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?

  • $\begingroup$ Do you get to choose what month they get off, or do they? If you get to choose then there's a very simple way to do it $\endgroup$ Dec 7, 2021 at 10:30
  • 1
    $\begingroup$ You want to ensure 33 people on deck all the time, right? Other important metrics that I can think of: can you reject a worker's off time request because too many of them have requested and ask them to relocate their vacation, your sick leave policy especiallywhat to do if one gets covid, leaves without prior notice due to death of a family member or otherwise, how catastrophic it would be if you are one or two hands short one or two days a year. $\endgroup$
    – ck1987pd
    Dec 7, 2021 at 10:30
  • 3
    $\begingroup$ Employ $36$ people altogether and make three of them have a holiday in each month. $\endgroup$
    – user700480
    Dec 7, 2021 at 10:34
  • 3
    $\begingroup$ You need 33*12 workers/month but all workers are available for 11 months, so 33*12/11=36. $\endgroup$
    – ck1987pd
    Dec 7, 2021 at 11:00
  • 1
    $\begingroup$ Thank you very much everyone, especially to @C.Koca, you made it look simple! $\endgroup$ Dec 7, 2021 at 11:08

1 Answer 1


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.

  • $\begingroup$ I'm just sorry I can only give 1 vote instead of 1000! This is very helpful, thanks a lot! $\endgroup$ Dec 7, 2021 at 11:41
  • $\begingroup$ Of course, glad I could be of service and that it wasn't too verbose $\endgroup$ Dec 7, 2021 at 11:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .