NEW THEORETICAL INVESTIGATIONS ON THE GAP OF THE SKIVING STOCK PROBLEM

Martinovic, John; Scheithauer, Guntram

doi:10.1590/0101-7438.2019.039.01.0001

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Articles • Pesqui. Oper. 39 (1) • Jan-Apr 2019 • https://doi.org/10.1590/0101-7438.2019.039.01.0001 copy

NEW THEORETICAL INVESTIGATIONS ON THE GAP OF THE SKIVING STOCK PROBLEM

Authorship SCIMAGO INSTITUTIONS RANKINGS

ABSTRACT

The one-dimensional skiving stock problem is a combinatorial optimization problem being of high relevance whenever an efficient and sustainable utilization of given resources is intended. In the classical formulation, a given supply of (small) item lengths has to be used to build as many large objects (specified by some target length) as possible. For this 𝒩𝒫-hard (discrete) optimization problem, we investigate the quality of the continuous relaxation by considering the additive integrality gap, i.e., the difference between the optimal values of the integer problem and its LP relaxation. In a first step, we derive an improved upper bound for the gap by focusing on the concept of residual instances. Moreover, we show how further upper bounds can be obtained if all problem-specific input data are considered. Additionally, we constructively prove the integer round-down property for two new classes of instances, and introduce several construction principles to obtain gaps greater than or equal to one.

Keywords:
Cutting and Packing; Skiving Stock Problem; Additive Integrality Gap

Phase I:
	1. Let $\tilde{n} = e^{⊥} b$ , with $e = {(1,...,1)}^{⊥} \in ℤ_{+}^{m}$ , be the total number of items. Sort them according to non-increasing lengths and number them consecutively: $1 > l_{1}^{} \geq l_{2}^{} \geq ... \geq l_{\tilde{n}}^{*}$ .
	2. If there are unallocated objects: Let $i \in {1,..., \tilde{n}}$ be the index of the first unallocated object. Choose the first bin that is able to accept this object and place it therein. If such a bin does not exist, add a new (empty) bin and place the object therein.
Phase II:
	1. If there is more than one nonempty bin that is not filled: consider the last of these bins, choose one of its items and allocate it to the first of these bins.
	2. If there is exactly one nonempty bin that is not filled: allocate its objects to the last bin that is filled.

ξ ₁	3	4	5	6	7	8	9	10	11	13	15
ξ ₂	3	4	5	6	7	8	9	10	11	13	15
2	$\frac{1}{42}$	X	$\frac{3}{110}$	X	$\frac{1}{42}$	X	$\frac{7}{342}$	X	$\frac{9}{506}$	$\frac{11}{702}$	$\frac{13}{930}$
3	X	$\frac{5}{156}$	$\frac{7}{240}$	X	$\frac{1}{42}$	$\frac{13}{600}$	X	$\frac{17}{930}$	X	X	X
4	X	X	$\frac{11}{420}$	X	$\frac{17}{812}$	X	X	X	X	X	X
5	X	X	X	$\frac{19}{930}$	X	X	X	X	X	X	X

ξ ₁	3	4	5	7	9
ξ ₂	3	4	5	7	9
2	$\frac{1}{21}$	X	$\frac{2}{55}$	$\frac{1}{35}$	$\frac{4}{171}$
3	X	$\frac{1}{26}$	$\frac{1}{30}$	X	X
4	X	X	$\frac{1}{35}$	X	X

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