# Given fixed number of bins with equal capacity and variable number of items how many bins will I need?

I'm slightly embarrassed to ask this question as I feel that the answer might be exceedingly simple but I sat down to think about it and can't seem come up with a formula.

I have a fixed number of bins and a variable number of items. How many total bins will I need to fit n number of items.

Example: 3 bins of capacity 3 and 5 items. How many bins are needed to fit 5 items? Obviously the answer is 2. 1 bin will be full and the other will have 1 empty slot.

What is the formula for this?

The information I need to know is:

1. How many completely empty bins will remain?
2. How many empty slots will remain in the partially full bin?

NOTE: I know the number of items will never exceed the capacity of all the bins combined.

• It is just division with remainder. Divide the number of items by the capacity of a bin. If the items are of different sizes, it is much harder. Bin packing is NP-hard Aug 31, 2020 at 22:09

One answer will involve a couple of functions you may not be familiar with: floor and modulo.

They're simple enough, though. The floor of $$x$$, denoted $$\lfloor x \rfloor$$, is simply the largest integer not greater than $$x$$. Broadly speaking, you just round down. So $$\lfloor 3.14 \rfloor = 3, \lfloor 6 \rfloor = 6, \lfloor -2.72 \rfloor = -3$$, and so forth.

Then $$a$$ modulo $$b$$, written $$a \bmod b$$, tells you the remainder after $$a$$ is divided by $$b$$. So, for example, $$8 \bmod 5 = 3, 27 \bmod 4 = 3, 6 \bmod 3 = 0$$, and so on.

OK, with that out of the way, suppose you have $$b$$ bins with capacity $$c$$ each, and you have $$n$$ items to put in the bins. Assuming the number of bins is in fact enough to hold the items, the answers are

1. The number of empty bins is $$\left\lfloor \frac{bc-n}{c} \right\rfloor$$.

2. The number of empty slots in the partially full bin is $$bc-n \bmod c$$.

For example, let's say you have $$7$$ bins, each capable of holding $$5$$ items. You have $$22$$ items. Then $$b = 7, c = 5, n = 22$$. That means

1. The number of empty bins is $$\left\lfloor \frac{7\times5-22}{5} \right\rfloor = \left\lfloor \frac{13}{5} \right\rfloor = 2$$.

2. The number of empty slots in the partially full bin is $$7 \times 5 - 22 \bmod 5 = 13 \bmod 5 = 3$$.

If this last result is $$0$$, that simply means there are no partially full bins.

• Brian, you're a life saver! My initial thought in the first 2 seconds was take the modulus but whatever I was doing wasn't working out and what I thought was a very simple question was making my brain explode. Aug 31, 2020 at 22:30