What kind of model, when store opening hours, durations of customer visits and the number of customers coming by car are known?
I need to estimate the amount of parking space that I need.
I've found that this is likely related to queueing theory, but having not background in it it seemed like the information that I'm given is not necessarily fit for models in queueing theory. Since those models seem to rely on e.g. the quantities "arrival rate" and "service rate", which I don't have.
On
This is a new version of the answer... As you see, there is not enough data to set this problem mathematically, you have opening hours but don't have closing hours(why would you have use of opening hours then?)
To estimate needed parking space again isn't clear. To be sure everyone will have where to park you need as many parking places as customers coming during the day(in case they all come at the same moment). So obviously you don't wanna be sure they will have enough places, you just want high probability of it.
Another dilemma may be: are the services being provided 1.simultaneously or 2.one by one? This also produces two different problems. If only one service is being provided in a period then you're dealing with the queueing, but then you also need total number of customers, not only those who coming by car, so we can assume the first
One more thing, has each period of the day got same traffic? You also need the distribution of customers' arriving time.
After all this, we can reformulate the question: Estimate amount of parking place needed, if during a day(S=12hrs) certain number of customers coming by car(n=1000 customers per day) uniformly distributed, service lasts certain amount of time(L=10min), so that a customer will have where to park with certain probability(p=0.9999)
If we mark each start of a service as $l_k\in (0,S-L)$ and needed number of places as $m$ then probability that visitor $l_n$ will have where to park is equal to probability that in last $L$ minutes there have less then $m$ customers came. Notice that for $l_n\in(0,L)$ there is some better probability for that. So $p=\frac{S-2L}{S-L}\sum_{k=0}^{m-1}{n-1\choose k}(\frac{L}{S-L})^k(\frac{S-2L}{S-L})^{n-k-1}+\int_0^L\sum_{k=0}^{m-1}{n-1\choose k}(\frac{x}{S-L})^k(\frac{S-L-x}{S-L})^{n-k-1}\frac{dx}{S-L}$ $$\sum_{k=0}^{m-1}{n-1\choose k}(\frac{L}{S-L})^k(\frac{S-2L}{S-L})^{n-k-1}<p<\frac{S-2L}{S-L}\sum_{k=0}^{m-1}{n-1\choose k}(\frac{L}{S-L})^k(\frac{S-2L}{S-L})^{n-k-1}+\frac L {S-L}$$ If we put the values from the example: $$0.99989857<\sum_{k=0}^{m-1}{999\choose k}(\frac{10}{710})^k(\frac{700}{710})^{999-k}<0.9999$$
This is binomial distribution which can be approximated by normal distribution $f(np,np(1-p))=f(999\frac 1 {71},999\frac 1 {71}\frac {70} {71})$ so for high probability $m$ has to be higher than $\frac {999} {71}+\frac 3 2σ$ so m>35 approximately
Maybe all you need is Little's Law: $L = \lambda w$, where $L$ is the average number of customers in the system, $\lambda$ is the arrival rate, and $w$ is the average time in the system. For arrival rate, you could take the number of customers arriving by car divided by the length of time the store is open, at least as an intial estimate. But I suspect that arrival rates vary by time of day, so you may need to estimate the rate during peak periods. https://en.wikipedia.org/wiki/Little%27s_law