A recursive branch-and-bound algorithm for the rectangular guillotine strip packing problem

Transcription

1 Computers & Operations Research 35 (2008) A recursive branch-and-bound algorithm for the rectangular guillotine strip packing problem Yaodong Cui, Yuli Yang, Xian Cheng, Peihua Song Department of Computer Science, Guangxi Normal University, Guilin , PR China Available online 16 October 2006 Abstract A heuristic recursive algorithm for the two-dimensional rectangular strip packing problem is presented. It is based on a recursive structure combined with branch-and-bound techniques. Several lengths are tried to determine the minimal plate length to hold all the items. Initially the plate is taken as a block. For the current block considered, the algorithm selects an item, puts it at the bottom-left corner of the block, and divides the unoccupied region into two smaller blocks with an orthogonal cut. The dividing cut is vertical if the block width is equal to the plate width; otherwise it is horizontal. Both lower and upper bounds are used to prune unpromising branches. The computational results on a class of benchmark problems indicate that the algorithm performs better than several recently published algorithms Elsevier Ltd. All rights reserved. Keywords: Strip packing; Cutting stock; Recursive algorithm; Branch and bound 1. Introduction The guillotine rectangular strip packing problem discussed in this paper appears widely in the manufacture of refrigerators, washing machines, and air-conditioners, where guillotine shears are used to divide the strip into rectangular items. It can be formally stated as: m small rectangular items are to be cut from a plate, where any cut that is made is restricted to be orthogonal guillotine cut, and the small items have to be laid out on the plate orthogonally. The ith item has size l i w i (length l i and width w i ), i = 1,...,m, and may be rotated by 90. The plate width W is fixed; the length is variable and has to be minimized. This problem is referred to as subtype RG of the two-dimensional strip packing problem (2D-SPP) that includes the following subtypes [1]: RF RG OF OG the items may be rotated by 90 (R) and no guillotine cutting is required (F); the items may be rotated by 90 (R) and guillotine cutting is required (G); the orientation of the items is fixed (O) and no guillotine cutting is required (F); the orientation of the items is fixed (O) and guillotine cutting is required (G). Corresponding author. Tel.: address: ydcui@263.net (Y. Cui) /$ - see front matter 2006 Elsevier Ltd. All rights reserved. doi: /j.cor

2 1282 Y. Cui et al. / Computers & Operations Research 35 (2008) They will be referred, respectively, as RF 2D-SPP, RG 2D-SPP, OF 2D-SPP, and OG 2D-SPP. For more detailed descriptions of the 2D-SPP, the reader is referred to [1 3]. Many authors have investigated the 2D-SPP problem, which belongs to a subset of classical cutting and packing problems and has been shown to be NP hard [4,5]. Hifi [6] presented exact algorithms for RG 2D-SPP. The algorithms are based upon branch-and-bound procedures, and can solve some small and medium problem instances. Lesh et al. [7] presented exhaustive approaches to OF 2D-SPP perfect packing that contains no waste space. They demonstrated a simple pruning approach that makes a branch-and-bound-based exhaustive search effective for problems with less than 30 rectangles. However, exact algorithms may be impractical for solving large-scale problems. To deal with problems with larger scale, most algorithms are heuristic in nature [2,8]. Zhang et al. suggested a heuristic recursive algorithm for RG 2D-SPP [9]. The algorithm considers a set of sequences of the items. For each sequence, it arranges the items using a recursive structure, and obtains the consumed plate length. The minimal length is determined after considering all sequences in the set. Martello et al. [10] proposed a relaxation that produces good lower bounds and gives information to obtain effective heuristic algorithms for OF 2D-SPP. The results were used in a branch-and-bound algorithm that was able to solve test instances from the literature involving up to 200 items. A near-optimal solution to RG 2D-SPP was reported in [11] without experimental tests. Hybrid algorithms combining non-deterministic and deterministic techniques were also reported [12 20]. Bortfeldt [12] suggested a genetic algorithm for 2D-SPP. The algorithm works without any encoding of solutions. Yeung and Tang [13] presented a hybrid genetic approach for OF 2D-SPP. With a combination of genetic algorithm and a heuristic allocation method, the packing problem is transformed into a simple permutation problem. The time complexity of the heuristic method is O(n) for the best and average cases. Tests on three larger problems (with number of items equal to 100, 300 and 500, respectively) indicate that the solutions are close to optimal with a difference less than 10%. Lee and Sewell [14] used a simulated annealing algorithm to solve 2D-SPP that appeared in a boat manufacturing firm. They demonstrated through experimental tests that the solutions are within 9.5% of the optimal solution, on average. Liu and Teng [15] suggested a genetic algorithm (for RF 2D-SPP) that considers the bottom-left-condition to reduce the number of possible orthogonal packing patterns. Dagli and Poshyanonda [16] presented two approaches for RF 2D-SPP. One is the neuro-optimization approach that integrates artificial neural networks and linear programming techniques; the other is the neuro-nesting approach that combines neural networks and genetic algorithms. Zhang et al. [17] presented a meta-heuristic algorithm for RG 2D-SPP. The algorithm is mainly based on the heuristic recursive strategy and simulated annealing algorithm. Wu et al. [18] presented a quasi-human heuristic, less flexibility first, for solving 2D-SPP. However, some of these non-deterministic algorithms are time consuming and are hence less practical for problems with large scale. An empirical investigation of meta-heuristic and heuristic algorithms for 2D-SPP was given by Hopper and Turton [19]. Kröger [20] presented a genetic algorithm for RG 2D-SPP. A guillotine constraint is directly reflected by the encoding mechanism and thus is ensured at any stage of the algorithm. The computational results indicate that the algorithm is superior to different approaches like random search or simulated annealing. Iori et al. [21] proposed a hybrid algorithm for OF 2D-SPP that is based on combining genetic and tabu search techniques. Although some of them can deal with large-scale problems, the solution may be relatively far from optimal, except that reported in [12,20]. This paper presents a heuristic algorithm for RG 2D-SPP. The computational results on some benchmark problems indicate that the algorithm is efficient both in material utilization and in computation time. The contents of the paper are organized as follows: Section 2 introduces the characteristics of the proposed patterns; Section 3 presents the approach; Section 4 gives the computational results; and Section 5 terminates the paper with conclusions. 2. Geometrical structure of the patterns The proposed pattern was also used in [9,17]. Assume that the sizes of the plate and the items are all integers, and the plate is positioned with its left-bottom corner at the origin of the rectangular coordinates (Fig. 1). The plate length is in the horizontal direction (X-direction). The plate is divided into several sections. The length of a section is equal to either the length or the width of the item placed at the bottom of the section. The pattern in Fig. 1 consists of five sections, where the numbers indicate the indexes of the items. Items 5, 7, 26, 19 and 17 are at the bottom of the sections. Patterns of this type are also referred as having a layer structure, with each section being a layer [3,12].

3 Y. Cui et al. / Computers & Operations Research 35 (2008) Fig. 1. The proposed pattern. Fig. 2. (a) The division of a segment; (b) the division of a block. A rectangular region of the plate is referred to either as a segment or a block. It is a segment if its width (measured vertically) is equal to the plate width; otherwise it is a block. The original plate can be also taken as a segment. If the current region is a segment (Fig. 2a), we select an item R, put it at the bottom-left corner of the segment, and divide the unoccupied region vertically into a block A and a segment B. It may be also interpreted as that the dividing cut divides the segment into a segment on the right and a section on the left, with item R at the bottom of the section. If the current region is a block (Fig. 2b), the unoccupied region is divided horizontally into two blocks A and B. A segment can only be divided vertically to keep the layer structure of the pattern, whereas a block may be divided horizontally, vertically, or both horizontally and vertically. Different dividing policies may lead to different results. This paper only considers dividing the block horizontally. 3. The approach As mentioned in the last section, to determine the layout of items in a rectangular region, the region is divided vertically if it is a segment, and horizontally if it is a block. This section first defines two recursive functions to determine the layouts, one for the segments, and the other for the blocks; then presents some heuristic rules to improve the computational performance; finally overall algorithm is presented.

4 1284 Y. Cui et al. / Computers & Operations Research 35 (2008) The recursive function for the blocks The value of an item is equal to the item area. The value of a block is the total area of the items packed in the block. Assume that the items have been arranged in descending order of their areas, and function RBlock(x,y,V f,u f, B, C,k) returns the value of block x y, where x y V f U f B C k block length. block width. the total value of the items that have been placed on the plate before the current block is considered. It is also referred to as the value of the father pattern. the upper bound of the unoccupied regions that have been generated on the plate before the current block is considered. That is to say, the total value of the items that can be packed in these regions later cannot be larger than U f. B={b 1,...,b m }, where b i = 1iftheith item has been placed in the father pattern, b i = 0 otherwise. C={c 1,...,c m }, where c i = 1iftheith item is placed in the current block, c i = 0 otherwise. only items with index larger than or equal to k will be considered to be placed in the current block. Recall that the items have been indexed according to the descending order of their areas. From Fig. 1 it is known that the maximum size of a section cannot be larger than L s W, where L s = max i (l i,w i ). Let S = m i=1 s i = m i=1 l i w i. Assume that V is the value of the best pattern obtained so far; v max is the best value of block x y obtained so far; k c is the minimum index of the items that have not been placed and have sizes not larger than the block. F(x,y) = xy is the upper bound for block x y, 0 x L s,0 y W. Box 1 shows the steps of recursive function RBlock(). To understand the steps, the reader may refer to the explanations that follow the box. Box 1. The steps of recursive function RBlock() Step 1: Return 0 if F(x,y) = 0; otherwise let i = k, v max = 0, and k c = 0. Step 2: If i>mthen go to Step 7. Step 3: If b i = 1 then let i = i + 1 and go to Step 2. Step 4: First, consider to place item i without rotation: Step 4.1: If l i >x or w i >y then go to Step 5; otherwise, if k c = 0 then let k c = i. Step 4.2: Place item i at the bottom-left corner of the block, and divide the unoccupied region horizontally (Fig. 2b). Let x A = x l i, y A = w i, x B = x, y B = y w i, where x A y A and x B y B are the sizes of regions A and B, respectively. Let the upper bound of the block be u b = s i + F(x A,y A ) + F(x B,y B ). Let the upper bound of the whole pattern be u p = V f + U f + u b.gotostep5ifu b v max or u p V. Step 4.3: Let b i = 1, and C A = 0. Let V fa = V f + s i, U fa = U f + F(x B,y B ), and v A = RBlock(x A,y A, V fa,u fa, B, C A,k c ). Step 4.4: Let u b = s i + v A + F(x B,y B ), u p = V f + U f + u b.ifu b v max or u p V then go to Step 4.9. Step 4.5: Let B = B + C A, C B = 0. Let V fb = V f + s i + v A and U fb = U f. Let v B = RBlock(x B,y B,V fb, U fb, B, C B,k c ). Let v = s i + v A + v B.Ifv v max then go to Step 4.8. Step 4.6: Let v max = v, C = C A + C B and c i = 1. If V f + v max V then go to Step 4.8. Step 4.7: Let V = V f + v max.ifv = S then terminate the recursive process. Step 4.8: Let B = B C A. Step 4.9: Let b i = 0. If v max = xy then go to Step 7. Step 5: Consider to place item i with rotation. The approach is similar to that in Step 4. Step 6: Let i = i + 1 and go to Step 2. Step 7: Return v max. Now we explain the steps in detail. For Step 1: Either x or y are zero when F(x,y)= 0; the block cannot hold any item and its value is zero. Otherwise, let the initial value of the block be v max = 0; let the index of the first item to be considered be i = k; let k c = 0 to denote that the minimum index has not been found.

5 Y. Cui et al. / Computers & Operations Research 35 (2008) For Step 2: All items have been considered when i>m. Go to Step 7 to return the block value. For Step 3: Item i has been placed when b i = 1. Go to Step 2 to consider the next item. For Steps 4 and 5: Steps 4 and 5 consider to place item i without and with rotation, respectively. Only Step 4 will be explained below. When l i >xor w i >y, the item cannot be placed in the block without rotation. Go to Step 5 to skip Step 4. Otherwise if k c = 0, let k c = i to record the first item that can be actually placed in the block (Step 4.1). For the child blocks that are branched from the current block, it is not necessary to consider items with indexes smaller than k c, because the sizes of the child blocks are smaller than that of the current block. Step 4.2 places item i at the left-bottom corner of the block, and divides the unoccupied region horizontally into two regions A and B (Fig. 2b). The upper bound of region A is F(x A,y A ), and that of region B is F(x B,y B ). Therefore, with item i at the corner, the upper bound of the block is u b = s i + F(x A,y A ) + F(x B,y B ). This branch should be discarded if u b v max. The upper bound of the whole pattern related to the current block consists of three parts: (1) V f, the value of the items placed in the father pattern; (2) U f, the upper bound of the unoccupied blocks in the father pattern; (3) u b, the upper bound of the current block. When u p = V f + U f + u b V, the upper bound of the whole pattern is not larger than that of the best pattern obtained so far, the current branch should be also discarded. Step 4.3 calls the recursive function to determine the value v A of region A. b i = 1 denotes that item i is placed in the father pattern of region A. C A is a vector denoting the items placed in the region and is initialized to 0. V fa = V f + s i is the value of the items that have been placed in the father pattern. U fa = U f + F(x B,y B ) is the upper bound of the unfilled blocks in the father pattern. It differs from U f by F(x B,y B ), because region B is unoccupied at the current time. In Step 4.4, both the upper bound of block x y and that of the whole pattern is improved, because the value of region A has been found and v A F(x A,y A ) holds. Go to Step 4.9 to discard the current branch if u b v max or u p V. Step 4.5 calls the recursive function to determine the value v B of region B. Vector B is renewed to include those items placed in region A. C B is a vector denoting the items placed in region B and is initialized to 0. V fb = V f + s i + v A is the value of the items placed in the father pattern of the region. U fb = U f is the upper bound of the unfilled blocks in the father pattern. v = s i + v A + v B is the value of block x y with item i at its corner. When v v max, discard the current branch and go to Step 4.8 to restore vector B. Step 4.6 renews the current best solution to block x y, where c i = 1 denotes that item i is placed in the current block. If the whole pattern has a value not larger than V, the current best solution should not be renewed. Therefore, go to Step 4.8 to restore vector B. Step 4.7 replaces the current best solution. When V = S, all items have been placed, the recursive process is terminated. Steps restore vectors B so as to consider the next item. When v max = xy, the maximum value of the block has been found, so go to Step 7 to exit from the recursive function The recursive function for the segments Assume that function RSegment(x,y,V f,u f, B, C,k) returns the value of segment x y, where y = W. The meanings of the parameters are the same as those of function RBlock(x,y,V f,u f, B, C,k), with the word block being replaced by segment. Assume that f(x)= xw is the value of the unconstrained solution for segment x W, 0 x L. The steps of the function are the same as those of function RBlock(), except the steps shown in Box 2. The main differences of the steps of the two recursive functions are: (1) The upper bounds of the blocks are kept in a two-dimensional array F(x,y); whereas those of the segments are kept in a one-dimensional array f(x) to save memory space, because all segments have the same width W. (2) The unoccupied region of a block is divided horizontally; whereas that of a segment is divided vertically (Step 4.2).

6 1286 Y. Cui et al. / Computers & Operations Research 35 (2008) (3) Inside the recursive function for the blocks, only function RBlock() is called; whereas inside that for the segments, both RBlock() and RSegment() are called. Box 2. Some steps of recursive function RSegment() Step 1: If f(x)= 0 then return 0; otherwise let i = k, v max = 0, and k c = 0. Step 4.2: Place item i at the bottom-left corner of the segment, and divide the unoccupied region vertically (Fig. 2a). Let the size of region A be x A y A, with x A = l i and y A = W w i. Let the length of region B be x B = x l i. Let the upper bound of the segment be u b = s i + F(x A,y A ) + f(x B ). Let the upper bound of the whole pattern be u p = V f + U f + u b.ifu b v max or u p V then go to Step 5. Step 4.3: Let b i = 1, and C A = 0. Let V fa = V f + s i, U fa = U f + f(x B ), and v A = RBlock(x A,y A,V fa, U fa, B, C A,k c ). Step 4.4: Let u b = s i + v A + f(x B ), u p = V f + U f + u b.ifu b v max or u p V then go to Step 4.9. Step 4.5: Let B = B + C A, C B = 0. Let V fb = V f + s i + v A and U fb = U f. Let v B = RSegment(x B,W,V fb, U fb, B, C B,k c ). Let v = s i + v A + v B.Ifv v max then go to Step Heuristic rules The following rules are helpful either for shortening the computation time or for improving the material utilization. (1) Use constraint on the maximum computation time for the current trial plate length. The searching process will be terminated as soon as possible if the computation time t for the current trial length has reached t max. Considering this rule, the following step may be added to the recursive functions: Step 0: Return 0 if t>t max. (2) Improving the solution to the current block/segment at most a number of times. Within the recursive function, v max is the best value of the block/segment obtained so far, and its initial value is zero. Assume that the solution to a segment will be improved at most N ms times, and that to a block will be improved at most N mb times. Let n m be the number of times at which v max has been improved. Considering this rule, Steps 4.6 and 4.9 in Section 3.1 can be modified, respectively, as: Step 4.6: Let n m = n m + 1. Let v max = v. Let C = C A + C B and c i = 1. If V f + v max V then go to Step 4.8. Step 4.9: Let b i = 0. If v max = xy or n m = N mb then go to Step 7. Steps 4.6 and 4.9 in Section 3.2 can be modified similarly. Different values will be used for N mb during the computation period for the current trial plate length. Assume that N mb0 and N mb1 are two positive integers and N mb0 N mb1. N mb will be set equal to N mb0 when t t max /2, and to N mb1 otherwise. (3) Underestimating the upper bounds. Only branches with higher upper bounds will be considered. Assume that L is the current trial plate length, and L 0 = S/W. Assume that r is a real number, 0 <r 1. Let r = 1 when t t max /2, and r = L 0 /L otherwise. In Step 4 (either Section 3.1 or 3.2), F(x,y) will be replaced with r F(x,y); and in Step 4 of Section 3.2, f(x)will be replaced with r f(x). That is to say, at the late age of the computation period for the current trial plate length, the upper bounds are underestimated, because r = L 0 /L < 1. (4) Fill the smaller region first. When an item has been placed at the corner of the current block, the unoccupied region is divided into two regions A and B. The recursive function in Section 3.1 always fills region A before filling region B. It should be modified to consider this rule The algorithm We use HRBB to denote the algorithm, where H stands for heuristic, R for recursive, and BB for branch-and-bound. Assume that the calculated minimal occupied plate length cannot be larger than L max. The steps of Algorithm HRBB are shown in Box 3 and are explained afterward.

7 Box 3. The steps of the algorithm Y. Cui et al. / Computers & Operations Research 35 (2008) Step 1: Arrange the items in descending order of their areas. Let L = L 0 = S/W. Let L max = αl 0, where α > 1. Step 2: Let the upper bound of block x y be F(x,y) = xy, 0 x L s and 0 y W. Let the upper bound of segment x W be f(x)= xw,0 x L max. Step 3: Let B = C = 0, V = S 0.1, and V = RSegment(L, W, 0, 0, B, C, 1). Step 4: If V = S then let L = L + 1 and go to Step 3. Step 5. Output L and the related pattern. Recall that V is the value of the best pattern obtained so far. In Step 3 it is initialized to S 0.1 instead of 0. The reason is that if the maximum value of the patterns with length L cannot be larger than S 0.1, it should be discarded as soon as possible, so as to try the next length. V is used as the global lower bound in the recursive functions. It is a very tight bound indeed. The algorithm tries all plate lengths between L 0 = S/W and L, where L is the calculated minimum plate length to hold all items. Some techniques can be used to shorten the searching space, such as the dichotomy method that is briefly described in Box 4. Box 4. The steps of the dichotomy method Step 1: Initially, let a =L 0, b =L max, c =a. Let B=C=0, V =S 0.1, and V =RSegment(c, W, 0, 0, B, C, 1). If V = S then go to Step 6. Step 2: If b a 1 then let c = b and go to Step 6; else let c = (a + b)/2. Step 3: Let B = C = 0, V = S 0.1, and V = RSegment(c, W, 0, 0, B, C, 1). Step 4: If V<Sthen let a = c else let b = c. Step 5: Go to Step 2. Step 6: Output c as the calculated minimum plate length. 4. The computational results This section presents the computational results of the test problems. The computations were performed on a computer with Pentium 4 CPU 2.80 GHz, and main memory 512 MB. There are two groups of benchmark problems used The computational results of the problems in the first group The first group includes 21 benchmark problems presented in [19]. They were generated such that the optimal patterns contain no waste. They can be classified into seven problem categories, each of which includes three problems. The number of item types ranges from 16 to 197 items. Table 1 shows some features of the problems, where the last column gives the occupied lengths of the optimal solutions (L opt ). The reader is referred to [19] for the problem details. Table 1 Features of the test problems Problem category Number of items: m Plate width: W Plate length: L opt C1 (C11 C13) 16 or C2 (C21 C23) C3 (C31 C33) 28 or C4 (C41 C43) C5 (C51 C53) C6 (C61 C63) C7 (C71 C73) 196 or

8 1288 Y. Cui et al. / Computers & Operations Research 35 (2008) Table 2 The computational results of the HRBB Set t max (s) t average (s) g (%) Table 3 The minimal plate lengths for each set ID L 1 L 2 L 3 L 4 L 5 L 6 C C C C C C C C C C C C C C C C C C C C C Table 4 Gap to the optimal solution (%) Approach C1 C2 C3 C4 C5 C6 C7 Average/best GA + BLF SA+BLF SPGAL JOIN HR SA+HR HRBB The rules in Section 3.3 are all considered in the algorithm, and the dichotomy method is used in searching the minimum feasible plate length. Six sets of parameters are used in the computations. N ms = N mb0 = 4, N mb1 = 1, and α = 1.05 (see Step 1 in Section 3.4) for all sets. For each set, the maximum computation time allowed for a trial plate length (t max ), and the average computation time t average of one problem, are shown in Table 2. Table 3 lists the minimum feasible plate lengths L i (i = 1,...,6). The gap to optimal solution for each set is also shown in Table 2. It is defined as g = (L L opt )/L opt, where L opt is obtained from Table 1, and L from Table 3. Table 4 gives the computational results obtained from different algorithms. The gaps of GA+BLF and SA+BLF are best values; the gaps of the other algorithms are average values. The solutions to Set 1 have been used for the HRBB.

9 Y. Cui et al. / Computers & Operations Research 35 (2008) Table 5 Average computation times Approach C1 C2 C3 C4 C5 C6 C7 Average GA + BLF SA + BLF HR JOIN SA+HR SPGAL 139 HRBB GA+BLF and SA+BLF stand for the two best meta-heuristics reported in [19], SPGAL for that in [12], JOIN for the heuristic in [10], HR for that in [9], and SA+HR for that in [17]. The first four algorithms generated non-guillotine patterns, and the last three generated guillotine patterns. The average computation times of the algorithms are shown in Table 5. First, evaluate the solution quality. HRBB is more efficient in improving material utilization than other algorithms. From Table 3, this statement is straightforward acceptable for the algorithms except SPGAL. The gap of SPGAL is 1.0%, and that of HRBB is 1.4%. The following comments should be noted in comparing HRBB with SPGAL: (1) the gap of SPGAL is obtained from non-guillotine patterns; it may be increased if only guillotine patterns are allowed. (2) The computation time of HRBB is much shorter than that of SPGAL (1.86 s on a computer with 2.8 GHz CPU vs. 139 s on a computer with 2.0 GHz CPU). (3) If the maximum computation time allowed for a trial plate length is increased, HRBB can yield solutions comparable to those of SPGAL. For example, when t max = 40 s, the gap of HRBB is 1.0%, and the average computation time is s. Both of them are comparable to those of SPGAL. Now evaluate the computation time. GA+BLF and SA+BLF were performed on a computer with Pentium Pro 200 MHz CPU and 65 MB of RAM, SPGAL on a Pentium PC with 2 GHz CPU, JOIN on a computer with Pentium III 800 MHz CPU, HR on a DELL GX260 with 2.4 GHz CPU, and SA+HR on a DELL GX270 with 3.0 GHz CPU. It is obvious that HRBB is faster than HR, much faster than SA+HR and SPGAL, because that the performances of the computers used are close. We cannot accurately estimate how faster HRBB is than GA+BLF, SA+BLF, and JOIN because the performances of the computers used to test the algorithms are quite different The computational results of the problems in the second group The second group includes 12 benchmark problems presented in [20]. The number of item types ranges from 25 to 60 items. The reader is referred to [20] for the problem details. Two sets of parameters are used in the computations. N ms = 5, N mb0 = 4, N mb1 = 1, and α = 1.05 for the two sets. The maximum computation time allowed for a trial plate length (t max ) is, respectively, 1 and 30 s for Sets 1 and 2. Table 6 lists the calculated minimum occupied plate lengths L i (i = 1, 2) and those obtained from other algorithms, in which SC stand for the algorithm in [22] KR for that in [20]. The average plate lengths for different algorithms, and the average computation time for the HRBB are shown in the last two rows of the table. From the computational results we know that the results obtained from the HRBB is better than those of SC, and slightly inferior to those of KR and SPGAL, with (L 4 L KR )/L KR = 0.6% and (L 4 L SPGAL )/L SPGAL = 0.9% in average. The HRBB can be used to obtain solutions close to the best ones quickly. For example, the average computation time for the first set is only s and the gap to the best solutions is not larger than 1.3% in average, namely (L 1 L SPGAL )/L SPGAL = 1.3%. 5. Conclusions Algorithm HRBB for the rectangular strip packing problem has been described in detail. The computational results of the problems in the first group indicate that HRBB is efficient both in computation time and in material utilization than the algorithms except SPGAL; its computation time is shorter than that of SPGAL, its average material utilization is the same as that of SPGAL (see set 5 in Table 2). The computational results of the problems in the second group

10 1290 Y. Cui et al. / Computers & Operations Research 35 (2008) Table 6 The computational results of the second group ID L 1 L 2 L SC L KR L SPGAL L average t average indicate although HRBB cannot yield material utilization as better as those of KR and SPGAL, it can be used to obtain solutions close to the best ones within a short time period. It should be indicated that if the solution quality is very important, the SPGAL may be the best among the algorithms tested. If the computation time is important, the HRBB may serve as a candidate for generating the patterns. The heuristic rules used in this paper can be combined with other policies to divide the unoccupied region of a block. The region is only divided horizontally in this paper. It can also be divided vertically, or both vertically and horizontally. Different dividing policies may lead to different results. This may be investigated in the future. References [1] Lodi A, Martello S, Vigo D. Heuristic and metaheuristic approaches for a class of two-dimensional bin packing problems. INFORMS Journal on Computing 1999;11: [2] Lodi A, Martello S, Monaci M. Two-dimensional packing problems: a survey. European Journal of Operational Research 2002;141: [3] Wascher G, Haussner H, Schumann H. An improved typology of cutting and packing problems. European Journal of Operational Research, forthcoming. [4] Hochbaum DS, Wolfgang M. Approximation schemes for covering and packing problems in image processing and VLSI. Journal of the Association for Computing Machinery 1985;32: [5] Leung J, Tam T, Wong CS, Young G, Chin F. Packing squares into square. Journal of Parallel and Distributed Computing 1990;10: [6] Hifi M. Exact algorithms for the guillotine strip cutting/packing problem. Computers & Operations Research 1998;25: [7] Lesh N, Marks J, McMahon A, Mitzenmacher M. Exhaustive approaches to 2D rectangular perfect packings. Information Processing Letters 2004;90:7 14. [8] Hopper E, Turton BCH. A review of the application of meta-heuristic algorithms to 2d strip packing problems. Artificial Intelligence Review 2001;16: [9] Zhang D, Kang Y, Deng A. A new heuristic recursive algorithm for the strip rectangular packing problem. Computers & Operations Research 2006;33: [10] Martello S, Monaci M, Vigo D. An exact approach to the strip-packing problem. INFORMS Journal on Computing 2003;15: [11] KenYon C, Remila E. A near-optimal solution to a two-dimensional cutting stock problem. Mathematics of Operations Research 2000;25: [12] Bortfeldt A. A genetic algorithm for the two-dimensional strip packing problem with rectangular pieces. European Journal of Operational Research 2006;172: [13] Yeung LHW, Tang WKS. Strip-packing using hybrid genetic approach. Engineering Applications of Artificial Intelligence 2004;17: [14] Lee HF, Sewell EC. The strip-packing problem for a boat manufacturing firm. IIE Transactions 1999;31: [15] Liu D, Teng H. An improved BL-algorithm for genetic algorithm of the orthogonal packing of rectangles. European Journal of Operational Research 1999;112: [16] Dagli CH, Poshyanonda P. New approaches to nesting rectangular patterns. Journal of Intelligent Manufacturing 1997;8: [17] Zhang D, Liu Y, Chen S, et al. A meta-heuristic algorithm for the strip rectangular packing problem, Lecture Notes in Computer Science, 2005, p [18] Wu YL, Huang W, Lau SC. et al. An effective quasi-human based heuristic for solving the rectangle packing problem. European Journal of Operational Research 2002;141:

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