A factory consists of N machines that produce products at a rate of 1 per hour. Each machine randomly chooses an hour to reset itself for maintenance, during that time the entire factory is shut down. What is the optimal N to maximize production rate?
Everything is discrete to full hour intervals, so for example if there are 2 machines, then there is 1/24 chance of collision in reset time, in which case the total production is going to be $2*23 = 46$ products. Otherwise, there will be 2 shut-downs and a total of $2*22 = 44$ products. So the expectation for N=2 will be $$E(2)=46\cdot\frac{1}{24}+44\cdot\frac{23}{24}$$
I want to find the probability function for k shut-downs given N machines and then calculate $$E(N)=\sum_{k=1}^{23}f(k,N)\cdot P(k|N)$$ where $f(k,N) = N\cdot(24-k)$ is the number of products for N machines and k shut-downs. Then I'll differentiate dN and finish the rest.
But I can't find $P$... How should I calculate it? Or is there a more elegant way to get E, say using its properties such as linearity? Thanks!
The probability that production occurs in a given hour is
$$ \left(\frac{23}{24}\right)^N\;. $$
Thus by linearity of expectation the expected product is
$$ 24N\left(\frac{23}{24}\right)^N\;. $$
To get the distribution of the number $k$ of shutdowns, note that this is equivalent to counting the number of strings of length $N$ over an alphabet of $24$ letters that use exactly $k$ different letters. There are $\binom{24}k$ ways to choose the $k$ hours, and by inclusion-exclusion there are
$$ \sum_{j=0}^k(-1)^j\binom kj(k-j)^N $$
ways for the $N$ machines to shut all of them down at least once, so the probability for exactly $k$ shutdowns is
$$ {24}^{-N}\binom{24}k\sum_{j=0}^k(-1)^j\binom kj(k-j)^N\;. $$