (27) for the lower bound in our mathematical model from Section 4.2. and find an optimal solution within a reasonable amount of time (Figure 4 and 5). The optimal solution itself is interesting in several aspects.

The items are placed freely in the space of either bin 1 or bin 2. From a theoretical point of view, we arrived at an optimal solution. However, our items are usually over one hundred kilogram which makes the solution is infeasible in practice. The items need support from either an item or the bin itself.

Figure 4: Bin 1, configuration for the numerical example

Figure 5: Bin 2, configuration for the numerical example

When given a set of items with certain physical dimensions and the physical dimensions of a bin, we receive a solution for the minimum number of bins which is necessary to process all the items. After running the mathematical model, the amount of idle space can easily be obtained with (28).

(∑ )

(29)

(21)

21 capacity.

Chen et al. (1995) theorized in their paper that finding an optimal solution with their original mathematical model is an inefficient procedure. We reduced the number of decision variables from 20 to 10 and the model (still) requires an immense amount of time to find an optimal solution. Considering that the simplified model from Section 4.2 places items freely in the space of a bin and that the model

(22)

22

5. Heuristic Approach

Our heuristic packs items in two stages in a bin. In the first stage, two-dimensional items are packed on a set of unlimited two-dimensional carts using a strip packing approach. In the second stage, we pack the carts in a one-dimensional bin, using a bin packing approach. The main reason for using two stages is that we want to control the placement of items in the bin. The items are usually over one hundred kilogram and must be supported by either the bin or another item directly below our item. Items cannot be placed freely in the space of a bin. Also, this configuration gives the option for linear programming (instead of dynamic programming) which saves a tremendous amount of computational time for larger instances. The heuristic is finished when all the items are assigned to bins. Throughout this section we use the notation from Section 3.

The first stage of the heuristic approach is different from to two-dimensional strip packing (recall Figure 3). However, the objective in the first stage of our heuristic is minimizing the sum of the lengths for all the carts. Within strip packing, the items are usually placed behind each other and the objective is to minimize the length of the strip. Our carts can differ in height, where the height of a bin is the limit.

5.1 First Stage of the Heuristic Approach

In the first stage, we want to minimize the sum of the lengths for all the carts. The sum of the length for all the carts is calculated when all the carts are placed behind each other (Figure 6).

(23)

23

Our goal is to efficiently stack all the items on the carts, without a losing idle space. We cannot change the dimensions of the items and will always have idle space (unless we are very lucky and all items have exact similar dimensions). The longest item assigned to the cart determines the length of cart. In advance, we do not know how many carts we need to use to pack all the items. As a consequence, the sum of the lengths of all the carts must be decided. For our objective function, we want to minimize the sum of the lengths for cart 1 until cart D (equation (26)).

∑

(30)

Equation (31) ensures that every item is packed to a cart.

∑

(31

If any item is assigned to a cart, equation (32) ensures that respective cart is considered used.

∑

(32)

Equation (33) defines the length of a cart. The longest item assigned to the respective cart, determines the length of the cart. The height of the cart is handled by equation (34). The sum of all the heights from the items assigned to the cart, determines the height of the cart.

(33)

∑

(34)

The height of cart a cart cannot be longer than a predetermined height. In our case the height of the cart is restricted by the height of the bin. Our bins are identical in their physical dimensions, for clarity, we write equation (35).

(35)

(24)

24

₍₃₆₎

After the first stage of the heuristic, the items are assigned to a set of carts and the length of the used carts is minimized. The lengths and the height of the carts are in the arrays and . These arrays are the input data for the second stage of the heuristic approach.

5.2 Second Stage of the Heuristic Approach

In the second stage, the carts are packed in bins. The goal in the second stage is to use a minimal number of bins. The objective function in the second stage of the heuristic approach (equation (37)) is equivalent to the objective function of the original mathematical model by Chen et al. (1995) in section 4.1 and the simplified model in section 4.2. This is not coincidental. We want to minimize the number of bins which are needed to pack all the carts from the first stage.

∑

(37)

Equation (38) ensures that every cart is assigned to a bin. Otherwise, all the carts (with the items assigned to it the first stage) are not packed.

∑

(38)

The length of all the carts assigned to the bin must be lower or equal to the length of the respective bin. Equation (39) handles this constraint.

∑

(39)

If any cart is assigned to a bin, equation (40) ensures that respective bin is considered used.

∑

(40)

(25)

25

₍₄₁₎

Additionally, after we finish the heuristic. We can calculate the utilization of the bins. We can do this similarly as in section 4.3 and using the same equation. For clarity, equation (29) is provided here

(∑ )

(29)

5.3 Numerical Example

In our numerical example, we use the same set of items and bins as in section 4.3 in Table 7 and 8. As a reminder, the length and height of the carts are not constant, but a decision variable. Our objective function is to minimize the number of bins needed to pack all the items.

Running the first stage of the model gives the following configuration for the set of carts (Figure 7).

Figure 7: Configuration of carts after the first stage of the heuristic approach

(26)

26

Cart # Cart Length (mm) Cart Height (mm)

1 1560 900

2 1620 900

3 1510 900

4 1540 900

5 1430 900

Table 9: Dimensions of the cart in the numerical example

In the second stage of the heuristic, we use the data presented in Table 9 as our input data. After running the stage we obtain the following solution (Figure 8 and 9).

Figure 8: Bin 1, numerical example of the heuristic

Figure 9: Bin 2, numerical example of the heuristic

The physical dimensions of the carts in the bins are presented in Table 10. Bin # Length (mm) Height (mm)

1 3180 900

2 4480 900

Table 10: Characteristics of the bins after the heuristic

Interestingly, our heuristic finds also the 2 as the minimum number of bins which is necessary to pack all the items. Comparing the solution of this numerical example with the numerical example presented in section 4.3, we can conclude that the heuristic found an optimal solution. The first stage of the heuristic required 2 hours of computational time and the second stage of the heuristic found a solution

(27)

27

(28)

28

6. Computational Experiments and Results

To get a better impression of the performance of the heuristic approach described in Section 5, we generated a set of artificial data and tested set of scenarios. This section starts with an exemplification of the used data set. Afterwards we describe the selected scenarios and the section ends with the result of the experiments. All experiments are performed with Xpress-IVE 7.5 using Mosel on a Windows XP

computer. The hardware configuration is a HP desktop with Intel(R) Core(TM)2 Duo CPU E7500 @ 2.992 GHz and 4.10 GB RAM.

6.1 Data Description

Our heuristic approach consists of two stages: (1) two-dimensional bin packing and (2) one-dimensional packing. Both stages are affected by the length and height of the items which we want to process. We use a set of distributions for the length (Table 11) and the height (Table 12).

Distribution # Length distribution of an item

1 Normal distribution (μ=1500mm, σ=100mm) 2 Normal distribution (μ=1500mm, σ=300mm) 3 Normal distribution (μ=1400mm, σ=100mm) 4 Normal distribution (μ=1400mm, σ=300mm) 5 25% 1200-1300mm, 25% 1600-1700mm, 50% 2000-2100mm 6 25% 1200-1300mm, 25% 1500-1600mm, 25% 1600-1700mm, 25% 2000-2100mm

Table 11: Distribution for different item lengths

(29)

29

Distribution # Height distribution of an item

1 50% 135 mm, 50% 210 mm

2 80% 135 mm, 20% 210 mm

3 30% 135 mm, 70% 210 mm

Table 12: Distribution for the heights of the items

The production data of the manufacturing company is used to generate artificial data for the height of each item. An item has either a height of 135mm or 210mm, no intervals are used. Three different distributions based on the production data decide the height of each item.

Combining the length distribution with the height distribution, we get 18 different scenarios (Table 13). For each single scenario we use 100 observations.

Scenario # Length distribution of an item Height distribution an item

1 Normal distribution (μ=1500, σ=100) 50% 135 mm, 50% 210 mm 2 Normal distribution (μ=1500, σ=100) 80% 135 mm, 20% 210 mm 3 Normal distribution (μ=1500, σ=100) 30% 135 mm, 70% 210 mm 4 Normal distribution (μ=1500, σ=300) 50% 135 mm, 50% 210 mm 5 Normal distribution (μ=1500, σ=300) 80% 135 mm, 20% 210 mm 6 Normal distribution (μ=1500, σ=300) 30% 135 mm, 70% 210 mm 7 Normal distribution (μ=1400, σ=100) 50% 135 mm, 50% 210 mm 8 Normal distribution (μ=1400, σ=100) 80% 135 mm, 20% 210 mm 9 Normal distribution (μ=1400, σ=100) 30% 135 mm, 70% 210 mm 10 Normal distribution (μ=1400, σ=300) 50% 135 mm, 50% 210 mm 11 Normal distribution (μ=1400, σ=300) 80% 135 mm, 20% 210 mm 12 Normal distribution (μ=1400, σ=300) 30% 135 mm, 70% 210 mm 13 25% 1200-1300mm, 25% 1600-1700mm, 50% 2000-2100mm 50% 135 mm, 50% 210 mm 14 25% 1200-1300mm, 25% 1600-1700mm, 50% 2000-2100mm 80% 135 mm, 20% 210 mm 15 25% 1200-1300mm, 25% 1600-1700mm, 50% 2000-2100mm 30% 135 mm, 70% 210 mm 16 25% 1200-1300mm, 25% 1500-1600mm, 25% 1600-1700mm, 25% 2000-2100mm 50% 135 mm, 50% 210 mm 17 25% 1200-1300mm, 25% 1500-1600mm, 25% 1600-1700mm, 25% 2000-2100mm 80% 135 mm, 20% 210 mm 18 25% 1200-1300mm, 25% 1500-1600mm, 25% 1600-1700mm, 25% 2000-2100mm 30% 135 mm, 70% 210 mm

Table 13: Scenario settings for the computational experiments

6.2 Experiment Settings

We are interested in experimenting with the number of items. The number of items {10, 25, 50} is based on the production data for the case company. We want to get an impression how well the heuristic performs with weekly production data.

(30)

30

experiment with the height of the bin. Currently, the height of a bin is fixed at 945 mm. However, we are interested if increasing the height of the bin results in a better performance. We use 975mm as our second experiment setting.

We expect that a runtime of 10 seconds for the first two experiments, 30 seconds for experiment three and four, and a runtime of 60 seconds is more than sufficient to find a near optimal solution in the first stage of the heuristic approach. For the second stage of the heuristic, we do not give a runtime and expect that heuristic can find a solution immediately. This gives the experiment settings provided in Table 14.

Experiment # # of items Bin height (mm) Runtime first stage (s)

1 10 945 10 2 10 975 10 3 25 945 30 4 25 975 30 5 50 945 60 6 50 975 60

Table 14: Experiment settings

6.3 Performance Criteria

In our experiments we are interested in two performance criteria.

1) The number of bins needed to process all the items in a scenario 2) The utilization percentage of a bin in a scenario.

The first criterion is from an economic perspective. The second criterion tells us how well the heuristic approach works, given the physical dimensions of a bin and the physical dimensions of the set of items.

6.4 Results

When running the computational experiments, we gathered 1.800 observations for each experiment, which results in 108 single scenarios with 10.800 single observations in total. The total runtime for all the six experiments in the first stage in the heuristic is exactly 100 hour of computational time. Each

experiment in the second stage of the heuristic found immediately a solution. None of the single experiments resulted in an error.

(31)

31

Figure 10 clearly shows that experiment 1 and 2 (# of items is 10) have the lowest utilization percentage and the highest variance. In contrast, experiment 5 and 6 (# of items is 50) scored the highest utilization percentage with the lowest variance. The results of experiment 3 and 4 (# of items is 25) score better than experiment 1 and 2, however worse than experiment 5 and 6.The heuristic packs a larger number of items into bins better than a lower number of items. This result is not unexpected, the larger the number of items which need to be packed, the more choices (and thus combinations) can be made.

Also we observe that the highest average score for a single scenario is scenario 8 in experiment 5: 0,87. This can be a consequence of the physical dimensions. Changing either the dimensions of the items or the dimensions of the bin can result in a utilization percentage closer to 1. However, it can be possible that we are close to the optimal solution with our heuristic approach and that higher capacity utilization is not possible. To check the optimal solution, we can use the mathematical model introduced in section 4.2. However, with the tight time constraint for this research, we suggest this as an interesting subject for future research.

(32)

32

From Figure 10 we observe that experiment 2, 4 and 6 have a slightly higher utilization percentage than experiment 1, 3 and 5. Using a Paired Samples T Test (Appendix A), we conclude that there is a statistically significant difference of 0.06 percent (Paired Samples Test, p = 0.001, α=0.05) between experiment 1 and experiment 2, there is no statistically significant difference between experiment 3 and experiment 4 (Paired Samples Test, p = 0.188, α=0.05) and statistically significant difference of 0.09 percent between experiment 5 and experiment 6 (Paired Samples Test, p = 0.009, α=0.05). This means that increasing the height of a bin from 945mm to 975 mm, the utilization percentage increases with 6% for 10 items and with 9% for 50 items.

From Table 15 we can see that experiment 2, 4 and 6 need fewer bins to pack all the items compared with experiment 1,3 and 5. Using a Paired Samples T Test (Appendix B), we conclude that there is a statistically significant difference of 0.12 bins (Paired Samples Test, p = 0.008, α=0.05) between experiment 1 and experiment 2, a statistically significant difference of 0.18 bins (Paired Samples Test, p = 0.001, α=0.05) between experiment 3 and experiment 4 and a statistically significant difference of 0.28 bins (Paired Samples Test, p = 0.000, α=0.05) between experiment 5 and experiment 6. This means that increasing the height of a bin from 945mm to 975 mm, decreases the number of bins for packing a set of items. This is not an unexpected result, since increasing the size of bin, gives more space to place the items. However, this result tells us that a very small increase of the height of a bin has already a significant influence on the number of bins needed to pack the items.

1 2 3 4 5 6 Scenario 1 1,04 1,01 2,95 2,52 4,91 4,63 Scenario 2 1,00 1,00 2,06 2,00 4,06 3,97 Scenario 3 1,32 1,07 2,97 2,45 5,14 4,80 Scenario 4 1,09 1,02 2,41 2,14 4,81 4,46 Scenario 5 1,00 1,00 2,02 2,00 4,17 3,96 Scenario 6 1,44 1,08 2,88 2,34 5,04 4,65 Scenario 7 1,00 1,00 2,00 2,00 4,00 3,70 Scenario 8 1,00 1,00 2,00 2,00 3,23 3,15 Scenario 9 1,00 1,00 2,00 2,00 4,02 4,00 Scenario 10 1,03 1,00 2,22 2,03 4,36 4,23 Scenario 11 1,00 1,00 2,01 2,00 3,91 3,71 Scenario 12 1,14 1,01 2,02 2,01 4,61 4,37 Scenario 13 1,19 1,01 2,97 2,95 5,06 4,96 Scenario 14 1,00 1,00 2,47 2,10 4,62 4,20 Scenario 15 1,63 1,12 3,00 2,99 5,72 5,06 Scenario 16 1,12 1,00 2,90 2,57 5,07 4,86 Scenario 17 1,00 1,00 2,21 2,04 4,40 4,07 Scenario 18 1,64 1,17 2,96 2,73 5,59 4,97 Average # Bins 1,15 1,03 2,45 2,27 4,60 4,32

(33)

33

Experiment 1 N Minimum Maximum Mean Std. Deviation

Scenario 1 100 .43 .95 .6620 .09146 Scenario 2 100 .55 .72 .6564 .03382 Scenario 3 100 .37 .95 .6713 .20639 Scenario 4 100 .39 .94 .7083 .10693 Scenario 5 100 .58 .85 .6978 .05520 Scenario 6 100 .38 .93 .6359 .17994 Scenario 7 100 .55 .93 .6726 .12068 Scenario 8 100 .56 .69 .6164 .02921 Scenario 9 100 .57 .94 .7856 .12650 Scenario 10 100 .38 .94 .6845 .09622 Scenario 11 100 .51 .80 .6609 .05466 Scenario 12 100 .38 .95 .7376 .14906 Scenario 13 100 .40 .88 .7446 .13052 Scenario 14 100 .64 .91 .7736 .05692 Scenario 15 100 .38 .89 .6198 .16613 Scenario 16 100 .41 .95 .7292 .10474 Scenario 17 100 .60 .82 .7456 .04125 Scenario 18 100 .38 .88 .5768 .15349 Average utilization .6877

Table 16: First experiment results for the utilization percentage

Scenario 1 100 .39 .74 .6648 .04290 Scenario 2 100 .56 .70 .6502 .02954 Scenario 3 100 .38 1.00 .6819 .10488 Scenario 4 100 .46 .97 .7248 .07199 Scenario 5 100 .56 .83 .6940 .05174 Scenario 6 100 .39 .99 .7464 .11353 Scenario 7 100 .55 .70 .6273 .03211 Scenario 8 100 .50 .68 .6123 .03307 Scenario 9 100 .58 .96 .6699 .08490 Scenario 10 100 .54 .84 .6833 .05620 Scenario 11 100 .52 .84 .6641 .05734 Scenario 12 100 .46 .96 .7329 .08272 Scenario 13 100 .52 .92 .8203 .06182 Scenario 14 100 .66 .91 .7782 .06219 Scenario 15 100 .48 .93 .8117 .12397 Scenario 16 100 .64 .99 .7703 .05190 Scenario 17 100 .59 .83 .7469 .04304 Scenario 18 100 .42 .99 .7447 .12809 Average utilization .7124

(34)

34

Table 18: Third experiment results for the utilization percentage

Scenario 1 100 .54 1.00 .7197 .13888 Scenario 2 100 .63 .88 .7101 .07030 Scenario 3 100 .55 .99 .7383 .13467 Scenario 4 100 .55 .95 .8375 .09691 Scenario 5 100 .65 .92 .7536 .06694 Scenario 6 100 .58 .98 .8129 .12707 Scenario 7 100 .64 .84 .7950 .02842 Scenario 8 100 .55 .81 .6614 .05990 Scenario 9 100 .60 .84 .7294 .07612 Scenario 10 100 .60 .95 .8289 .05822 Scenario 11 100 .61 .84 .6973 .05130 Scenario 12 100 .55 .90 .7653 .06898 Scenario 13 100 .56 .95 .6708 .06207 Scenario 14 100 .57 .91 .8023 .07040 Scenario 15 100 .62 .93 .6855 .04450 Scenario 16 100 .56 .96 .7538 .15082 Scenario 17 100 .60 .94 .7729 .06686 Scenario 18 100 .58 .96 .7266 .12536 Average utilization .7479

(35)

35

Table 20: Fifth experiment results for the utilization percentage

Scenario 1 100 .59 .86 .7132 .07307 Scenario 2 100 .68 .93 .7736 .04720 Scenario 3 100 .68 .93 .7366 .06731 Scenario 4 100 .68 .96 .8019 .06947 Scenario 5 100 .68 .97 .7815 .05843 Scenario 6 100 .71 .97 .8247 .07033 Scenario 7 100 .71 1.00 .8384 .09608 Scenario 8 100 .63 .99 .8489 .08049 Scenario 9 100 .75 .91 .8178 .02084 Scenario 10 100 .66 .93 .7954 .06796 Scenario 11 100 .67 .97 .7928 .08241 Scenario 12 100 .69 .96 .8361 .06661 Scenario 13 100 .73 .95 .8092 .04132 Scenario 14 100 .66 .91 .8232 .05907 Scenario 15 100 .73 .94 .8491 .04099 Scenario 16 100 .67 .92 .7771 .05307 Scenario 17 100 .66 .94 .8037 .05695 Scenario 18 100 .66 .95 .8137 .05450 Average utilization .8021

(36)

36

7. Conclusion

In this paper we investigated a dimensional bin packing problem, which involves packing two-dimensional items in two-two-dimensional bins. This problems involves have rectangular items, which are slightly different from each other, and identical bins. Dyckhoff (1990) typifies this problem as the problem. The research question in this paper is the following:

“How are rectangular items optimally packed in a set of identical bins in two dimensions?”

For answering this question, we provided two different solution approaches: 1) a mathematical model where the practical loading constraints are neglected and,

2) a heuristic which solves our problem in a reasonable amount of time considering practical loading constraints.

By simplifying a mathematical model for 3D-container loading (Chen et al., 1995), we derived a model which obtains an optimal solution. For large instances, the computation time for our simplified version of the model is still immense. For practical use, we suggest to use lower and/or upper bounds as constraints. Finding these bounds is outside the scope of this thesis and might be an interesting suggestion for future research.

The items in our case are usually over one hundred kilogram. The items in the mathematical model are placed freely in the space of the bin and lack therefor support from other items or the bin itself. This lack of support in the mathematical model requires either a different solution approach such as dynamic programming or an approximation approach.

To our best knowledge, there is no exact algorithm or approximation algorithm which solves our problem. This research is among the first papers which tackle a two-dimensional bin packing problem with the additional loading constraint. Two-dimensional strip packing and bin packing are close variants of our packing problem; in our case we use a combination of both packing problems in order to deal with the loading constraint. The first stage of the heuristic is a strip packing model, the second stage a bin packing model. Using a combination of both packing models ensures that the linear programming models remain relatively simple.

(37)

37

within sixty seconds a reasonable solution and the second stage finds immediately a solution. For a large set of instances (50 items) we conclude that we can find a reasonable solution within a reasonable amount of time. For future research we suggest to use a larger number of items with our heuristic.

In this research we used distributions based on the production data to generate artificial data. As a consequence, our research has a direct practical contribution; however the data set is also a limitation. To increase the theoretical contribution of our research, it can be interesting to use the data set provided by Lodi, Martello and Vigo (1999) or Hopper and Turton (2001). In both papers, the researchers perform an approximation algorithm on a two-dimensional set of items and bins. They place their items freely in the space of the bin. As a remark, our heuristic is based on an additional constraint. Items cannot be placed freely in the space of a bin and must be supported by another item or the bin itself.

Additionally, the main assumption in this paper is that the items are relatively long compared to their height, which is a limitation for placing our research in the current theoretic field of work.

Nonetheless, research on two-dimensional packing problems using this type of items is scarce and our paper can be seen as novel.

(38)

38

References

Chen, M. and W. Huang (2007). A two-level search algorithm for 2D rectangular packing problem.

Computers & Industrial Engineering 53, 123-136

Chen, C.S., S.M. Lee, Q.S. Shen (1995). Theory and Methodology An analytical model for the container loading problem. European Journal of Operational Research 80, 68-76.

Dowsland, K.A. and W.B. Dowsland (1992). Packing problems. European Journal of Operational

Research 56, 2-14.

Dyckhoff, H. (1990). A typology of cutting and packing problems. European Journal of Operational

Research 44, 149-159.

Fernández, A. C. Gil, R. Baños and M. G. Montoya. A parallel multi-objective algorithm for two-dimensional bin packing with rotations and load balancing. Expert Systems with Applications 40, 5169– 5180

Gilmore, P.C. and R.E. Gomory (1965). Multistage cutting problems of two and more dimensions.

Operations Research 13, 94–119.

Hopper, E., and B. Turton (2001). An empirical investigation of meta-heuristic and heuristic algorithms for a 2D packing problem..European Journal of Operational Research 128, 34–57.

Junqueira, L., R. Morabito and D.S. Yamashita (2012). MIP-based approaches for the container loading problem with multi-drop constraints. Annals of Operations Research 199, 51-75

Lenstra, J.K., A.H.G. Rinnooy Kan and P. Brucker (1977). Complexity of machine scheduling problems.

Annals of Discrete Mathematics 1, 343-362.

Lodi, A., Martello, S., and M. Monaci (2002). Two-dimensional packing problems: A survey. European

Journal of Operational Research 141, 241–252.

Lodi, A., Martello,S. and D. Vigo (1999a). Heuristic and metaheuristic approaches for a class of two-dimensional bin packing problems. Journal of Computing 11, 345–357.

Martello, S., D. Pisinger and D. Vigo (2000). The three-dimensional bin packing problem. Operations

(39)

39

de Queiroz, T. A. and F.K. Miyazawa (2013). Two-dimensional strip packing problem with load balancing, load bearing and multi-drop constraints. International Journal of Production Economics 2, 511-530

Valério de Carvalho, J.M. (2002). LP models for bin packing and cutting stock problems. European

(40)

40

Appendix A: Paired sampled t test for the utilization percentage

Paired Samples Statistics

Mean N Std. Deviation Std. Error Mean Pair 1 Experiment 1 .6877 18 .05731 .01351 Experiment 2 .7479 18 .05313 .01252 Pair 2 Experiment 3 .7285 18 .05340 .01259 Experiment 4 .7479 18 .05313 .01252 Pair 3 Experiment 5 .7808 18 .04213 .00993 Experiment 6 .8021 18 .03634 .00857

Paired Samples Correlations

N Correlation Sig. Pair 1 Experiment 1 &

Experiment 2

18 .286 .250 Pair 2 Experiment 3 &

Experiment 4

18 .367 .134 Pair 3 Experiment 5 &

Experiment 6

18 .703 .001

Paired Samples Test

(41)

41

Appendix B: Paired sampled t test for the number of bins

Paired Samples Statistics

Mean N Std. Deviation Std. Error Mean Pair 1 Experiment1 1.1467 18 .21639 .05100 Experiment2 1.0272 18 .04968 .01171 Pair 2 Experiment3 2.4472 18 .43224 .10188 Experiment4 2.2706 18 .34671 .08172 Pair 3 Experiment5 4.5956 18 .63783 .15034 Experiment6 4.3194 18 .52545 .12385

Paired Samples Correlations

N Correlation Sig. Pair 1 Experiment1 &

Experiment2

18 .961 .000 Pair 2 Experiment3 &

Experiment4

18 .901 .000 Pair 3 Experiment5 &

Experiment6

18 .973 .000

Paired Samples Test

A Heuristic for Two Dimensional Packing of Rectangular Items in Large Identical Bins

A Heuristic for Two Dimensional Packing

of Rectangular Items in Large Identical

Bins

Bas Minkes

s1462091

Master’s Thesis Technology and Operations Management

Supervisor RUG: Dr. ir. D. Catanzaro

Co-assessor RUG: Dr. E. Ursavas

Supervisor Philips: W. Mulder

A Heuristic for Two Dimensional Packing

of Rectangular Items in Large Identical

Bins

Bas Minkes

s1462091

June 30, 2014

Abstract

Contents

List of Figures

List of Tables

1. Introduction

2. Theoretical Background

3. Notation and Problem Formulation

4. Mathematical Model

4.1 Original Mathematical Model

4.2 Simplification of the Mathematical Model

4.3 Numerical Example

5. Heuristic Approach

5.1 First Stage of the Heuristic Approach

5.2 Second Stage of the Heuristic Approach

5.3 Numerical Example

6. Computational Experiments and Results

6.1 Data Description

6.2 Experiment Settings

6.3 Performance Criteria

6.4 Results

7. Conclusion

References

Appendix A: Paired sampled t test for the utilization percentage

Appendix B: Paired sampled t test for the number of bins