How to solve variable sized bin packing when the items have equal weights, or is it still NP-hard?
I'll change the description to avoid artificial sums. 

We have n bins $bin_i$. Each bin has a capacity $C_i$ and a weight $f_i$. We have m items $item_i$. Each item has a weight $w_j$. We want to pick a set B of bins, and for each $bin_i$ in B a set $I_i$ such that the weights of all items in $I_i$ is at most $C_i$, and the sum of the weights $f_i$ of the bins in B is minimised. Each item must be in exactly one of the bins $bin_i$ in B. We assume $f_i &gt; 0$ and $w_i &gt; 0$. 

If all $w_i = w$ are the same, then we can divide each $C_i$ by w and replace $w_i$ by 1 without changing the problem. Now the number of items in $I_i$ must be ≤ $C_i$. Since the number of items in $I_i$ is an integer, we can round each $C_i$ down to the nearest integer without changing the problem. 

There are m items, and each item is exactly in one bin, so the sum of the $C_i$ for bins $bin_i$ in B is ≥ m for a solution. On the other hand, if the sum of $C_i$ for bins in B is ≥ m, then we can fit the m items into those bins (because the $C_i$ are integers). So now we can completely ignore the items, and the problem is: Find a set B of bins so that the sum of $C_i$ for bins in B is ≥ m, while the sum of $f_i$ for those bins is minimised. If we find the set B, then putting items into the bins is trivial. 

Let B' be the set of bins not in B, and let C be the sum of all $C_i$. Instead of finding B, we can find B' such that the sum of $C_i$ for bins in B' is ≤ C - m, and the sum of $f_i$ for bins in B' is maximised. We rename "bin" to "item" and call B' a "knapsack", and voila - a knapsack problem. Very much easier to solve than the original problem if the number of bins is significantly smaller than the number of items, but still NP-complete.