My question is about how to interpret 'arbitrary customer' in the following scenario (see question 2. listed below):

"At a single server service station two types of arrivals occur. According to a Poisson process individual customers arrive at the station with a rate of 5 customers per hour. Next to that, according to a Poisson process groups of two customers arrive at the station with a rate of 3 groups per hour. It takes an exponentially distributed time with a mean of 225 seconds to serve one single customer. Customers are served in order of arrival.

  1. Determine the distribution of the number of customers in the system.
  2. What is the mean sojourn time of an arbitrary customer?
  3. What is the mean sojourn time of a customer who arrived in a group of two customers and who is served after the other customer arriving at the same time?"

So, to me, there are two possible interpretations for this: from (1.), we know the distribution, so we can calculate $E[L]$, the average number of customers in the system and apply Little's formula to obtain $E[S]$, the mean sojourn time (in this case, we have $E[L] = 11\cdot E[S]$, as we have on average 11 customer arrivals in one hour).

However, I felt pretty unsatisfied with this, as the 'arbitrary customer' could be the unlucky one and arrive as the second in a group, in which case, we could calculate $E(S)$ by splitting it up in the three possible arrival possibilities of a customer, where we would answer (3.) at the same time: \begin{align*} E[S] &= E[S\mid\text{arriving customer not in group}] \cdot P(\text{arriving customer not in group})\\ &\quad+ E[S\mid\text{arriving customer first in group}] \cdot P(\text{arriving customer first in group})\\ &\quad+ E[S\mid\text{arriving customer second in group}] \cdot P(\text{arriving customer second in group}). \end{align*}

My gut feeling tells me the first possibility is the right one, as it would be stupid to ask question (3.) after (2.), but it seems a little crazy not to talk about the position the customer arrives when we talk about the mean sojourn time of an arbitrary customer.

Hope you can help me with this!


Your second approach is correct. Imagine the customers came in as groups of eleven, one group per day. It would be (almost) certain that when the group arrives the store is empty, but the poor eleventh customer has to wait for ten others before he is served. The distribution of the number of customers in the system will reflect this-it will have substantial probability of $11$ in the store, while your problem does not.

  • $\begingroup$ Thanks! To check, I also argued that an 'arbitrary' customer has to wait for all customers in the system at arrival + its own service time + (3/11) * service time, where the term "(3/11) * service time" reflects the three customers that arrive in an 'unlucky' place. Luckily, both answers were the same, but is this argument correct? $\endgroup$ – Ghostface Apr 3 '16 at 21:03
  • 1
    $\begingroup$ Yes. As you say, $3/11$ of the customers know they are arriving just after one other, so they have to wait for that customer. $\endgroup$ – Ross Millikan Apr 3 '16 at 23:29

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