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arXiv 2104.05093cs.DS
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基于动态球仓集的负载均衡

Load Balancing with Dynamic Set of Balls and Bins

Anders Aamand, Jakob Bæk Tejs Knudsen, Mikkel Thorup

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Accepted at STOC'21
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摘要：

在动态负载均衡中，我们希望在一个既可以添加又可以删除球和箱子的环境中将球分配到箱子中。我们希望最小化任何箱子的最大负载，但我们也希望在添加或删除球或箱子时最小化受影响的球和箱子的数量。我们想要一个散列式的解决方案，我们给定一个球的ID可以有效地找到它的bin。我们给出了平衡参数$c=1+ε$，其中$ε\in (0,1)$。使用$n$和$m$表示当前的球数和料箱数，我们不希望料箱的负载高于$C=\lceil c n/m\rceil$，也就是料箱的容量。我们提出了一个计划，我们可以找到一个球检查$1+O(\log 1/ε)$箱的期望。当插入或删除球时，我们希望移动$O(1/ε)$球，当插入或删除bin时，我们希望移动$O(C/ε)$球。以前的界限偏离了一个因子$1/ε$。当$C=O(1)$时，这些界限是最好的，但对于更大的$C$，我们可以做得更好：让$f=εC$如果$C\leq \log 1/ε$，$f=ε\sqrt{C}\cdot \sqrt{\log(1/(ε\sqrt{C}))}$如果$\log 1/ε\leq C<\tfrac{1}{2ε^2}$，$C=1$如果$C\geq \tfrac{1}{2ε^2}$。我们表明，我们希望移动$O(1/f)$球时，插入或删除一个球，$O(C/f)$球时，插入或删除一个bin。对于具有较大$C$的边界，我们首先必须解决一个简单得多的概率问题。将$n$球放入容量为$C$的$m$箱中，每次一个球。每个球挑选一个均匀随机的非满仓。我们表明，在预期和高概率，非满仓的分数是$Θ(f)$。然后，一个新的球要找到一个未满的球所需访问的垃圾箱的预期数量是$Θ(1/f)$。事实证明，我们在更复杂的方案中获得了相同的复杂性，其中球和箱都可以添加和删除。

英文摘要：

In dynamic load balancing, we wish to distribute balls into bins in an environment where both balls and bins can be added and removed. We want to minimize the maximum load of any bin but we also want to minimize the number of balls and bins affected when adding or removing a ball or a bin. We want a hashing-style solution where we given the ID of a ball can find its bin efficiently. We are given a balancing parameter $c=1+ε$, where $ε\in (0,1)$. With $n$ and $m$ the current numbers of balls and bins, we want no bin with load above $C=\lceil c n/m\rceil$, referred to as the capacity of the bins. We present a scheme where we can locate a ball checking $1+O(\log 1/ε)$ bins in expectation. When inserting or deleting a ball, we expect to move $O(1/ε)$ balls, and when inserting or deleting a bin, we expect to move $O(C/ε)$ balls. Previous bounds were off by a factor $1/ε$. These bounds are best possible when $C=O(1)$ but for larger $C$, we can do much better: Let $f=εC$ if $C\leq \log 1/ε$, $f=ε\sqrt{C}\cdot \sqrt{\log(1/(ε\sqrt{C}))}$ if $\log 1/ε\leq C<\tfrac{1}{2ε^2}$, and $C=1$ if $C\geq \tfrac{1}{2ε^2}$. We show that we expect to move $O(1/f)$ balls when inserting or deleting a ball, and $O(C/f)$ balls when inserting or deleting a bin. For the bounds with larger $C$, we first have to resolve a much simpler probabilistic problem. Place $n$ balls in $m$ bins of capacity $C$, one ball at the time. Each ball picks a uniformly random non-full bin. We show that in expectation and with high probability, the fraction of non-full bins is $Θ(f)$. Then the expected number of bins that a new ball would have to visit to find one that is not full is $Θ(1/f)$. As it turns out, we obtain the same complexity in our more complicated scheme where both balls and bins can be added and removed.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

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