Basically we consider two levels of mapping (the first is called partition and second mapping strategy) of balls into bins. And try to find the best partition strategy (the first level of mapping) to minimised the max bin sizes.
To state the problem precisely, it is a bit complex. Suppose there is a set of $m$ bins, partition number $k$ and $n$ balls. Balls are ordered with index $i$, $(1\leq i\leq n)$, and ball $i$ has weight $w_i$.
Firstly, the set of balls is partitioned into $k$ disjoint subset in their original order (here $m \leq n$ and $k \leq n$). For instance, if there are $n=6$ balls, with weights $3, 1, 3, 2, 2, 1$ and $k=4$, one way to do partition is $(3), (1, 3), (2), (2, 1)$.
Then, each partition is mapped to one of $m$ bins by certain mapping strategy $M$. Afterwards, each bin $j$ has some partitions and has total weight $bw_j$. Following previous example, assume $n=4$, one mapping strategy is to map each partition in $(3), (1,3), (2), (2,1)$ to a distinct bin, that is, $[(3)], [(1,3)], [(2)], [(2,1)]$. The maximal bin weight is $bw_2=3+1=4$.
Our problem is how to partition the set of balls so that the maximal bin weight (i.e., $max(bw_j)$) is minimised. Here, we suppose we are provided with the best possible mapping strategy $M$.
We follow previous example for better illustrating the problem. Now, if $k$ is changed to $5$, the partitioning we seek is $(3), (1), (3), (2), (2, 1)$. We consider all mapping strategy $M$ and the best mapping is $[(3)], [(1), (2)], [(3)], [(2, 1)]$, yielding max bin weight being just $3$. The problem is how to design an algorithm to find such partitioning, like $(3), (1), (3), (2), (2, 1)$.