Queues

The two tellers in a bank each take an exponentially distributed time to deal with any customer; their parameters are $\lambda$ and $\mu$ respectively. You arrive to find exactly two customers present, each occupying a teller.

Customers arrive at a desk according to a Poisson process of intensity $\lambda$. There is one clerk, and the service times are independent and exponentially distributed with parameter $\mu$. At time 0 there is exactly one customer, currently in service. Show that the probability that the next customer arrives before time $t$ and finds the clerk busy is

Vehicles pass a crossing at the instants of a Poisson process of intensity $\lambda$; you need a gap of length at least $a$ in order to cross. Let $T$ be the first time at which you could succeed in crossing to the other side. Show that $\mathbb {E}\left[T\right] = \left(e^{a \lambda }-1\right) / \lambda$, and find $\mathbb {E}\left[e^{\theta T}\right]$.

Suppose there are two lanes to cross, carrying independent Poissonian traffic with respective rates $\lambda$ and $\mu$. Find the expected time to cross in the two cases when:

Customers arrive at the instants of a Poisson process of intensity $\lambda$, and the single server has exponential service times with parameter $\mu$. An arriving customer who sees $n$ customers present (including anyone in service) will join the queue with probability $(n+1) /(n+2)$, otherwise leaving for ever. Under what condition is there a stationary distribution? Find the mean of the time spent in the queue (not including service time) by a customer who joins it when the queue is in equilibrium. What is the probability that an arrival joins the queue when in equilibrium?

Customers enter a shop at the instants of a Poisson process of rate 2. At the door, two representatives separately demonstrate a new corkscrew. This typically occupies the time of a customer and the representative for a period which is exponentially distributed with parameter $1$, independently of arrivals and other demonstrators. If both representatives are busy, customers pass directly into the shop. No customer passes a free representative without being stopped, and all customers leave by another door. If both representatives are free at time $0$, show the probability that both are busy at time $t$ is $\frac{2}{5}-\frac{2}{3} e^{-2 t}+\frac{4}{15} e^{-5 t}$.