Online Unit Profit Knapsack with Untrusted Predictions

J. Boyar, Lene M. Favrholdt, Kim S. Larsen
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引用次数: 4

Abstract

A variant of the online knapsack problem is considered in the settings of trusted and untrusted predictions. In Unit Profit Knapsack, the items have unit profit, and it is easy to find an optimal solution offline: Pack as many of the smallest items as possible into the knapsack. For Online Unit Profit Knapsack, the competitive ratio is unbounded. In contrast, previous work on online algorithms with untrusted predictions generally studied problems where an online algorithm with a constant competitive ratio is known. The prediction, possibly obtained from a machine learning source, that our algorithm uses is the average size of those smallest items that fit in the knapsack. For the prediction error in this hard online problem, we use the ratio $r=\frac{a}{\hat{a}}$ where $a$ is the actual value for this average size and $\hat{a}$ is the prediction. The algorithm presented achieves a competitive ratio of $\frac{1}{2r}$ for $r\geq 1$ and $\frac{r}{2}$ for $r\leq 1$. Using an adversary technique, we show that this is optimal in some sense, giving a trade-off in the competitive ratio attainable for different values of $r$. Note that the result for accurate advice, $r=1$, is only $\frac{1}{2}$, but we show that no algorithm knowing the value $a$ can achieve a competitive ratio better than $\frac{e-1}{e}\approx 0.6321$ and present an algorithm with a matching upper bound. We also show that this latter algorithm attains a competitive ratio of $r\frac{e-1}{e}$ for $r \leq 1$ and $\frac{e-r}{e}$ for $1 \leq r
在线单位利润背包与不可信的预测
在可信和不可信预测的设置中考虑了在线背包问题的一种变体。在单位利润背包中,物品具有单位利润,并且很容易找到离线的最佳解决方案:将尽可能多的最小物品装入背包。对于在线单位利润背包,竞争比是无界的。相比之下,先前关于具有不可信预测的在线算法的工作通常研究具有恒定竞争比的在线算法已知的问题。我们的算法使用的预测(可能是从机器学习源获得的)是背包中最小物品的平均尺寸。对于这个硬在线问题中的预测误差,我们使用比值$r=\frac{a}{\hat{a}}$,其中$a$是这个平均大小的实际值,$\hat{a}$是预测值。该算法实现了$\frac{1}{2r}$对$r\geq 1$和$\frac{r}{2}$对$r\leq 1$的竞争比。使用对手技术,我们证明这在某种意义上是最优的,给出了不同r值可达到的竞争比率的权衡。请注意,准确建议的结果$r=1$,只有$\frac{1}{2}$,但是我们表明,没有算法知道$a$的值可以获得比$\frac{e-1}{e}\约0.6321$更好的竞争比,并提出了一个具有匹配上界的算法。我们还证明了后一种算法对于$r\ leq 1$和$\frac{e-r}{e}$对于$1 \leq r达到了$r\frac{e- r}{e}$的竞争比
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