Relations between capacity utilization and minimal bin number

Transcription

1 Als Manuskript gedruckt Technische Universität Dresden Herausgeber: Der Rektor A note on Relations between capacity utilization and minimal bin number Torsten Buchwald and Guntram Scheithauer MATH-NM August 01

2 Contents 1 Introduction: problem formulation and goal 1 Preliminaries and notations 3 The theorem 6 4 Conclusions and outlook 1

3 A note on Relations between capacity utilization and minimal bin number Torsten Buchwald and Guntram Scheithauer Technische Universität Dresden July 01 Abstract In the paper Relations between capacity utilization and minimal bin number [HS1] the following two-dimensional bin packing problem (BPP) is considered: given a set K of rectangular items of sizes l i w i and bins of size L W. Rotation of the items is not permitted. It is shown, if every item fits into the given bin and if the total area of the items does not exceed the area of the bin then at most three bins are needed to pack all items. The bound three is tight. Furthermore, it is shown that if l i L/3 and w i W/3 for all i K then at most two bins are necessary to pack all items. In this note we give another, simplerproof of the latter resultwhich is based on properties of the FFDH-heuristic (instead of using NFDH-heuristic as in [HS1]), and we improve it to if l i L/ and w i W/3 for all i K then at most two bins are necessary. Key words: Cutting and Packing, Bin Packing Problem, Strip Packing Problem 1 Introduction: problem formulation and goal In this paper we consider as in [HS1] the two-dimensional bin packing problem (BPP) and the strip packing problem (SPP) with non-rotatable rectangular items: Let be given a list L(K) := {R i : i K} of small rectangles R i (items) of width l i > 0 and height w i > 0, i K := {1,...,n}. In the BPP the items have to be packed into a minimum number of rectangular bins of given width L and height W such that the items do not overlap each other and each of them fits completely into one of the bins, whereas in the SPP the height W of the bin is variable and has to be minimized. In [HS1] it is shown, if every item fits into the given bin, i.e. l i L and w i W for all i K, and if the total area of the items does not exceed the area of the bin, i.e. i K l iw i LW, then at most three bins are needed to pack all items. The bound three is tight. Furthermore, it is shown that if l i L/3 and w i W/3 for all i K then 1

4 A note on Relations between capacity utilization and minimal bin number at most two bins are necessary to pack all items. In order to prove these statements, properties of the well-known NFDH-heuristic (next fit, decreasing height, [CGJT80]) are used. In this paper we give another, simpler proof of the latter result which is based on properties of the FFDH-heuristic(first fit, decreasing height, [CGJT80]) and we improve it to if l i L/ and w i W/3 for all i I then at most two bins are necessary. Throughout the paper, for any set S K, or any list of rectangles L(S) := {R i : i S} with R i = l i w i, let A(L(S)) = A(S), l(s) and w(s) denote the total area, width, or height, i. e. A(S) := l i w i, l(s) := l i, w(s) := i. i S i S i Sw Moreover, let l max (S) := max{l i : i S} and w max (S) := max{w i : i S}. Preliminaries and notations For the sake of completeness we refer to Theorem 1 (Steinberg (1997) [Ste97]). If for a list L(K) of rectangles and a bin of size L W the following inequalities hold, l max (K) L, w max (K) W, A(K) LW (l max (K) L) + (w max (K) W) + then it is possible to pack all items into the bin. (As usual, (x) + = max{0,x}.) Note, in case of l max (K) L/ or w max (K) W/ then all items can be packed if their total area is at most 50% of the bin area. Consequently, if A(K) LW then all items fit in a rectangular bin of size L W or L W, but we cannot conclude that all items can be packed into two bins of size L W since the guillotine property is not guaranteed. In the following theorems we apply the well-known FFDH-heuristic in order to pack a single bin, denoted as FFDH 1 -heuristic, which works as follows: Let w j, j = 1,,..., s denote the heights of the (horizontal) strips obtained by the FFDH-heuristic applied for SPP with strip width L. Hence, w j w j+1 for j = 1,,..., s 1. Furthermore, let w s+1 := 0. In case not all items can be packed into a single bin by the FFDH-heuristic then s w j > W. We define s := argmax{ s w j W} as the number of strips packed by the FFDH 1 -heuristic. The FFDH 1 -heuristic stops just before the first item has to be packed into the (s + 1)st strip. Let K denote the item set packed by the FFDH 1 -heuristic. Additionally to the FFDH 1 -heuristic, we also consider the First-Fit Decreasing-Width 1 - heuristic (FFDW 1, transposed FFDH 1 -heuristic) where all items are sorted by nonincreasing width, and then they are packed sequentially in the left-most vertical strip in which they fit until the first item cannot be packed into the bin. As shown in [HS1], the (upper) bound three is tight in the case p = q = 1 when items with l i L/p, w i W/q for all i K and A(K) LW have to be packed into bins of

5 T, Buchwald, G. Scheithauer. August 3, 01 3 size L W. This bound is also tight for p = 1 and any q Z + = {1,,...}, and vice verse. Theorem. Let p = 1 and q Z +. Then there exist lists L(K) of items with l i L/p, w i W/q for all i K and A(K) LW such that at least three bins are needed to pack all items. Proof: The statement follows from the following example. Consider q + 1 items with l i = L+ε and w i = W+ε for some sufficient small positive ε and K = {1,...,q +1}. q+1 Then we have for a sufficient small positive ε: A(K) = l i w i = (q +1)( L+ε )( W +ε q +1 ) = (LW +ε(l+w +ε)) LW. q +1 q + i K Since only q items can be packed in one bin, at least three bins are necessary to pack all items. Lemma 3. Let q Z +, W/q w 1 w w n > 0 and n i=1 w i > W. Let s = argmax{ s i=1 w i W}. Then q s q +1 W w j W. Proof: Let us assume s w j < q W. Then w q+1 s+1 > W/(q +1). On the other hand, because of assumption, s q and due to the non-increasing sorting of the w i -values, we have s w j > s W q+1 = q W q+1 which leads to a contradiction. Corollary 4. Because of the non-increasing strip heights w j, j = 1,...,s generated by the FFDH 1 -heuristic, it follows: If not all items of K are packed by the FFDH 1 -heuristic, then we have for the total height used: q s q +1 W w j W. A similar result holds for the FFDW 1 -heuristic: Let l j, j = 1,...,s be the widths of the vertical strips obtained by the FFDW 1 -heuristic. In case not all items are packed then we have for the total width used: p s p+1 L l j L. In the FFDH 1 -heuristic we need to consider regular and non-regular items. If item r is packed into strip j and strip j is the highest non-empty strip at that time when item r is packed then r is called a regular item, otherwise a non-regular or fall-back item. Let K ij, j = 1,...,i, i = 1,...,s, denote the items set packed into strip j at the time before strip i + 1 has to be opened, i.e. when in strip i all regular items have been packed and probably some fall-back items in strips below strip i. Hence, K = s K sj. In case of K K the first item not packed into the bin, is denoted by k 0. CorrespondingtoK ij, j = 1,...,i, i = 1,...,s,wedefinetheusedwidth l ij := j K ij l j and area A ij := A(K ij ) = j K ij l j w j. Furthermore, for i = 1,...,s, let i := max{0, p p+1 L l ii }.

6 4 A note on Relations between capacity utilization and minimal bin number Theorem 5. Let p Z +, q Z +, and let l i L/p and w i W/q for all i K, and A(K) > 1 LW. Furthermore, it is assumed that not all items can be packed by the FFDH 1 -heuristic, i.e. K K, and k 0 is the first item not packed. If there does not exist any subset K K {k 0 } with A( K) 1 LW, that can be packed into the bin, then it holds w k0 < p+1 pq W. Proof: At first we show by induction i l ij p p+1 L i i, i = 1,...,s. (1) Initialization step i = 1: In this case, formula (1) reduces to l 11 p L max{0, p L l p+1 p+1 11 }, which is a valid inequality. Induction step: Let i {,...,s}, and let inequality (1) be fulfilled for i 1. If i 1 = 0 then i 1 l ij i 1 l i 1,j p p+1 L (i 1). Similar to the initialization step one can show the validity of inequality (1) also for i. In case of i 1 = p p+1 L l i 1,i 1 > 0 we distinguish two subcases: (i) Strip i does not contain any item k K ii with width l k < 1 L. p+1 1 Since the width of the first (regular) item in strip i is at least L+ p+1 i 1, and since at least p regular items are packed into strip i, we have l ii p L+ p+1 i 1. Since inequality (1) is valid for i 1 we obtain i l ij l ii + i 1 l i 1,j p L i+ p+1 i 1 i 1. Consequently, formula (1) is also fulfilled for i. (ii) Strip i contains a regular item t 0 with width w t0 < 1 L. p+1 This can only happen when further fall-back items with total width larger than i 1 are packed into strip i 1 beforeitem t 0 is packed into strip i. Hence, l i,i 1 l i 1,i 1 i 1. 1 Sincethefirstitemofstripihaswidthatleast L+ p+1 i 1 itfollowsbecauseofinduction i l ij l ii + i 1 l i 1,j + l i,i 1 l i 1,i 1 p L i p+1 i. Consequently, inequality (1) is fulfilled, and we obtain s l sj p L s p+1 s. If s = 0 then s l sj p p+1 L s. Otherwise, if s > 0, then we have l ss = p L p+1 s, and hence s 1 l sj p L s p+1 s l ss = p L (s 1). p+1 In order to construct a packing with s l sj p L s we replace strip s by another p+1 one as follows. Sort the items of K ss {k 0 } according to non-increasing width and pack them in this sequence into the s-th strip. Because of Corollary 4 the new packing of strip s has width l s p L. Consequently, the obtained new bin packing fulfills p+1 s 1 l sj + l s p L s. p+1

7 T, Buchwald, G. Scheithauer. August 3, 01 5 The rest of the proof is done by contradiction. Let w k0 p+1 W. Since at least q strips pq are packed into the bin, it follows for the total area A of the packed items: A s i=1 j K si l j w j s p+1 i=1 j K si l j W s p+1 pq i=1 l sj W s p p+1 L W pq p+1 pq q p p+1 L W = 1 LW which contradicts the assumption. p+1 pq Corollary 6. Let p Z +, q Z +, and let l i L/p and w i W/q for all i K, and A(K) > 1LW. If all items of K can be packed into a single bin, or if w k 0 p+1 W then pq there exists a subset K K with A( K) 1 LW which can be packed into a single bin. Proof: Since at least one of the assumptions of Theorem 5 cannot be fulfilled, either all items of K can be packed into a bin, or there exists a subset K with A( K) 1LW which can be packed into one bin. Since A(K) 1 LW, the statement follows. Corollary 7. Let p Z +, q Z +, and let l i L/p and w i W/q for all i K, and A(K) > 1LW. If all items of K can be packed into a single bin, or if l k 0 q+1 L (where pq k 0 denotes the first item which is not packed by the FFDW 1 -heuristic) then there exists a subset K K with A( K) 1 LW which can be packed into a single bin. Proof: The statement follows from the previous corollary applied on an instance with switched l i and w i, respectively L and W. Theorem 8. Let p Z +, and let l i L/p for i K. Then for the total area of items packed by the FFDH-heuristic it holds: A(K) = s A sj p L s w p+1 j+1. Proof: For i = 1,..., s we show by induction: i A ij p p+1 L i Initialization step i = 1: In this case, inequality () reduces to A 11 w ( p L max{0, p L l p+1 p+1 11 }), which is a valid inequality because of A 11 w l 11. Induction step: Let 1 < i s, and let () be fulfilled for i 1. If i 1 = 0 then i 1 A ij i 1 A i 1,j p p+1 L i 1 w j+1, w j+1 i w i+1 () and, similar to the initialization step, the validity of () can be shown also for i. In case of i 1 > 0 we distinguish two subcases: (i) Strip i does not contain any item with width smaller than 1 L. p+1 1 Since the width of the first (regular) item in strip i is at least L+ p+1 i 1, and since at least p regular items are packed into strip i, it follows that l ii p L+ p+1 i 1. Hence,

8 6 A note on Relations between capacity utilization and minimal bin number A ii ( p p+1 L+ i 1) w i+1 +( 1 p+1 L+ i 1) ( w i w i+1 ) p p+1 L w i+1 + i 1 w i. Using the induction hypothesis one obtains i A ij A ii + i 1 A i 1,j p L i p+1 w j+1 + i 1 w i i 1 w i and therefore inequality () holds also for i. (ii) Strip i contains a regular item t 0 with width smaller than 1 L. p+1 This can only happen when further, fall-back items have been packed into strip i 1 with total width larger than i 1 before item t 0 is packed into strip i. Therefore, 1 A i,i 1 A i 1,i 1 i 1 w i+1. Sincethefirstiteminstripihaswidthatleast L+ p+1 i 1, it follows with the induction hypothesis: i A ij A ii + i 1 A i 1,j +A i,i 1 A i 1,i 1 ( p p+1 L i) w i+1 +( 1 p+1 L+ i 1)( w i w i+1 )+ p p+1 L i 1 w j+1 i 1 w i + i 1 w i+1 p p+1 L i w j+1 i w i+1. Therefore inequality () is valid, and it follows with w s+1 = 0: A = s A sj p p+1 L s w j+1 s w s+1 = p p+1 L s w j+1. Corollary 9. As a consequence of this theorem it follows for the total height needed by the FFDH-heuristic for SPP to pack a set( K of items ) with A(K) LW: s w j p W. p q Proof: From Theorem 8 it follows: ( ) s w j p+1 A(K)+ w pl 1 p+1 LW + W = p W. pl q p q Corollary 10. Let A(K) LW and let l i L/3 and w i W/3 for all i K, i.e. p = q = 3. Then all items of L(K) can be packed into at most two bins. Proof: Because of the previous corollary, the FFDH-heuristic for SPP packs all items of K with total height not larger than s w j 5 W. On the other hand, using the 3 FFDH 1 -heuristic to fill a single bin covers at least 3 W of height because of Corollary 4. 4 Therefore, at most two bins are needed to pack all items of K. 3 The theorem Theorem 11. Let be given bins of size L W and a list of rectangular items L(K) with l max (K) L/, w max (K) W/3, i.e. p = and q = 3. If A(K) = i K l i w i LW then all items of L(K) can be packed into at most two bins.

9 T, Buchwald, G. Scheithauer. August 3, 01 7 Proof: Without loss of generality, we may assume that A(K) = LW, and we consider the following partition of the set of items: K 1 := {i K : l i 1 3 L} and K := {i K : l i > 1 3 L}. Case 1: Let A(K 1 ) 1 LW and A(K ) 1 LW. Because of Steinberg s theorem, item set K 1 as well as K can be packed into a bin so that at most two bins are needed to pack K. Case : Let A(K 1 ) 1 LW and A(K ) > 1 LW. Pack items of K according to the FFDW 1 -heuristic into the first bin. Here, and in the following, let B 1 denote those items packed into the first bin (either with FFDW 1, FFDH 1 or somehow else). Since l i > 1 L = q+1 L for all items of K 3 pq, it follows with Corollary 7 that there exists a subset K K with A( K ) 1 LW, which can be packed into the first bin. Because A(K \B 1 ) = LW A(B 1 ) 1 LW, these items can be packed into the second bin using Steinberg s theorem. Case 3: Let A(K 1 ) > 1 LW and A(K ) 1 LW. Pack items of K 1 according to the FFDH 1 -heuristic into the first bin, hence, now B 1 K 1, and let s denote again the number of (horizontal) strips. In case of A(B 1 ) 1 LW then all remaining items can be packed into the second bin due to Steinberg s theorem. In the following, let A(B 1 ) < 1LW and w i < p+1 W = W, i K pq 9 1\B 1 according to Theorem 5 applied on K 1, i.e. p = q = 3. Suppose, the last condition is not fulfilled, then because of Corollary 6 a subset K of K 1 would exists with A( K) 1 LW which can be packed into a bin. Hence, all items of K could be packed into two bins. Because of the sorting in FFDH 1 we have w i < 9 W for all items i K 1\B 1. With formula () and Theorem 5 we obtain A(B 1 ) p L s w p+1 j+1 s w s+1 > 3L(W w 4 1) s w s+1 (3) > 3 4 LW 3 (3L 4 3 L)W = 1 LW 1 LW Now, sort the items of K according to non-increasing height, and pack them leftjustified on top of the previous, as long as possible, until the bin height is reached, into the second bin (Fig. 1, left). Let W := w(k ). We distinguish three subcases: Subcase (i) Let W > W. Because of Corollary 4 the total height occupied in the second bin is at least 3 W. Since 4 A(K ) 1LW andl i > 1Lwe have w(k 3 ) < 3W. Let W := W (W 3W) = 7W W. 4 4 Pack the remaining, non-packed items of K right-justified, starting at the bottom of the bin on top of each other into the second bin. Above these items it remains an unused rectangularareawithsize 1L W (Fig. 1, right). Moreover, A(K ) > 1Lw(K 3 ) = 1LW. 3 We show that the items of K 1 \ B 1 can be packed into the unused area of the second bin. For the total area of the non-packed items of K 1 we have:

10 8 A note on Relations between capacity utilization and minimal bin number Figure 1: Packing of many items of K A(K 1 \B 1 ) = A(K) A(B 1 ) A(K ) < LW ( 1 LW LW ) LW = 1 LW LW + LW Therefore, A(K 1 \B 1 ) < ( 1 LW LW + L LW W) = LW + LW = LW + LW L W + 7LW L W = 35 LW L W + L (7W W). 4 Two subcases have to be investigated: (a) Let 4W 9 (7 W W) 0, then 4 A(K 1 \B 1 ) < 35 LW LW + L 16 6 (7W W) 4 = 35 LW LW + L 16 6 (7W W) (l 4 max (K 1 \B 1 ) L) +( 4W 9 (7W W)) 4 + < L (7W W) (l 4 max(k 1 \B 1 ) L) +(w max (K 1 \B 1 ) ( 7W W)) 4 + = L W (l max (K 1 \B 1 ) L ) +(w max (K 1 \B 1 ) W) +. Consequently, due to Steinberg s theorem, the items of K 1 \B 1 can all be packed into a rectangle of size 1 L W. (b) Let 4 W 9 (7 W W) > 0, then 4 A(K 1 \B 1 ) < LW L 6 W + L (7W 4 W) = LW L 6 W + L (7W 4 W)+ L 6 (w max(k 1 \B 1 ) ( 7 4 W W)) + L 6 (w max(k 1 \B 1 ) ( 7 4 W W)) LW L 6 W+L (7W 4 W)+L 6 ( 9 W (7 4 W W)) + L 6 (w max(k 1 \B 1 ) ( 7 4 W W)) + = LW L 6 W + L (7W 4 W)+ L 6 (4 9 W 7 4 W +W) L 6 (w max(k 1 \B 1 ) ( 7 4 W W)) + = LW + L (7W 4 W)+ L W L 6 (w max(k 1 \B 1 ) ( 7 4 W W)) + < L (7W 4 W) L 6 (w max(k 1 \B 1 ) ( 7 4 W W)) + = L (7W 4 W) ( 1 3 L L ) +(w max (K 1 \B 1 ) ( 7 4 W W)) + L (7W 4 W) (l max (K 1 \B 1 ) L ) +(w max (K 1 \B 1 ) ( 7 4 W W)) + = L W (l max (K 1 \B 1 ) L ) +(w max (K 1 \B 1 ) W) + Again, due to Steinberg s theorem, also in this subcase the items of K 1 \B 1 can all be packed into a rectangle of size 1 L W. Subcase (ii) Let W W and A(K ) LW 4 + LW 54. Because of W W all items of K are already packed on the left side of the second bin. For the total area of the not yet packed items of K 1 it holds: A(K 1 \B 1 ) := A(K) A(B 1 ) A(K ) LW ( 1 LW LW 54 ) (LW + LW) = LW

11 T, Buchwald, G. Scheithauer. August 3, 01 9 Because of l i 1 L i K there exists a non-occupied area of size 1 L W in the second bin. (cf. Fig 1, left). Due to Steinberg s theorem, all items of K 1 \B 1 can be packed into that rectangle. Subcase (iii) Let W W and A(K ) < LW + LW Therefore A(K 1 ) = A(K) A(K ) > LW ( LW + LW) = 3LW 1 LW We consider a partition of K 1 into K 1 := {i K 1 : w i 5 81 W} and K1 := {i K 1 : w i < 5 81 W}. and the following two subcases: (a) Let A( K 1 ) 1LW. Empty the first bin (i.e. set B 1 := ) and pack items of K 1 using the FFDH 1 -heuristic into the first bin (now B 1 K 1 ). In case not all items of K 1 are packed into the first bin and there exists no subset Ǩ1 K 1 with A(Ǩ1) 1 LW that can be packed into one bin then, because of formula () and Theorem 8, A(B 1 ) p L s p+1 w j+1 s w s+1 > 3L(W w 4 1) s w s+1 > 3 4 L56W 81 (3L 4 3 L)W = 14LW 1 LW = 1LW Since A(B 1 ) 1LW all items of K \ B 1 can be packed into the second bin due to Steinberg s theorem. (b) Let A( K 1 ) < 1 LW, then A(K 1 ) = A(K) A(K ) A( K 1 ) > LW ( LW + LW ) 1 LW = LW LW = 5 LW. Let M = {1,...,m} := K 1 \B 1 (with B 1 as defined at the beginning of Case 3) be sorted according to non-increasing item area. Because of Theorem 5 we have l i w i < 1 3 LW = 9 LW for all i M. Furthermore, let B 7 1 := B 1 = K 1 \M. Because of Theorem 5, all items of K 1 are packed in the first bin, i.e. K 1 B 1. Several subcases can occur: (1) Let l(k 1 ) L. Let j M denote the smallest index with j i=1 l iw i 1 54 LW. Let M := {1,...,j}. LW. 7 Moreover, because of the assumed sorting in M, we have A(M) = j i=1 l iw i < Next we show that all items of M can also be packed into the first bin besides those of B 1. In that case, then A(B 1 M) 1 LW 1 LW + 1 LW = 1 LW, and consequently, the remaining items can be packed into the second bin due to Steinberg s theorem. In order to show this we distinguish subcases: (A) Let A(B 1 ) 1 LW LW. Empty the first bin (i.e. B 1 = ) and pack all items of K 1 along the bottom side of the bin, i.e. B 1 := K 1. Above these items a non-used rectangular area of size L 3 W (Fig., left) remains. It holds, A(B 1 \K 1 )+A(M) 1LW 1 5 LW LW LW = LW = 1LW Hence, all items of (B 1 \K 1 ) M can be packed into a rectangle of size L 3 W. (B) Let A(B 1 ) > 1 LW 1 LW. 108 If A(K ) LW + LW then, similar to subcase (ii), a packing of K into two bins can be constructed. Otherwise,

12 10 A note on Relations between capacity utilization and minimal bin number k Figure : Packing of items of K 1 A(K 1 ) = A(K) A(K ) A( K 1 ) > LW ( LW 4 + LW 108 ) 1 LW = LW 4 LW 108 = LW Analogously to subcase (A), one obtains A(B 1 \K 1 )+A(M) 1 6 LW LW LW = LW = 1LW Therefore, also in this Subcase, the items of M (B 1 \K 1 ) can be packed into a rectangle of size L W. 3 () Let l(k 1 ) > L. Sort the items of K 1 according to non-increasing width, and pack them left-justified side by side along the bottom of the bin until the next item does not fit. Let K 1 denote the set of packed items. Because of Corollary 4, l( K 1 ) 3L and therefore A( K 4 1 ) 5 L. 108 Two cases are distinguished: (A) Let l( K 1 ) 7 L, then 9 A( K 1 ) 7 9 L5W = 175 LW. We consider the following partition of M: M 1 := {i M : l i w i LW} and M := {i M : l i w i > LW}. Two cases are distinguished: (α) Let A(M 1 ) 1 LW. 54 Let M 1 = {1,...,m 1 } be sorted according to non-increasing item area. Let j 1 M 1 denote the smallest index with j 1 i=1 l iw i 1 LW. Let M 54 1 := {1,...,j 1 }. Moreover, because of the assumed sorting in M 1, we have A(M 1 ) = j 1 i=1 l iw i LW. Analogously to the previous case we show that all items of (B 1 \ K 1 ) M 1 can be packed into the unused area of size L W. For their total area we have 3 A(B 1 \ K 1 )+A(M 1 ) LW LW + LW = LW = 486 LW = 1LW Hence, A(B 1 ) = A(B 1 M 1 ) 1 LW, and all remaining items can be packed into the second bin. (β) Let A(M 1 ) < 1 LW, then 54 A(M ) = A(K) A(K ) A(B 1 ) A(M 1 ) > LW ( LW + LW ) LW LW = LW = 3 LW. Because of this and l iw i 9 L1W = 8 LW for all i M, the set M contains at least three items. Pack three items of M side by side

13 T, Buchwald, G. Scheithauer. August 3, on the top of the first bin (Fig. 3). Because of w i < W for all i M 9 there exist a unused area of size L 4W. We show that there exists a set Ǩ 9 1 K 1 \ ( K 1 M ) with A( K 1 M Ǩ1) 1 LW that can be packed into this unused area. Let M = {1,..., m} := K 1 \ ( K 1 M ). So we have A( M) = A(K 1 \ ( K 1 M )) = A(K 1 ) A( K 1 ) A(M ) ( )LW = LW = 19 LW > 1LW. Let j M denote the smallest index with j i=1 l iw i 1 j LW. Because of the sorting one obtains 9 i=1 l iw i LW. Hence, 9 these items can be packed in the unused area and we obtain A(B 1 ) 175 LW LW LW = LW = 833 LW > 1 LW. Due to Steinberg s theorem all remaining items can be packed into the second bin. Figure 3: Packing of items of K 1 and M (B) Let l( K 1 ) < 7 L. 9 Because of construction of K 1 it holds that l i > L i K 9 1 since otherwise a further item of K 1 could be packed besides the items of K 1. Empty again the first bin (i.e. B 1 = ), and pack items of K 1 using the FFDW 1 - heuristic, into the bin. If now A(B 1 ) 1 LW then all remaining items can be packed into the second bin. If however A(B 1 ) < 1 LW then, because of Theorem 5, all items of K 1 are packed, i.e. K1 B 1. Let k denote the area-largest item of K 1 \B 1. Because of Theorem 5 and the sorting of items with respect to non-increasing width one has l k < L. We consider two subcases: 9 (α) Let l k w k 1 LW. 54 Empty the first bin and pack items of K 1 \{k} using the FFDH 1 -heuristic into that bin. Let B 1 K 1 \{k} denote those items now packed into the first bin. If A( B 1 ) 1LW then we are done. Otherwise, if A( B 1 ) < 1LW then it holds A( B 1 ) 1 LW 1 LW because of inequality (3). Furthermore, because of Theorem 5, all items 54 of K 1 are packed in the first bin, i.e. K1 B 1. We show that the items B 1 {k} can bepacked into one bin. Empty again the first bin and pack anew items of K 1 along the bottom side of bin one. Since the total width is at most 7 L, item k fits besides the items 9 of K 1. Then, for the total area of not yet packed items of B 1 we have A( B 1 \ K 1 ) 1 LW LW = LW < 1 3 LW. Therefore, these items can be packed into the unused area of size L 3 W (Fig., right). Because of inequality (3), the total area of this packing is A( B 1 )+l k w k 1 LW.

14 1 A note on Relations between capacity utilization and minimal bin number Consequently, all items not packed into the first bin can be packed into the second due to Steinberg s theorem. (β) Let l k w k < 1 LW. 54 Let M be defined similar to above. Now 1 1 LW A(M) LW Empty the first bin and pack items of K 1 \M using the FFDH 1 -heuristic. Let B 1 denote the set of items which are packed now. Then A( B 1 ) 1LW 1 LW. If A( B 54 1 ) < 1LW, empty again the first bin and pack items of K 1 side by side along the bottom of the first bin. Then A( B 1 \ K 1 )+A(M) 1 5 LW LW LW = LW = 33 LW < 1LW Hence, all these items can be packed into the first bin, and we have A(B 1 ) = A(( B 1 M) 1 LW. Therefore, all remaining non yet packed items fit in the second bin. Case 4: LetA(K 1 ) > 1 LW and A(K ) > 1 LW. Because of A(K) = A(K 1 ) + A(K ) > 1 LW + 1 LW = LW the knapsack condition A(K) LW is violated. Consequently, this case cannot occur. 4 Conclusions and outlook Let be given positive integers p and q, bins of size L W and a set K of rectangles with l i L/p, w i W/q for all i K and A(K) LW. In [HS1] it is shown if p = q = 1 then all items can be packed into at most three bins, and if p = q = 3 at most two bins are needed. These bounds are tight. In this paper, we show that only two bins are needed in case of p =,q = 3, or vice verse. Moreover, we give worst-case instances for p = 1 and q Z + where three bins are needed. There is still a lack between these statements: the case p = q = is still open, which will be part of future research. References [CGJT80] E. G. Coffman, M. R. Garey, D. S. Johnson, and R. E. Tarjan. Performance bounds for level oriented two dimensional packing algorithms. SIAM J. on Computing, 9(4):808 86, [HS1] [Ste97] K. Hoffmann and G. Scheithauer. Relations between capacity utilization and minimal bin number. Preprint MATH-NM-03-01, Technische Universität Dresden, 01. A. Steinberg. A strip-packing algorithm with absolute performance bound. SIAM J. on Computing, 6(): , 1997.