Combinatorial investigations on the maximum gap for skiving stock instances of the divisible case

Abstract

We consider the one-dimensional skiving stock problem which is strongly related to the dual bin packing problem: find the maximum number of objects, each having a length of at least L, that can be constructed by connecting a given supply of $m \in \mathbb {N}$ smaller item lengths $l_1,\ldots ,l_m$ with availabilities $b_1,\ldots , b_m$. For this $\mathcal {NP}$-hard discrete optimization problem, the (additive integrality) gap, i.e., the difference between the optimal objective values of the continuous relaxation and the skiving stock problem itself, is known to be strictly less than 3 / 2 if $l_i \mid L$ is assumed for all items, hereinafter referred to as the divisible case. Within this framework, we derive sufficient conditions that ensure the integer round-down property, i.e., a gap smaller than one, of a given instance. As a main contribution, we present improved upper bounds for the gap (of special subclasses) of the divisible case by means of combinatorial and algorithmic approaches. In a final step, possible generalizations of the introduced concepts are discussed. Altogether, the results presented in this paper give strong evidence that 22 / 21 represents the best possible upper bound for the gap of divisible case instances.

A Proof of Proposition 3

For the sake of contradiction, let us assume that there is no such pattern. In particular, E does not contain an exact pattern with $a \le b$, i.e., each item of length 1 / q (for some $q \in \mathbb {N}$) may appear at most $q-1$ times. Due to (7) we can state

$$\begin{aligned} \sum _{i \notin \Gamma } b_il_i = l^\top b - \sum _{i\in \Gamma } b_il_i \ge K+\frac{1}{N}-\left( K+\frac{1}{N}- \frac{1}{6} \right) = \frac{1}{6} , \end{aligned}$$

(17)

i.e., there is a vector $\tilde{a} \in \mathbb {Z}^m_{+}$ with $\tilde{a}\le b$, $l^\top \tilde{a} \ge 1/6$ and $\tilde{a}_i=0$ for $i \in \Gamma$.

Part 1

There is a vector $a \in \mathbb {Z}^m_{+}$ with $a \le b$, $a_i=0$ for $i \notin \Gamma$ and $l^\top a \ge 5/6$.

For the sake of contradiction, let us assume that this statement is wrong. Then we obtain

$$\begin{aligned} \sum _{i \in \Gamma } b_il_i < \frac{5}{6} \end{aligned}$$

which leads to

$$\begin{aligned} \sum _{i \notin \Gamma } b_il_i = l^\top b - \sum _{i \in \Gamma } l_i b_i \ge K+ \frac{1}{N}- \frac{5}{6}>K- \frac{5}{6} \ge 1+ \frac{1}{6}. \end{aligned}$$

But then the vector $a^\star \in \mathbb {Z}^m_{+}$ with $a^\star _i=0$ for $i \in \Gamma$ and $a^\star _i=b_i$ for $i \notin \Gamma$ satisfies $a^\star \le b$ and $l^\top a^\star > 7/6$. Moreover, $a^\star$ contains only items of length less than or equal to $1/N \le 1/6$. Hence, by removing appropriate items, $a^\star$ can be reduced to a pattern (for simplicity also called $a^\star$) with $a^\star \le b$ and $1 \le l^\top a^\star \le 1+1/N$ which gives the contradiction to the initial assumption.

Part 2

There is a vector $a \in \mathbb {Z}^m_{+}$ with $a \le b$, $a_i=0$ for $i \notin \Gamma$ and $1 \ge l^\top a \ge 5/6$.

For the sake of contradiction, let us assume that this statement is wrong. Due to Part 1 of this proof, we can find a vector $a \in \mathbb {Z}^m_{+}$ satisfying $a\le b$, $a_i=0$ for $i \notin \Gamma$ and $l^\top a \ge 5/6$. According to the assumption of Part 2, also $l^\top a > 1$ has to be true. Note that, in this case, the vector a represents a packing pattern of E, and can be chosen minimal without loss of generality. Consequently, a can only contain items of length 1 / 2, 1 / 3, 1 / 4 or 1 / 5, since any item of length $\le 1/ 6$ could be removed such that a vector with $5/6 \le l^\top a \le 1$ is obtained (which would contradict to the assumption of Part 2). Since E does not contain any exact pattern, the item of length

• 1 / 2 may appear at most once,

• 1 / 3 may appear at most twice,

• 1 / 4 may appear at most once (if the item 1 / 2 is present), otherwise it may appear at most three times,

• 1 / 5 may appear at most four times.

As the case may be, the only minimal patterns that can be build are the following ones:

• $l^\top a = 1/4+ 4/5=21/20$, but then an item of length 1 / 5 can be removed leading to a vector with $l^\top a=17/20 \in [5/6,1]$,

• $l^\top a = 2/4+ 3/5=11/10$, but then an item of length 1 / 5 can be removed leading to a vector with $l^\top a=9/10 \in [5/6,1]$,

• $l^\top a = 3/4+ 2/5=23/20$, but then an item of length 1 / 5 can be removed leading to a vector with $l^\top a=19/20 \in [5/6,1]$,

• $l^\top a = 1/3+ 4/5=17/15$, but then an item of length 1 / 5 can be removed leading to a vector with $l^\top a=14/15 \in [5/6,1]$,

• $l^\top a = 1/3+ 1/4+3/5=71/60$, but then an item of length 1 / 5 can be removed leading to a vector with $l^\top a=59/60 \in [5/6,1]$,

• $l^\top a = 1/3+ 2/4+1/5=31/30$, but then an item of length 1 / 5 can be removed leading to a vector with $l^\top a=5/6 \in [5/6,1]$,

• $l^\top a = 1/3+ 3/4=13/12$, but then an item of length 1 / 4 can be removed leading to a vector with $l^\top a=5/6 \in [5/6,1]$,

• $l^\top a = 2/3+ 2/5=16/15$, but then an item of length 1 / 5 can be removed leading to a vector with $l^\top a=13/15 \in [5/6,1]$,

• $l^\top a = 2/3+1/4+ 1/5=67/60$, but then an item of length 1 / 5 can be removed leading to a vector with $l^\top a=11/12 \in [5/6,1]$,

• $l^\top a = 2/3+ 2/4=7/6$, but then an item of length 1 / 4 can be removed leading to a vector with $l^\top a=11/12 \in [5/6,1]$,

• $l^\top a = 1/2+ 3/5=11/10$, but then an item of length 1 / 5 can be removed leading to a vector with $l^\top a=9/10 \in [5/6,1]$,

• $l^\top a = 1/2+ 1/4+ 2/5=23/20$, but then an item of length 1 / 5 can be removed leading to a vector with $l^\top a=19/20 \in [5/6,1]$,

• $l^\top a = 1/2+1/3 +1/5=31/30$, but then an item of length 1 / 5 can be removed leading to a vector with $l^\top a=5/6 \in [5/6,1]$,

• $l^\top a = 1/2+1/3+1/4=13/12$, but then an item of length 1 / 4 can be removed leading to a vector with $l^\top a=5/6 \in [5/6,1]$,

In any case, we obtained a contradiction; hence the statement of Part 2 has to be true.

Part 3

Finding a contradiction to the initial assumption.

Now consider a vector a from Part 2 of this proof. By combining this vector with the vector $\tilde{a}$ from the beginning, we obtain a vector $\hat{a}:=a+\tilde{a} \in \mathbb {Z}^m_{+}$ with $\hat{a} \le b$ and

$$\begin{aligned} l^\top \hat{a} = l^\top a + l^\top \tilde{a} \ge \frac{5}{6}+ \frac{1}{6} = 1, \end{aligned}$$

i.e., a pattern of E. If necessary, by successively removing items belonging to $I\setminus \Gamma$ (which all have a length $\le 1/N$), we obtain a pattern that satisfies $1 \le l^\top \hat{a} \le 1+1/N$ which gives the contradiction. Hence this proposition is proved.