How many employees do I need to fill 33 positions? Real world problem thanks for reading this thread!
This is a problem which I don't know how to put in a formula properly:
We're filling 33 positions with workers across a full year (12 months).
Each worker will take 1 month off during the year.
We will need to fill the positions with other extra workers. Extra workers also will take 1 month off during the year.
The problem is, how many extra workers do we need to fill these 33 positions (taking into account off-times for all workers)?
The number may contain decimals, but will be rounded-up.
[Updated:]
Notes:
(1) We have a say in the decision for timing of the vacation, so yes we can offer time-off when we decide.
(2) Also I understand that we need 3 other employees for a total of 36. I actually solved this using MS Excel Solver add-on, but I don't know how to derive/explain this mathematically. Is there any sort of formula for this?
 A: This may be getting a bit too involved for what you realistically need, but if you were wondering this is how I would formulate a more formal solution to this problem.
First, we have to begin with the pigeonhole principle: essentially, if I have $n$ balls to put into $m$ boxes, then at least one box has to have at least $\lceil \frac{n}{m} \rceil$ balls, where $\lceil()\rceil$ represents the ceiling function, which rounds a decimal up to the next highest integer. The justification for this is that otherwise all $m$ boxes must contain less than $\frac{n}{m}$ balls, so altogether they would have to have less than $n$ balls, which contradicts our setup which states we need to have $n$ balls total.
So, if we have $n$ employees, and each one must take one of $12$ months off, then at least one month will have at least $\lceil \frac{n}{12} \rceil$ employees taking a vacation. This means that given $n$ employees, the largest number of positions which we could possibly constantly fill is $n - \lceil \frac{n}{12} \rceil.$
Now in order to see how many employees we need to constantly fill $33$ positions, we can solve the equation $$n - \lceil\frac{n}{12}\rceil = 33$$ To solve this, we can consider a substitution $n = 12a + b,$ for integers $a$ and $b$ and $0 < b \leq 12.$ I've chosen to bound $b$ this way because if $12a < n \leq 12a + 12$ we must have that $a < \frac{n}{12} \leq a+1$ so $\lceil \frac{n}{12} \rceil = a+1.$ So, $$12a + b - (a + 1) = 33 \Rightarrow 11a + b = 34$$ Considering both sides mod $11,$ we get that $b \equiv 1 \pmod{11}$ so either $b = 1$ or $b = 12.$ If $b = 1$ then $11a = 33 \Rightarrow a = 3$ and $n = 12(3) + 1 = 37,$ and if $b = 12$ then $11a = 22 \Rightarrow a = 2$ and $n = 12(2) + 12 = 36,$ so the smallest number of employees $n$ we can hire to constantly fill $33$ positions is $36.$
The cool thing about this method is that it works for any number of positions: let's say I instead needed to fill $50$ positions. Jumping ahead I get $11a + b = 51,$ so $b \equiv 7 \pmod{11}$ and $b = 7,$ so $11a = 44 \Rightarrow a = 4$ and $n = 12(4) + 7 = 55.$ We can also rework the equation to work for other lengths of breaks by changing the $12$ in the denominator.
