A shop is selling tickets for a football match. They have set up two parallel counters. For each counter, the time to sell one ticket (i.e. take the money, give the ticket, register the buyer) is exponential with rate μ. Customers arrive to buy the tickets according to a Poisson process with rate λ. If a customer arrives and finds both the desks busy, they leave without buying a ticket (this customer is said to be lost). If a customer arrives and finds at least one free desk, the ticket purchasing process immediately starts, and the customer leaves immediately after completing the transaction.
a) If both counters are presently busy, find the expected time until the next customer starts the process of buying a ticket.
b) Starting with both counters empty, find the expected time until both desks are busy.
My attempted solution:
a) Let's call the two counters $c_1$ and $c_2$.
I think the expected time should be the lower of time taken by the two processes since a customer can only be served once one of the two counters is free.
So, $E[t] = E[min(c_1,c_2)] = \frac1{2\mu}$
b) I'm not sure how to go about this.
This is a $M/M/2/2$ queueing system - that is, the arrival process is Poisson, the service times are i.i.d. with exponential distribution, there are $2$ servers, and there is room for $2$ customers in the system. Let $X(t)$ be the number of customers in the system at time $t$. Then $\{X(t):t\geqslant0\}$ is a continuous-time Markov chain on state space $E=\{0,1,2\}$ with generator matrix $$ Q=\pmatrix{ -\lambda&\lambda&0\\ \mu&-(\lambda+\mu)&\lambda\\ 0&2\mu&-2\mu& }. $$ For part a), the random time in question consists of a transition from state $2$ to state $1$, then either a transition from state $1$ to state $2$ or a transition from state $1$ to state $0$ and then a transition from state $0$ to state $1$. Hence, the expected time is given by $$ \frac1{2\mu}+\left(\frac\lambda{\lambda+\mu}\right)\frac1\lambda + \left(\frac\mu{\lambda+\mu}\right)\left(\frac1\mu+\frac1\lambda\right) = \frac{1}{\lambda +\mu }+\frac{1}{\lambda }+\frac{1}{2 \mu }. $$ For part b), the expected time to reach state $2$ starting in state $0$ satisfies the recursion $$ t = \frac{1}{\lambda }+\left(\frac{\lambda}{\lambda +\mu }\right)\frac1\lambda+\left(\frac{\mu}{\lambda +\mu }\right)\left(\frac1\mu+t\right), $$ which yields $$ t= \frac{3}{\lambda }+\frac{\mu }{\lambda ^2}. $$