Advanced Algorithm Design: Load Balancing

During last lecture, we showed that using the simple randomized strategy when assigning n balls to n bins, max load = O log n log log n. We now consider a more intelligent strategy of assigning balls to bins and show that it results in a suprising improvement over the above result. 1.1 Power of two choices In our new strategy, we choose two bins uniformly at random and place the ball in the bin with the smaller existing load. Following this strategy results in the following bound: Theorem 1. In this 2-choice strategy, the max load is O(log log n). We first consider a rough sketch of why this new bound makes sense and then prove it using another method which involves using properties of randomized graphs. Suppose for a ball b in B (the set of balls), height(b) = load of bin in which b is placed. Then, E[fraction of balls in B w/ height(b) ≥ 2] ≤ 1 2 , since the total number of balls in all the bins equals n. Notice then that E[fraction of balls in B w/ height(b) ≥ 3] ≤ 1 2 2 , as it is necessary that for each ball, both the bins being considered have at least two balls.