Load balancing with dynamic set of balls and bins

ABSTRACTIn 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 that are 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 user-specified balancing parameter c=1+ε, where ε∈ (0,1). Let n and m be the current number of balls and bins. Then we want no bin with load above C=⌈ c n/m⌉, referred to as the capacity of the bins. We present a scheme where we can locate a ball checking 1+O(log1/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. The above bounds are best possible when C=O(1) but for larger C, we can do much better: We define f=ε C when C≤ log1/ε, f=ε√C· √log(1/(ε√C)) when log1/ε≤ C<1/2ε2, and f=1 when C≥ 1/2ε2. 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. Moreover, when C≥ log1/ε, we can search a ball checking only O(1) bins in expectation. 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, this is also the complexity of an insertion in our more complicated scheme where both balls and bins can be added and removed.