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You arrive to your favorite restaurant\bar only to realize that there is no empty table :(

Assuming there are $n$ number of tables, on average a group at a table spends $\bar t$ time and finally that you are far way from the opening or closing time of the establishment1, how long would you have to wait?

I have given this much thought and ended up convincing myself that if the arrivals do not follow any distribution (are random), the average time one would have to wait is simply ${\bar t}/n$.


Example:

$15$ tables and $90$ min of dining on average $\Rightarrow 6$ min of waiting

Question:

Is it really so or is my intuition mistaken? Is there more math to this problem?


1 No bias on arrivals or departures of groups.

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    $\begingroup$ Nice intuition. Read about Little's law: en.wikipedia.org/wiki/Little%27s_law $\endgroup$ – Ethan Bolker May 7 '18 at 11:54
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    $\begingroup$ @Ethan: I was thinking about that, but how do you apply Little's law in this case? What are $L$, $\lambda$, and $W$ here? $\endgroup$ – Rahul May 7 '18 at 12:04
  • $\begingroup$ @EthanBolker Thanks about the link and info. It definitely looks a lot like Little's law but I cannot transform his equations to mine.. $\endgroup$ – Ev. Kounis May 7 '18 at 13:32
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Without Little's law:

Assuming people leave when done at random, with one table in the establishment the departure rate would be one departure per $t$ minutes. With $n$ tables, customers would leave at a rate of $n$ per $t$ minutes, so on average you'd wait $n/t$ minutes for the next free table.

With Little's law (essentially the same argument).

occupancy = throughput rate * service time

Since the occupancy is $n$ and the service time is $t$ the throughput is $n/t$ customers per minute. So you wait on average that long for the next customer to depart.

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  • $\begingroup$ I see, you apply Little's law applied to the tables alone, not to the whole restaurant (whose current occupancy is $n+1$). $\endgroup$ – Rahul May 8 '18 at 5:09

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