Integer Rounding and Modi ed Integer Rounding for the ...scheith/ABSTRACTS/PREPRINTS/15-MIRDP.pdf · Integer Rounding and Modi ed Integer Rounding for the Skiving Stock Problem John

Als Manuskript gedruckt

Technische Universitat DresdenHerausgeber: Der Rektor

Integer Rounding and Modified Integer Rounding forthe Skiving Stock Problem

John Martinovic Guntram Scheithauer

MATH-NM-02-2015

January 2015

Contents

1 Introduction and Preliminaries 1

2 A generalization of Zak’s theorem 3

3 The MIRDP of the divisible case 7

4 A construction principle for gaps greater than one 11

5 Some investigations on the maximum gap for the divisible case 21

6 Conclusion and outlook 25

Integer Rounding and Modified Integer Rounding forthe Skiving Stock Problem

John Martinovic Guntram Scheithauer

Technische Universitat Dresden

January 28, 2015

Abstract

We consider the one-dimensional skiving stock problem which is also known as thedual bin packing problem: find the maximum number of items with minimum lengthL that can be constructed by connecting a given supply of m ∈ N smaller itemlengths l1, . . . , lm with availabilities b1, . . . , bm. For this optimization problem, weinvestigate the quality of the continuous relaxation by considering the gap, i.e., thedifference between the optimal objective values of the continuous relaxation and theskiving stock problem itself. In a first step, we derive an upper bound for the gapby generalizing a result of E. J. Zak. As a main contribution, we prove the modifiedinteger round-down property of the divisible case. In this context, we also presenta construction principle for non-IRDP instances of the divisible case and investigatethe maximum gap that can be obtained based on these instances.

Key words: Skiving Stock Problem, (Modified) Integer Round Down Property, Gap,Divisble Case, Continuous Relaxation

1 Introduction and Preliminaries

In this paper, we consider the one-dimensional skiving stock problem (SSP) which is alsoknown as the dual bin packing problem (DBPP) in literature. In the classical formulation,m ∈ N := {1, 2, . . .} different item lengths l1, . . . , lm with availabilities b1, . . . , bm are given,the so-called item supply. We aim at maximizing the number of products with minimumlength L that can be constructed by connecting the items on hand.

Such computations are of high interest in many real world applications, e.g. industrial pro-duction processes (see [Zak03] for an overview) or politico-economic problems (cf. [AJKL84]and [LLM95]). Furthermore, also neighboring tasks, such as dual vector packing problem[CFGR91] or the maximum cardinality bin packing problem [BD85], [PD06], are often as-sociated or even identified with the dual bin packing problem. These formulations are ofpractical use as well since they are applied in multiprocessor scheduling problems [ARGA04]or surgical case plannings [VPS+13].

1

2 Integer Rounding and Modified Integer Rounding for the Skiving Stock Problem

Throughout this paper, we will use the abbreviation E := (m, l, L, b) for an instance of theSSP with l = (l1, . . . , lm)> and b = (b1, . . . , bm)>. Without loss of generality, we assume allinput-data to be positive integers with L > l1 > . . . > lm > 0.

The classical solution approach is due to [Zak03] and based on the formulation of Gilmoreand Gomory in the context of one-dimensional cutting [GG61]. Any feasible arrange-ment of items leading to a final product of minimum length L is called (packing) patternof E. We always represent a pattern by a nonnegative vector a = (a1, . . . , am)> ∈ Zm+where ai ∈ Z+ denotes the number of items of type i ∈ I := {1, . . . ,m} being containedin the considered pattern. For a given instance E, the set of all patterns is defined byPE :=

{a ∈ Zm+

∣∣ l>a ≥ L}

. In [MS14], the authors slightly improve Zak’s formulation byonly considering so-called minimal patterns obtaining a finite model, hereinafter referredto as the standard model, of the skiving stock problem. A pattern a ∈ PE is called minimalif there exists no pattern a ∈ PE such that a 6= a and a ≤ a hold (componentwise). Theset of all minimal patterns is denoted by P ?

E. Let xj ∈ Z+ denote the number how often

the minimal pattern aj = (a1j, . . . , amj)> ∈ Zm+ (j ∈ J) of E is used, where J = {1, . . . , n}

represents an index set of all minimal patterns. Then the skiving stock problem can beformulated as

z?(E) = max

{∑j∈J

xj

∣∣∣∣∣∑j∈J

aijxj ≤ bi, i ∈ I, xj ∈ Z+, j ∈ J

}.

A common (approximate) solution approach consists in considering the continuous relax-ation

z?c (E) = max

{∑j∈J

xj

∣∣∣∣∣∑j∈J

aijxj ≤ bi, i ∈ I, xj ≥ 0, j ∈ J

}and the application of appropriate heuristics.

Practical experience and computational simulations, cf. [Zak03], have shown that there isonly a small gap ∆(E) := z?c (E) − z?(E) for any instance E. Based on the contributionsof Baum and Trotter [BT81] for general linear maximization problems, these observationshave initiated the following definitions. A set P of instances has the integer round-downproperty (IRDP) if ∆(E) < 1 holds for all E ∈ P . An instance E with ∆(E) ≥ 1 is callednon-IRDP instance. Furthermore, a set P of instances has the modified integer round-downproperty (MIRDP) if ∆(E) < 2 is true for all E ∈ P . It is conjectured in [Zak03] that theone-dimensional skiving stock problem possesses the MIRDP.

In this paper, we investigate the gap of the skiving stock problem from a theoretical pointof view. To this end, we first take the result of [Zak03, Theorem 4], i.e., the proof ofthe IRDP for all instances with m = 2 item lenghts, as an initial point and generalizethis inequality to arbitrary values of m. This case and other special cases have also beenstudied by Marcotte [Mar83] in the context of one-dimensional cutting.

In Section 3, we focus intensively on the divisible case of the skiving stock problem whereli|L holds for each item length li (i ∈ I). For these instances, we prove the MIRDP bymeans of the first fit decreasing (FFD) heuristic presented in [AJKL84] for the dual binpacking problem. Afterwards, we introduce a construction priciple for an infinte number ofnon-equivalent non-IRDP instances of the divisible case. In a final step, we investigate themaximum gap that can be obtained on the basis of these instances, give some conclusionsand an outlook of future research.

J. Martinovic, G. Scheithauer. January 28, 2015 3

2 A generalization of Zak’s theorem

To our best knowledge, there is only few work concerning theoretical aspects of the skivingstock problem’s gap. One of the only results being known in literature is Theorem 4 in[Zak03] where the IRDP for instances with m = 2 item lengths is proved. In the followingtheorem, we generalize this statement such that it provides an upper bound for the gap ofarbitrary instances E = (m, l, L, b).

Theorem 1. Let m ∈ N with m ≥ 2 be given and let E = (m, l, L, b) be an arbitraryinstance with m item lengths. Then ∆(E) < m− 1 holds.

Proof. Let n := |J | denote the number of all minimal patterns of E. Then, consider theoptimization problem

c>x→ max s.t.(A I

)x = b, x ≥ 0 (1)

where c ∈ Zn+m+ is defined by cj = 1 for j ∈ {1, . . . , n} and cj = 0 for j ≥ n+ 1. Note that

the columns of A consist of the minimal patterns of E, and I represents the appropriateidentity matrix. Let j1, . . . , jm ∈ {1, . . . , n + m} denote the (pairwise different) indices ofthose columns of

(A I

)that belong to the basis matrix of an arbitrary but fixed solution

xLP ∈ Rn+m+ of (1). Thus, we have xLPj = 0 for all j ∈ {1, . . . , n+m} \ {j1, . . . , jm}.

i) Case 1: There is an index jk ∈ {j1, . . . , jm} with jk ≥ n+ 1.Let A ⊆ {j1, . . . , jm} contain all these indices. Then, we obtain an integer solutionby means of

xj :=

{⌊xLPj

⌋for j ∈ {j1, . . . , jm} \ A,

0 otherwise.

Due to |A| ≥ 1,

z?c (E)− z?(E) ≤n+m∑j=1

xLPj −n+m∑j=1

xj

=∑

j∈{j1,...,jm}\A

(xLPj − xj

)︸︷︷︸<1

< |{j1, . . . , jm} \ A| = m− |A| ≤ m− 1

follows which proves the assertion.

ii) Case 2: jk ≤ n holds for all jk ∈ {j1, . . . , jm}.Without loss of generality, we set j1 = 1, . . . , jm = m. Then, the equality

m∑j=1

aijxLPj = bi (2)

holds for all i ∈ {1, . . . ,m}. If there were an index k ∈ {1, . . . ,m} with xLPk ∈ Z+,we would obtain an integer solution by means of

xj :=

{⌊xLPj

⌋for j ∈ {1, . . . ,m},

0 otherwise


proving the assertion due to

z?c (E)− z?(E) ≤n+m∑j=1

xLPj −n+m∑j=1

xj =m∑j=1

(xLPj − xj

)=

m∑j=1,j 6=k

(xLPj − xj

)+(xLPk − xk

)︸︷︷︸=0

=m∑

j=1,j 6=k

(xLPj − xj

)︸︷︷︸<1

< m− 1.

Thus, only the case where xLPj (for all j ∈ {1, . . . ,m}) is not an integer remains. Tothis end, we consider the m-dimensional hypercube W (m) (see Fig. 1) described by

W (m) ={y ∈ Rm

+

∣∣ ∀ j ∈ {1, . . . ,m} : yj ∈[⌊xLPj

⌋,⌈xLPj

⌉]}.

1LPx

2LPx

1LPx

2LPx

A B

CD

(1)

(2)

Figure 1: The cube W (2) with the areas for the subsequent case-by-caseanalysis. Triangle ABD (without the segment BD) corresponds to Case2.1, triangle BCD corresponds to Case 2.2.

Due to the non-integrality assumption, the projection of xLP onto the first m com-ponents belongs to the interior of W (m).

a) Case 2.1: Letm∑j=1

(xLPj −

⌊xLPj

⌋)< 1.

Then,

m∑j=1

(xLPj −

⌊xLPj

⌋)< 1

⇐⇒m∑j=1

xLPj −m∑j=1

⌊xLPj

⌋< 1

=⇒ z?c (E)− z?(E) ≤m∑j=1

xLPj −m∑j=1

⌊xLPj

⌋< 1 ≤ m− 1

follows since

xj :=

{⌊xLPj

⌋for j ∈ {1, . . . ,m},

0 otherwise


represents an integer solution. Thus, the assertion has been proved for Case 2.1.

b) Case 2.2: Letm∑j=1

(xLPj −

⌊xLPj

⌋)≥ 1.

Defining

∆xj := xLPj −⌊xLPj

⌋> 0

for all j ∈ {1, . . . ,m} and

a?i := bi −m∑j=1

aij⌊xLPj

⌋for all i ∈ {1, . . . ,m}, we obtain, by means of (2),

∀ i ∈ {1, . . . ,m} : a?i =m∑j=1

aij∆xj.

Then, a? := (a?1, . . . , a?m)> ∈ Zm+ describes a packing pattern since we obtain

l>a? =m∑i=1

lia?i =

m∑i=1

li

m∑j=1

aij∆xj

=m∑j=1

∆xj

m∑i=1

liaij =m∑j=1

∆xj ·(l>aj

)≥

m∑j=1

∆xj · L = L ·m∑j=1

∆xj ≥ L

with the help of this case’s assumption.

Without loss of generality, we assume a? to be a minimal pattern. Otherwise,some entries can be repeatedly reduced until this situation is achieved. Thus, a?

has to be a column of A. If a? does not match to one of the patterns a1, . . . , am,we can set am+1 = a? obtaining an integer solution by means of

xj :=

⌊xLPj

⌋for j ∈ {1, . . . ,m},

1 for j = m+ 1,

0 otherwise

since

[(A I

)x]i

=m∑j=1

aij⌊xLPj

⌋+

m∑j=1

aij∆xj =m∑j=1

aij(⌊xLPj

⌋+ ∆xj

)=

m∑j=1

aijxLPj = bi


holds for all i ∈ {1, . . . ,m}. Otherwise, there exists k ∈ {1, . . . ,m} with ak = a?.In this case, we define

xj :=

⌊xLPj

⌋for j ∈ {1, . . . ,m} \ {k},⌊

xLPj⌋

+ 1 for j = k,

0 otherwise

also representing an integer solution since

[(A I

)]i

=m∑

j=1,j 6=k

aij⌊xLPj

⌋+ aik ·

(⌊xLPk

⌋+ 1)

=m∑j=1

aij⌊xLPj

⌋+ aik

(aik=a?i )=

m∑j=1

aij⌊xLPj

⌋+

m∑j=1

aij∆xj =m∑j=1

aij(⌊xLPj

⌋+ ∆xj

)=

m∑j=1

aijxLPj = bi

is true for all i ∈ {1, . . . ,m}.In both cases, we obtain an objective value of z =

∑mj=1

⌊xLPj

⌋+ 1 implying that

z?c (E)− z?(E)(z?(E)≥z)≤

m∑j=1

xLPj −

(m∑j=1

⌊xLPj

⌋+ 1

)

=m∑j=1

(xLPj −

⌊xLPj

⌋)︸︷︷︸

<m

−1 < m− 1.

Thus, the assertion has been proved for Case 2.2.

This theorem provides first insights on instances having the IRDP and MIRDP, respectively.In particular, Theorem 4 from [Zak03] is contained as a special case.

Corollary 2. An instance E = (m, l, L, b) of the skiving stock problem with

i) m = 2 has the IRDP.

ii) m = 3 has the MIRDP.

Remark 3. Note that instances with m = 3 item lengths, in general, do not possess theIRDP. Considering E = (m, l, L, b) = (3, (21, 14, 6), 42, (1, 2, 6)), we notice that

∆(E) = z?c (E)− z?(E) =85

42− 1 =

43

42> 1.

Thus, E is a non-IRDP instance.


3 The MIRDP of the divisible case

In this section, we consider a special class of instances of the skiving stock problem. If aninstance E = (m, l, L, b) satisfies li|L for all i ∈ I, then E is an instance of the divisiblecase. In the context of one-dimensional cutting, those instances have turned out to possessthe modified integer round-up property (MIRUP), see [ST92] or [Rie03, page 24ff.]. In thefollowing, we prove that such a result can be obtained for the skiving stock problem, too.Thereby, the a priori knowledge of the continuous relaxation’s optimal objective value hasbeen proved beneficial.

Lemma 4. Let E = (m, l, L, b) be an instance of the divisible case. Then

z?c (E) =l>b

L(3)

holds.

Proof. Let ki := L/li be defined for all i ∈ I and let ei ∈ Zm+ denote the i-th unit vector.Considering the patterns ai = ki · ei and applying them xi = bi/ki times (i ∈ I), we obtaina feasible solution since

m∑j=1

aijxj = ki · xi = ki ·biki

= bi

holds for all i ∈ I. Moreover, the objective value results to

z =m∑j=1

xj =m∑j=1

biki

=m∑i=1

biliL

=l>b

L.

Thus, we have found a feasible solution whose objective value equals the upper boundl>b/L proving the optimality of the given feasible solution.

Note that the patterns ai (i ∈ I) from the proof satisfy l>ai = L. Those patterns are calledexact patterns.

For instances E = (m, l, L, b) of the divisible case, we firstly apply a scaling of the involvedlengths such that L is set to L′ = 1, and the (modified) item lengths belong to the set{1/2, 1/3, 1/4, . . .}. Then, the resulting instance is called standard form of E.

Lemma 5. Let E = (m, l, L, b) be an instance of the divisible case. Then, E is equivalent

to E = (m,κ, 1, b) where the entries of κ = (1/k1, . . . , 1/km)> are given by ki := L/li forall i ∈ I.

Proof. It suffices to show that E and E possess the same patterns:

a is a pattern of E

⇐⇒ l>a ≥ L, a ∈ Zm+

⇐⇒ 1

Ll>a ≥ 1, a ∈ Zm+

⇐⇒(

1

Ll

)>︸︷︷︸

=κ>

a ≥ 1, a ∈ Zm+

⇐⇒ a is a pattern of E


Hence, both corresponding optimization problems are equivalent.

Let E = (m, l, L, b) be an instance of the divisible case in standard form, i.e., L = 1. Atfirst, we only consider the case where E does not contain any exact pattern a ∈ P ?

E witha ≤ b. Thus, in particular,

li · bi < 1 for all i = 1, . . . ,m (4)

holds. Now, the first fit decreasing (FFD) algorithm, proposed in [AJKL84] (cf. Algorithm1) for the dual bin packing problem, represents an initial point of our further investigations.Thereby, bins of capacity lB = L = 1 are filled successively. A bin B is called filled if theobjects allocated to it possess a total length of at least lB. If this total length equals lB,the bin B is said to be filled exactly. Furthermore, we say that a bin B can accept anobject if, after adding this item, the total length of all objects in B does not exceed lB.

Algorithm 1 First Fit Decreasing

Phase I:

i) Let n = e>b, with e = (1, . . . , 1)> ∈ Zm+ , be the total number of items. Sort themaccording to decreasing lengths and number them consecutively:

1 > l?1 ≥ l?2 ≥ . . . ≥ l?n.

ii) If there are unallocated objects: Let i ∈ {1, . . . , n} be the index of the first unal-located object. Choose the first bin that is able to accept this object and place ittherein. If such a bin does not exist, add a new (empty) bin and place the objecttherein.

Phase II:

i) If there is more than one nonempty bin that is not filled: consider the last of thesebins, choose one of its items and allocate it to the first of these bins.

ii) If there is exactly one nonempty bin that is not filled: allocate its objects to thelast bin that is filled.

Note that after Phase II each non-empty bin contains objects with a total length of atleast L = 1. Hence, each bin can be interpreted as a uniquely determined packing patternof E. Actually, observe that all bins except for the very last one correspond to a minimalpattern. However, also the last non-empty bin can easily be reduced to a minimal pattern.

Lemma 6. Let E = (m, l, 1, b) be an instance of the divisible case in standard form notpossessing any exact pattern a ∈ P ?

E with a ≤ b. Furthermore, let q ∈ N denote the numberof existing bins after Phase I of Algorithm 1. Then, for all j ∈ {1, . . . , q − 1}, the j-th binBj contains at least j + 1 objects after the completion of Phase I.

Proof. Let j ∈ {1, . . . , q − 1} be given and let lji ∈ {l1, . . . , lm} denote an object lengththat is contained in bin Bj. Then, we claim that

lji ≤1

j + 1(5)


holds: Due to (4), Phase I of Algorithm 1 needs at most j − 1 bins for the allocation of allobjects with l?i ≥ 1/j. Hence, all objects with lengths li ≥ 1/j are contained in the binsB1, . . . , Bj−1 implying that the j-th bin Bj only possesses objects with lengths ≤ 1/(j+1).Since we have j < q, there exists a bin Bj+1. Based on the same considerations, all ofits objects’ lengths are ≤ 1/(j + 2). Let l? be the largest of these objects lengths, i.e.,l? ≤ 1/(j + 2) holds particularly. Due to Phase I of Algorithm 1, the remaining capacityR(j) of Bj results to

R(j) < l? ≤ 1

j + 2(6)

since, otherwise, the object with length l? would have been assigned to Bj.Assuming that there are at most j objects in Bj, then, due to (5), their total length wouldbe at most j · 1

j+1, i.e., the remaining capacity of Bj would result to

R(j) ≥ 1− j · 1

j + 1=

1

j + 1>

1

j + 2

which gives the contradiction, see (6). Thus, our assumption was wrong and the lemmahas been proved.

The following lemma states a lower bound for the number q of bins that exist after PhaseI of Algorithm 1. Thereby, the assumption that E does not contain any exact patterna ∈ P ?

E with a ≤ b plays an important role.

Lemma 7. Let E = (m, l, 1, b) be an instance of the divisible case in standard form. If Edoes not contain any exact pattern a ∈ P ?

E with a ≤ b and satisfies⌊l>b⌋

= K

for some K ∈ N, then the inequality q ≥ K + 1 holds.

Proof. We assume that there were at most K bins after the completion of Phase I. Then,the total length of all objects would be strictly smaller than K since no bin is filled exactly.This gives the contradiction.

Now we have all necessary preliminaries to prove our main result.

Theorem 8. Let E be an instance of the divisble case in standard form not containing anyexact pattern a ∈ P ?

E with a ≤ b. Then E has the MIRDP.

Proof. Let q denote the number of existing bins after Phase I. Then, certainly, q ≥⌊l>b⌋+1

holds. Let aij be the number of items of type i ∈ {1, . . . ,m} that have been assigned tothe bin j ∈ {1, . . . , q}. Due to the assumption of this theorem,

m∑i=1

liaij < 1

holds for all j ∈ {1, . . . , q}.


Let lqi ∈ {l1, . . . , lm} be the length of an arbitrary object of bin Bq. Then, we obtain

∀ j ∈ {1, . . . , q − 1} : 1−m∑i=1

liaij < lqi (7)

since, otherwise, the considered object would have been assigned to a previous bin.

i) Case 1: Bq contains at least q − 1 objects.Due to (7), each one of these q − 1 objetcs can be assigned to one of the binsB1, . . . , Bq−1 leading to a total length of at least L = lB in each of these bins. Since

q ≥⌊l>b⌋

+ 1

holds, we found a feasible solution for the ILP with objective value z = q−1 ≥⌊l>b⌋.

In this case, we can even state z?(E) =⌊l>b⌋

leading to the IRDP of E which impliesthe MIRDP, too.

ii) Case 2: Bq contains k < q − 1 objects.By means of the same procedure as in the first case, the bins B1, . . . , Bk can be filled.Now, we consider the bin Bq−1. Let lq−1

i be an arbitrary item length that is containedin Bq−1. Then, in analogy to (7), we obtain

1−m∑i=1

liaij < lq−1i

for all j = k + 1, . . . , q − 2 since, otherwise, this object would have been assignedto a previous bin by Phase I of the algorithm. Due to Lemma 6, Bq−1 containsat least q objects. As described in the first case, they can be used to fill the binsBk+1, . . . , Bq−2. Since

q ≥⌊l>b⌋

+ 1

holds, we found a feasible solution for the ILP with objective value z = q − 2 ≥⌊l>b⌋− 1. Hence, we obtain z?(E) ≥

⌊l>b⌋− 1 proving the MIRDP.

The previous investigations could only cope with special instances of the divisble casesince we required the absence of exact patterns with a ≤ b for our argumentations. In thefollowing, we show how Theorem 8 can be applied to arbitrary instances of the divisiblecase.

Theorem 9. Let E = (m, l, 1, b) be an instance of the divisible case in standard form.Then E has the MIRDP.

Proof. We divide E into two subinstances E1 = (m, l, 1, α) and E2 = (m, l, 1, β). There-fore, we remove successively exact patterns a ∈ P ?

E with a ≤ b from E and assign theircorresponding items to E2. The remaining instance, without exact patterns satisfyinga ≤ b, is called E1. Note that this partition is not unique. For the sake of simplicity, weallow αi and βi to equal zero for some i ∈ {1, . . . ,m} thus obtaining m item types in bothsubinstances.


i) subinstance E1:Here, we can apply Theorem 8 since E1 matches all of its prerequisites, and obtainz1 ≥

⌊l>α⌋− 1.

ii) subinstance E2:Since E2 consists of (disjoint) exact patterns, we have

⌊l>β⌋

= l>β leading to z2 =l>β.

Since E1 and E2 share no common objects, a feasible solution of E can be obtained bycombining the subinstances’ feasible solutions. This results to an objective value of

z = z1 + z2

≥(⌊l>α⌋− 1)

+ l>β

(l>β ∈Z+)=

⌊l>α + l>β

⌋− 1

(α+β=b)=

⌊l>b⌋− 1,

i.e., the optimal objective values satisfies z?(E) ≥⌊l>b⌋− 1 proving the MIRDP.

Obviously, this result cannot be improved since there are instances of the divisible case notpossessing the IRDP, cf. Remark 3.

4 A construction principle for gaps greater than one

In the previous explanations, we proved the MIRDP for the divisible case, but only indi-cated a single example of a non-IRDP instance. Thus, the question arises whether there aremany of those instances at all. In this section, we present a construction principle for aninfinite number of non-equivalent non-IRDP instances and calculate their gap explicitely.As an initial point, we state the following necessary condition.

Lemma 10. Let E = (m, l, 1, b) be an instance of the divisible case in standard form. If∆(E) ≥ 1 holds, then

K := K(E) :=⌊l>b⌋≥ 2

is true.

Proof. Assuming that the claim is wrong, there are two cases left: If K = 0 holds, theoptimal objective value of the ILP equals zero thus obtaining a contradiction by

∆(E) = z?c (E)− z?(E) = l>b− 0 = l>b < 1.

If, in the other case, K = 1 holds, the optimal objective value of the ILP equals one thusobtaining a contradiction by

∆(E) = z?c (E)− z?(E) = l>b− 1 < 2− 1 = 1.

Hence, the assumption was wrong and the lemma is proved.


Consequently, we can restrict our investigations to instances with K ≥ 2. So, let K ∈ Nwith K ≥ 2 be given and let P ⊂ N denote the set of all prime numbers, then we define

p := p(K) := inf

q ∈ P∣∣∣∣∣∣∣

q∑p=2,p∈P

p− 1

p∈ [K − 1, K)

. (8)

Obviously, p is well-defined due to the divergence of the series. Since all denominators ofthe considered summands {

1

p

∣∣∣∣ p ∈ P, p ≤ p

}are relatively prime, the common denominator

h := h(K) :=

p∏p=2,p∈P

p (9)

is given by the product of all these primes. Hence, the construction of p immediatelyimplies that there exists s := s(K) ∈ Z+ with s < h and

p∑p=2,p∈P

p− 1

p= K − 1 +

s

h. (10)

Lemma 11. Let h and s be given as in (9) and (10). Then, the following statements aretrue:

i) s and h are coprime.

ii) s > 0.

iii) h− s > 1.

Proof. We want to prove each assertion by contradiction:

i) Obviously,p∑

p=2,p∈P

p− 1

p=

p∑p=2,p∈P

(p− 1

h· rp)

holds with

rp :=h

p=

p∏q=2,

q∈P,q 6=p

q (11)

for all p ∈ Λ := Λ(K) := {p ∈ P | p ≤ p}.

Assuming that there is a common prime factor c ∈ P of s and h, this number has


to satisfy c ∈ Λ. Then, actually, c is a divisor of all rp with p ∈ Λ and p 6= c.Consequently, c is also a divisor of

(K − 1)h+ s−p∑

p=2,p∈P,p 6=c

(p− 1)rp

since all summands possess the prime factor c. Due to (10), we obtain

(K − 1)h+ s−p∑

p=2,p∈P,p 6=c

(p− 1)rp = (c− 1)rc

stating that c also divides (c− 1)rc. Since c ∈ P holds we have c ≥ 2, and c cannotdivide c − 1. Hence, c divides rc = h/c implying that c is, at least, a double primefactor of h which contradicts to (9).

ii) Assuming that s = 0 holds, then

1

h

p∑p=2,p∈P

(p− 1)rp =

p∑p=2,p∈P

p− 1

p= K − 1 ∈ N

follows directly from (10) und (11). Hence, the numerator on the left hand side hasto be a multiple of h. In particular, this numerator has to be divisible by each primefactor of h.

Now, we continue similar to the first part of this proof. Let t be a prime factor of h,then t ∈ Λ holds. All of the numerator’s summands of type (p− 1)rp with p ∈ Λ andp 6= t are divisible by t, but the summand (t− 1)rt is not. Thus, the numerator is nomultiple of t which gives the contradiction.

iii) Due to s < h, only the case h− s = 1 has to be excluded. Let h− s = 1, then (10)results to

p∑p=2,p∈P

p− 1

p= K − 1 +

h− 1

h= K − 1

h.

Now, we subtract the term (p− 1)/p on the left hand side obtaining the sum

p−1∑p=2,p∈P

p− 1

p=

p∑p=2,p∈P

p− 1

p− p− 1

p= K − 1

h− p− 1

p

≥ K − 1

h− h− 1

h= K − 1

where p ≤ h has been used, according to (9). But then, p is not minimal which givesthe contradiction.


In what follows, we would like to construct an instance E = (m, l, 1, b) of the divisible casein standard form satisfying bl>bc = K. Due to (10), our current item supply is not yetsufficient. Thus, we have to add some items in a smart way.

Proposition 12. There are positive integers v, w ∈ N satisfying

K − 1 +s

h+v

w= K +

1

hw. (12)

Proof. The assertion (12) can be transformed equivalently into

K − 1 +s

h+v

w= K +

1

hw⇐⇒ sw + hv − hw = 1⇐⇒ hv = (h− s)w + 1. (13)

Now, we aim at constructing a number g ∈ N with

g ≡ 0 mod h,

g ≡ 1 mod (h− s), (14)

g ≥ 2

since this number allows the definition

v :=g

h, w :=

g − 1

h− s. (15)

Then, (14) would lead to

hv = h · gh

= g,

(h− s)w + 1 = (h− s) · g − 1

h− s+ 1 = g

which proves (12) due to the equivalence (13).

In the following, we show how the number g can be constructed. As a consequence ofLemma 11 (i), the numbers h und h− s are coprime. By means of the extended Euclideanalgorithm, we can find ρ1, ρ2 ∈ Z such that

ρ1h+ ρ2(h− s) = gcd(h, h− s) = 1

holds. Setting γ := ρ1h, we obtain

γ ≡ ρ1h ≡ 0 mod h

by construction. Due to γ = ρ1h = 1− ρ2(h− s), also

γ ≡ 1− ρ2(h− s) ≡ 1 mod (h− s)

follows since h− s > 1 is true according to Lemma 11.

If γ ≥ 2 holds, set g := γ, otherwise consider

g := γ + δ lcm(h, h− s)


with

δ := inf {ξ ∈ N | γ + ξ lcm(h, h− s) ≥ 2} = min {ξ ∈ N | γ + ξ lcm(h, h− s) ≥ 2} .

Since h and h − s are coprime, lcm(h, h − s) = h(h − s) holds stating that g satisfies theconditions (14) because g ≥ 2 and

g ≡ γ + δ lcm(h, h− s) ≡ γ + δh(h− s) ≡ 0 mod h,

g ≡ γ + δ lcm(h, h− s) ≡ γ + δh(h− s) ≡ 1 mod (h− s)

hold. Defining v and w as in (15), the claim has been proved.

In the following lemma, we state some properties of v and w that will be needed for theconstruction of non-IRDP instances.

Lemma 13. The following statements are true:

i) Let C ∈ N be a given constant. Then, w in (12) can be chosen such that w > Cholds.

ii) The numbers w in (12) and h in (9) are coprime.

iii) The numbers v and w in (12) can be chosen such that v < w holds.

iv) There exists a sequence {(v(K), w(K))}K≥2 in N2 with:

(a) The element (v(K), w(K)) satisfies (12) for all K ≥ 2 .

(b) The sequence {w(K)}K≥2 is monotonically increasing with w(2) > 1.

(c) v(K) < w(K) holds for all K ≥ 2.

Proof.

i) Let C ∈ N with w ≤ C be given. Then, we set

g′ := g + lcm(h, h− s) = g + h(h− s)

=⇒ v′ =g′

h= v + (h− s),

w′ =g′ − 1

h− s= w + h

for the corresponding g ∈ N from the proof of Proposition 12. Hence, w′ ≥ w + 2holds and the condition (12) is still satisfied due to

K − 1 +s

h+v′

w′= K +

1

hw′(13)⇐⇒ hv′ = (h− s)w′ + 1

⇐⇒ h(v + h− s) = (h− s)(w + h) + 1

⇐⇒ hv = (h− s)w + 1

(13)⇐⇒ K − 1 +s

h+v

w= K +

1

hw

where the last line is true by construction of v and w. Since w increases by at least2 in every step, w > C is obtained after a finite number of steps.


ii) Assuming that w and h possess a common prime factor p ∈ P, then

gcd(hv, w(h− s)) ≥ p ≥ 2

holds. Due to (15), this implies gcd(g, g − 1) ≥ p ≥ 2 which gives the contradiction.

iii) Note that the method of part (i) of this proof cannot be applied directly since vincreases in each step, too. Assuming that v ≥ w, we have to consider two cases:

(a) Case 1: v = wIn this case, we obtain

hv = (h− s)w + 1(v=w)⇐⇒ hv = (h− s)v + 1⇐⇒ sv = 1⇐⇒ s = v = 1

by means of (12) and (13) implying that w = 1 holds due to v = w. In analogyto the construction in part (i), we set v′ = v + (h− s) and w′ = w + h. Due toh− s < h, we have v′ < w′, and the condition (12) is still satisfied.

(b) Case 2: v > wIn this case, we obtain

hv = (h− s)w + 1 = hw − sw + 1(v>w)< hv − sw + 1⇐⇒ sw < 1

by means of (12) and (13). Thus, at least one of the variables s and w has toequal zero. By definition and Lemma 11(ii), respectively, w = 0 and s = 0 areboth impossible stating that this second case cannot occur.

iv) This claim follows directly from the previous parts of this proof. Due to Propo-sition 12, for each K ≥ 2 there are natural numbers v(K) and w(K) satisfy-ing (12). Furthermore, according to Lemma 13(iii), these numbers can be chosensuch that v(K) < w(K) holds for all K ≥ 2. Considering the resulting sequence{(v(K), w(K))}K≥2, the properties (iv)(a) and (iv)(c) are satisfied by construction.

Due to Lemma 13(i), for K = 2 we can choose w(2) such that w(2) > 1 is true. Bythe same reason, w(K) ≥ w(K − 1) can be attained for each K ≥ 3. Note that thenecessary transformations (see part (i)) do not change the relation v(K) < w(K)since h − s < h holds. Thereafter, the sequence possesses all properties (a)-(c) ofassertion (iv).

Definition 1. A sequence possessing the properties of Lemma 13(iv) is called M-sequence.

With the help of these preliminary studies, we can formulate and prove the main result ofthis section.

Theorem 14. Let K ≥ 2 and an M-sequence{(v(K), w(K)

)}K≥2

be given. Setting

m = |Λ(K)|+ 1,

l =

(1

2,1

3,1

5, . . . ,

1

p(K),

1

w(K)

),

L = 1,

b = (1, 2, 4, . . . , p(K)− 1, v(K)) ,

we obtain an instance E = (m, l, L, b) possessing the following properties:


i) E is an instance of the divisible case in standard form with⌊l>b⌋

= K.

ii) There is no pattern a ∈ Zm+ with a ≤ b and

1 ≤ l>a < 1 +1

h(K)w(K). (16)

iii) The gap can be calculated by

∆(E) = 1 +1

h(K)w(K), (17)

i.e., E is a non-IRDP instance.

Proof.

i) Due to L = 1, E respresents an instance of the divisible case in standard form.Besides, we have

l>b =

p(K)∑p=2,p∈P

p− 1

p+v(K)

w(K)

(10)= K − 1 +

s(K)

h(K)+v(K)

w(K)

(12)= K +

1

h(K)w(K)

implying that bl>bc = K.

ii) Let a ∈ Zm+ be a pattern satisfying (16). Obviously, the common denominator of allcomponents of the vector l is given by h(K)w(K). Hence, a pattern can only have atotal length of

1 +ξ

h(K)w(K)

for some ξ ∈ Z+. According to the presumption (16), only the case ξ = 0 is possible.Thus, it suffices to show that an exact pattern a ∈ P ?

E with a ≤ b does not exist.

Assuming that a ∈ Zm+ is an exact pattern with a ≤ b, i.e., l>a = 1 holds. Let

I(a) = {i ∈ {1, . . . ,m} | ai > 0}

denote the set of active indices. Then,

l>a = 1 ⇐⇒m∑i=1

aili = 1 ⇐⇒ 1

h(a)

∑i∈I(a)

airi(a) = 1

holds where

h(a) :=∏j∈I(a)

1

lj

denotes the common denominator of all lengths appearing in a ∈ Zm+ and ri(a) isgiven by

ri(a) :=∏

j∈I(a),i 6=j

1

lj= lih(a) ∈ N


for all i ∈ I(a), similar to (11).

In order to obtain l>a = 1, the numerator∑i∈I(a)

airi(a)

has to be a multiple of every denominator in the set

N(a) :=

{1

li

∣∣∣∣ i ∈ I(a)

}.

Let t ∈ N(a) be given, then there is j ∈ I(a) with t = l−1j . Consequently, t divides all

summands of the numerator with i 6= j since, in these cases, t divides ri(a). However,t is not a divisor of ajrj(a) since, on the one hand, aj ≤ bj < l−1

j = t holds, and, on

the other hand, rj(a) and l−1j are coprime. Thus, the numerator is not a multiple of

t implying that there does not exist any exact pattern a ∈ Zm+ with a ≤ b.

iii) Due to Lemma 4 and part (i) of this theorem, the considered instance E has theoptimal objective value

z?c (E) = K +1

h(K)w(K).

Hence, it suffices to show that there is no pattern a ∈ Zm+ with a ≤ b and

1 ≤ l>a ≤ 1 +1

h(K)w(K).

According to part (ii) of this theorem, it only remains to check whether

l>a = 1 +1

h(K)w(K)(18)

is possible or not. Therefore, we consider two cases depending on the value of K.

(a) Case 1: K = 2In this case, p(2) = 3 holds due to the equation

1

2+

2

3= (K − 1) +

1

6.

Furthermore, we have m(2) = 3, h(2) = 6, s(2) = 1 and h(2) − s(2) = 5.According to our construction principle from Proposition 12, we aim at findinga natural number g ≥ 2 satisfying

g ≡ 0 mod 6,

g ≡ 1 mod 5.

Obviously, the solutions of this system are given by g = 6 + 30ξ with ξ ∈ Z+.Setting ξ = 0, i.e., g = 6, we obtain

w(2) =g − 1

h(2)− s(2)=

5

5= 1


contradicting property (b) of an M-sequence. Consequently, we have g = 6+30ξwith ξ ∈ N leading to

w(2) =g − 1

h(2)− s(2)=

5 + 30ξ

5= 1 + 6ξ ≥ 7,

=⇒ 1

w(2)≤ 1

7and

1

h(2)w(2)=

1

6 · (1 + 6ξ)≤ 1

42.

In order to attain a total length as in (18), every single item length li (i = 1, 2, 3)has to be taken at least once since, otherwise, the common denominator woulddiffer from h(2)w(2). Observing that

1

2+

1

3+

1

w(2)≤ 1

2+

1

3+

1

7=

41

42< 1,

we conclude that one item length has to appear at least twice in the pattern a.Obviously, this is not possible for l2 = 1

3since then the permitted total length

would be exceeded due to

1

2+

2

3+

1

w(2)>

1

2+

2

3= 1 +

1

6> 1 +

1

42≥ 1 +

1

h(2)w(2).

Hence, only l3 = w(2)−1 can appear more than once. Let α ∈ N with 2 ≤ α ≤v(2) denote the number of items with length l3 appearing in a. Then, we obtain

1 +1

6w(2)=

1

2+

1

3+

α

w(2)

⇐⇒ α

w(2)=

6w(2) + 1− 3w(2)− 2w(2)

6w(2)

⇐⇒ 6α = w(2) + 1

and, due to w(2) = 1 + 6ξ for some ξ ∈ N, this is equivalent to

⇐⇒ 6α = 2 + 6ξ

⇐⇒ α− ξ =1

3

Since α− ξ ∈ Z has to hold there is no pattern a with property (18) in the caseK = 2.

(b) Case 2: K ≥ 3In this case, we have

p(K)∑p=2,p∈P

p− 1

p≥ 2,

i.e., the sum on the left hand side contains at least the fractions

1

2+

2

3+

4

5+

6

7.


Due to (9), the inequality h(K) ≥ h(K − 1) holds. Furthermore, we havew(K) ≥ w(K − 1) according to property (b) of an M-sequence implying thath(K)w(K) ≥ h(K − 1)w(K − 1). In particular, this leads to

h(K)w(K)(K≥3)

≥ h(3)w(3)(w(3)>1)> 2 · 3 · 5 · 7 = 210. (19)

We assume that there is a pattern a ∈ Zm(K)+ with a ≤ b satisfying (18). In

order to attain this total length, every single item length li (i = 1, . . . ,m(K))has to be taken at least once since, otherwise, the common denominator woulddiffer from h(K)w(K). But then, due to (19), the total length of these itemssatisfies

1

2+

1

3+

1

5+

1

7+

1

w(K)> 1 +

37

210> 1 +

1

210≥ 1 +

1

h(K)w(K),

contradicting (18). Hence, there is no such pattern.

Consequently, the optimal objective value of the ILP cannot equal K, i.e., z?(E) ≤K − 1 holds. Due to Theorem 9, we also have ∆(E) < 2 entailing z?(E) ≥ K − 1.Hence, the gap ∆(E) results to

∆(E) = z?c (E)− z?(E) =

(K +

1

h(K)w(K)

)− (K − 1) = 1 +

1

h(K)w(K)

proving (17). In particular, E is a non-IRDP instance.

We would like to point out that this construction can be used in a more general way toobtain a multitude of further non-IRDP instances.

Remark 15. Let K ∈ N be a natural number with K ≥ 2, P ⊆ P a subset of the primenumbers and f : P→ N a mapping with f(p) ≤ p− 1 for all p ∈ P such that the series∑

p∈P

f(p)

p

diverges. Defining, in analogy to (8),

pf,P := pf,P(K) := inf

q ∈ P∣∣∣∣∣∣∣

q∑p=2,

p∈P

f(p)

p∈ [K − 1, K)

the whole theory presented in this section can be applied in a straightforward way to obtainnon-IRDP instances.

Note that we used P = P as well as f(p) = p− 1 for all p ∈ P. Hence, we took the greatestpossible denominators and also their maximum availability. Thus, for some fixed K ≥ 2,the instances from Theorem 14 contain the minimum number m of item types among allinstances that can be generated by this method.


5 Some investigations on the maximum gap for the

divisible case

In this section, based on the instances of Theorem 14, we would like to investigate the gapof the divisible case in more detail. In the context of one-dimensional cutting, the greatestknown gap is given by 137/132, see [Sch08, p. 69]. As a main result, we show that this gapcan be surpassed in the skiving scenario by modifying the considered instances slightly.

As an initial point, remember that for an M-sequence {(v(K), w(K))}K≥2 the sequence

{h(K) · w(K)}K≥2

is monotonically increasing. Thus, the greatest gap is provided in the case K = 2 by theinstance

E? =

(3,

(1

2,1

3,1

7

), 1, (1, 2, 6)

)which is known from Remark 3. Obviously, this gap results to

∆(E?) = 1 +1

42.

By adding new item lengths in a smart way, it is possible to increase the gap.

Theorem 16. Let E = (m, l, 1, b) be an instance of the divisble case in standard form withl>b = K + r for some K ∈ N with K ≥ 2 and some r ∈ R with 0 ≤ r < 1. Let ε1, ε2 > 0exist satisfying

z(E) := min{l>a

∣∣ a ∈ Zm+ , l>a ≥ 1, a ≤ b}

= 1 + r + ε1, (20)

z(E) := max{l>a

∣∣ a ∈ Zm+ , l>a ≤ 1, a ≤ b}

= 1− ε2 (21)

then define

ε := min{ε1, ε2}.

Now, let d ∈ N,

λ =

(1

λ1

, . . . ,1

λd

)and β = (β1, . . . , βd) ∈ Nd

be given with different λ1, . . . , λd ∈ N satisfying

∀ j ∈ {1, . . . , d} :1

λj/∈ {l1, . . . , lm} (22)

andd∑i=1

βiλi< ε. (23)

Then, the following assertions are true:

i) ∆(E) = 1 + r.


ii) Let E denote the instance being formed by adding the lengths λ and availabilities βto E. Then, we have

∆(E) = 1 + r +d∑i=1

βiλi. (24)

In particular, ∆(E) > ∆(E) holds.

Proof.

i) Due to (20), it is not possible to find K (disjoint) patterns among the given itemsupply, i.e., z?(E) ≤ K − 1 has to hold. According to Theorem 9, we also havez?(E) ≥ K − 1 leading to ∆(E) = 1 + r.

ii) Let d ∈ N,

λ =

(1

λ1

, . . . ,1

λd

)and β = (β1, . . . , βd) ∈ Nd

with different λ1, . . . , λd ∈ N be given satisfying

∀ j ∈ {1, . . . , d} :1

λj/∈ {l1, . . . , lm}

andd∑i=1

βiλi< ε.

Obviously, it suffices to show that E contains no pattern (a, α)> ∈ Zm+d+ with

(a, α)> ≤ (b, β)> and

1 ≤(lλ

)>(aα

)≤ 1 + r +

d∑i=1

βiλi. (25)

We assume that there exists such a pattern (a, α)> ∈ Zm+d+ . Without loss of gener-

ality, (a, α)> can be considered as a minimal pattern. Let

I(α) = {1 ≤ i ≤ d |αi > 0}

denote the set of the new objects’ active indices. Then, I(α) 6= ∅ holds since,otherwise, a ∈ Zm+ would be a pattern of E with a ≤ b satisfying the inequal-ity 1 ≤ l>a < 1 + r + ε, due to (25) and (23). But this is not possible sincez(E) = 1 + r + ε1 ≥ 1 + r + ε has to hold.

The minimality of (a, α)> leads to

1 ≤(lλ

)>(aα

)< 1 + min

{1

λi

∣∣∣∣ i ∈ I(α)

}.


Removing all item lengths belonging to λ from the pattern (a, α)> (setting αi = 0for i ∈ I(α)), we obtain a cutting pattern of E because of

l>a < 1 + min

{1

λi

∣∣∣∣ i ∈ I(α)

}−∑i∈I(α)

αiλi︸︷︷︸

≤0

≤ 1.

This cutting pattern’s total length satisfies

l>a =

(lλ

)>(aα

)−∑i∈I(α)

αiλi≥ 1−

d∑i=1

βiλi︸︷︷︸

<ε≤ε2

> 1− ε2

in contradiction to the presumption z(E) = 1−ε2. Hence, there is no packing pattern(a, α)> ∈ Zm+d

+ with (a, α)> ≤ (b, β)> and the property (25).

The following lemma states an important observation for our intended construction ofinstances with large gaps.

Lemma 17. Let E = (m, l, 1, b) be an instance of the divisible case in standard formsatisfying l>b = 2 + r for some r ∈ R with 0 ≤ r < 1. Furthermore, let ε > 0 exist suchthat

z(E) := min{l>a

∣∣ a ∈ Zm+ , l>a ≥ 1, a ≤ b}

= 1 + r + ε (26)

holds. Then,z(E) := max

{l>a

∣∣ a ∈ Zm+ , l>a ≤ 1, a ≤ b}

= 1− ε (27)

is also true.

Proof. According to the presumption (26), the shortest packing pattern a? ∈ Zm+ witha ≤ b of E satisfies l>a? = 1 + r + ε. Choosing those objects that do not belong toa?, we obtain a feasible solution for the maximization problem (27) with objective value2 + r − (1 + r + ε) = 1− ε which leads to

z(E) ≥ 1− ε.

Assuming that z(E) > 1− ε holds, then there would exist a cutting pattern α? ∈ Zm+ withl>α? > 1 − ε. Considering those objects that do not belong to α?, we obtain a packingpattern a ∈ Zm+ satisfying a ≤ b and

l>a = 2 + r − l>α? < 2 + r − (1− ε) = 1 + r + ε

in contradiction to the optimality of a?. Thus, z(E) = 1− ε has to hold.

As a direct consequence of this lemma, we have ε1 = ε2 = ε in Theorem 16 for the caseK = 2. This result plays an important role in the proof of the following constructionmethod.


Corollary 18. Let δ ∈ R with 0 ≤ δ < 1/42 be given. Then, there exists an instance E ofthe divisible case in standard form satisfying

1 +1

21> ∆(E) ≥ 1 +

1

42+ δ. (28)

Proof. Consider the instance E? from the beginning of this section. There, we have K = 2,r = 1/42 and

z(E?) =44

42= 1 +

1

21

stating that ε = 1/42 holds. Due to the density of the rational numbers in R, there aredifferent lengths λ1, . . . , λd ∈ N and availabilities β1, . . . , βd ∈ N with

δ ≤d∑i=1

βiλi<

1

42.

Let E? denote the instance that is created by adding these objects to E?. According toTheorem 16, its gap results to

∆(E?) = 1 +1

42+

d∑i=1

βiλi≥ 1 +

1

42+ δ.

Besides, we also have

∆(E?) = 1 +1

42+

d∑i=1

βiλi< 1 +

1

42+

1

42= 1 +

1

21

proving the assertion.

Hence, we presented a construction principle for instances of the divisible case whose gapscan be arbitrarily close to ∆(E) = 22/21. Due to

22

21− 137

132=

3

308> 0,

the maximum known gap of the cutting stock problem’s divisible case can be surpassed ifδ ∈ R is chosen in such a way that

1

42− 3

308< δ <

1

42

holds.


6 Conclusion and outlook

In this paper, we gave an introduction to the gap of the skiving stock problem from atheoretical point of view. As a starting point, we generalized a result of [Zak03] in orderto obtain a general upper bound for the gap. Thereby, we also stated a first sufficientcondition for instances possessing the MIRDP.

Thereupon, we studied the divisible case intensively. For these instances, we have proved,as a main result, the MIRDP by means of the FFD heuristic for the dual bin packingproblem. Moreover, we presented a contruction principle obtaining an infinite number ofnon-equivalent non-IRDP instances of the divisible case and showed how this method canbe used, in a more general way, to create further classes of examples. Additionally, we in-vestigated the gap of these instances also stating a technique to increase the gap by meansof adding new item lengths. As a main contribution, this method was shown to lead toinstances with gaps arbitrarily close to 22/21 which surpasses the best known gap for thedivisible case of the related one-dimensional cutting stock problem.

Except for the continuous relaxation studied in this paper, there are other possible relax-ations that are of theoretical and practical interest. One of these is the proper relaxationwhere, as an additional restriction, only minimal patterns a ∈ P ?

E with a ≤ b, so-calledproper patterns, are considered. This relaxation has also been studied in one-dimensionalcutting scenarios [NST99], [KRS13], so its practicability in terms of the skiving stock prob-lem will be investigated in the near future.

Since we mainly focused on the divisible case it will be part of our future research to studythe gap of arbitrary instances of the skiving stock problem. Therefore, we aim at improv-ing the upper bound m − 1 introduced in this paper in order to get a bit closer to theMIRDP-conjecture. Another main objective is given by the investigation of further specialclasses of instances also considering those ones that have turned out to possess the MIRUPin the context of one-dimensional cutting.

References

[AJKL84] Assmann, S.F., D.S. Johnson, D.J. Kleitman and J.Y.-T. Leung: Ona dual version of the one-dimensional Bin Packing Problem. Journal of Algo-rithms, 5:502–525, 1984.

[ARGA04] Alvim, A.C.F., C.C. Ribeiro, F. Glover and D.J. Aloise: A hybridimprovement heuristic for the one-dimensional bin packing problem. Journal ofHeuristics, 10(2):205–229, 2004.

[BD85] Bruno, J.L. and P.J. Downey: Probabilistic Bounds for Dual Bin-Packing.Acta Informatica, 22:333 – 345, 1985.

[BT81] Baum, S. and L.E. Trotter: Integer rounding for polymatroid and branch-ing optimization problems. SIAM Journal of Algebraic and Discrete Methods,2:416–425, 1981.


[CFGR91] Csirik, J., J.B.G. Frenk, G. Galambos and A.H.G. Rinnooy Kan:Probabilistic Analysis of Algorithms for Dual Bin Packing Problems. Journalof Algorithms, 12:189–203, 1991.

[GG61] Gilmore, P.C. and R.E. Gomory: A Linear programming approach to thecutting-stock problem (Part I). Operations Research, 9:849–859, 1961.

[KRS13] Kartak, V.M., A.V. Ripatti and G. Scheithauer: Minimal propernon-IRUP instances of the one-dimensional Cutting Stock Problem. PreprintMATH-NM-08-2013, Technische Universitat Dresden, 2013.

[LLM95] Labbe, M., G. Laporte and S. Martello: An exact algorithm for the dualbin packing problem. Operations Research Letters, 17:9–18, 1995.

[Mar83] Marcotte, O.: Topics in combinatorial packing and covering. Cornell Uni-versity, 1983. (Technical Report No.568).

[MS14] Martinovic, J. and G. Scheithauer: ILP models for the skiving stockproblem. Preprint MATH-NM-05-2014, Technische Universita Dresden, 2014.

[NST99] Nitsche, C., G. Scheithauer and J. Terno: Tighter relaxations for thecutting stock problem. European Journal of Operational Research, 112(3):654– 663, 1999.

[PD06] Peeters, M. and Z. Degraeve: Branch-and-price algorithms for the dualbin packing and maximum cardinality bin packing problem. European Journalof Operational Research, 170(2):416–439, 2006.

[Rie03] Rietz, J.: Untersuchungen zu MIRUP fur Vektorpackprobleme. PhD thesis,TU Bergakademie Freiberg, 2003.

[Sch08] Scheithauer, G.: Zuschnitt- und Packungsoptimierung – Problemstellungen,Modellierungstechniken, Losungsmethoden. Vieweg+Teubner, Wiesbaden, 1.edition, 2008.

[ST92] Scheithauer, G. and J. Terno: About the gap between the optimal value ofthe integer and continuous relaxation one-dimensional cutting stock problem. InOperations Research Proceedings, pages 439–444, Berlin und Heidelberg, 1992.Springer Verlag.

[VPS+13] Vijayakumar, B., P. Parikh, R. Scott, A. Barnes and J. Gallimore:A dual bin-packing approach to scheduling surgical cases at a publicly-fundedhospital. European Journal of Operational Research, 224(3):583–591, 2013.

[Zak03] Zak, E.J.: The skiving stock problem as a counterpart of the cutting stockproblem. International Transactions in Operational Research, 10:637–650, 2003.

Documents

Integer Rounding and Modi ed Integer Rounding for the ...scheith/ABSTRACTS/PREPRINTS/15-MIRDP.pdf · Integer Rounding and Modi ed Integer Rounding for the Skiving Stock Problem John