论文标题

在线单位利润背包,不受信任的预测

Online Unit Profit Knapsack with Untrusted Predictions

论文作者

Boyar, Joan, Favrholdt, Lene M., Larsen, Kim S.

论文摘要

在受信任和不受信任的预测的设置中考虑了在线背包问题的变体。在单位利润背包中,这些物品具有单位利润,并且很容易离线找到最佳的解决方案:将尽可能多的最小物品包装到背包中。对于在线单位利润背包,竞争比率是无限的。相比之下,以前关于不受信任预测的在线算法的工作通常研究了具有持续竞争比率的在线算法。我们的算法使用的预测可能是从机器学习来源获得的,是适合背包中最小的物品的平均大小。对于此困难在线问题的预测错误,我们使用比率$ r = \ frac {a} {\ hat {a}} $,其中$ a $是此平均大小的实际值,$ \ hat {a} $是预测。提出的算法达到了$ \ frac {1} {2r} $的竞争比率,用于$ r \ geq 1 $和$ \ frac {r} {2} $ for $ r \ r \ leq 1 $。使用对手技术,我们表明这在某种意义上是最佳的,可以使$ r $的不同值可以实现竞争比率的权衡。请注意,准确建议的结果是$ r = 1 $,仅是$ \ frac {1} {2} $,但我们表明,没有算法知道$ $ a $的竞争比率比$ \ frac {e-frac {e-1} {e-1} {e} {e} {e} \ of Algorithm and a algorithm and a algorithm and argorithm bumpter a匹配。我们还表明,后者的算法达到$ r \ frac {e-1} {e-1} {e} $ for $ r \ leq 1 $和$ \ frac {e-r} {e-r} {e-r} {e-r} $ $ 1 \ leq leq r <e $,而没有algorithm可以更好地占$ r <1 $ 1 $ 1 $ 1 $ $ r <e $ r <e。

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 $1 \leq r < e$, and no algorithm can be better for both $r<1$ and $1\leq r<e$.

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