M($\lambda$)/M($\mu$)/n queue, recurrence and busy periods Question
A surfing company has a very high number of surfboards to rent. The owner makes the following estimates. A new individual customer enters the shop on average every 5
minutes. Then, each customer decides to rent a surfboard with a probability $\frac{1}{2}$. If they make
this choice, they use it on average for 30 minutes before returning it to the shop.
(i) Give an irreducible continuous-time Markov chain describing the number of surfboards in use, assuming the number of surfboards in the shop is infinite.
(ii) Is the chain recurrent, and hence is it explosive?
(iii) Estimate the proportion of time over a long period when no surfboard is in use
(iv) The owners call the busy period the maximal interval of time when at least one surfboard is in use. How long is said busy time on average?
(v) In the long run, what is the probability that two customers return their surfboard within
a time slot of 20 minutes?

Attempt
(i)
We can model this example as an $M(\lambda) / M(\mu) / n$ queue, $n \in\{1,2, \ldots\} \cup\{\infty\}$. Which is an irreducable continuous time Markov Chan with the following diagram:

With $\lambda=\frac{1}{10}$ and $\mu=\frac{1}{30}$. Since the number of surfboards ("servers") is infinite, we can use this to model the number of surfboards being rented.
Note for future reference:
$
q_{x y}=\left\{\begin{array}{cc}
\frac{1}{10}, & y=x+1 \\
\frac{x}{30}, & y=x-1 \\
-\frac{1}{10}-\frac{x}{30}, & x=y
\end{array}\right.
$
$P_{x, y}=\left\{\begin{array}{cl}\frac{\frac{1}{10}}{\frac{1}{10}+\frac{x}{30}}=\frac{3}{3+x} & , y=x+1 \\ \frac{\frac{x}{30}}{\frac{1}{10}+\frac{x}{30}}=\frac{x}{3+x}, & y=x-1 \\ 1 & , x=0;y=1 \\ 0 & , x=0 ; y>0\end{array}\right.$
(ii) This part I am unsure with...
We know for $\lambda,\mu \gt 0,n\lt\infty$, the M($\lambda$)/M($\mu$)/n queue is recurrent iff $\lambda\le n\mu$. It is clear to see that for $n\ge3$ this is true and since the chain is irreducible all states are recurrent. And hence the chain is not explosive. $\square$
(iii) We now note that for a measure $\pi$ on state-space $S$ the detailed balance equation satisfies the following:
$\pi(x)q_{x,y}=\pi(y)q_{y,x} \quad \forall x,y\in S$, and that if said chain is irreducible, we also have the following: $\pi(y)=\frac{1}{q_{y}\mathbb E[T_{y,1}]}$ where $\mathbb E[T_{y,1}]$ is the 1st expected time to return to state $y$.
$\Rightarrow \pi(x)q_{x,x+1}=\pi(x+1)q_{x+1,x}$
$\Rightarrow \frac{1}{10}\pi(x)=\frac{x+1}{30}\pi(x+1)$
$\Rightarrow \pi(x+1)=\frac{3\pi(x)}{x+1} \quad \Rightarrow \pi(x+1)=\frac{3}{x+1}\left\{\frac{3}{x}\left\{\frac{3}{x-1}...\left\{\frac{3}{0+1}\pi(0)\right\}...\right\}\right\}$
It is clear to see and easy to prove via induction that:
$\pi(x)=\frac{3^x}{x!}\pi(0).\quad$ Also note that for an invariant measure the following is true $\sum_{x\in S}\pi(x)=1$ i.e We have that $\sum_{x\in S}\pi(x)=\pi(0)\sum_{x=0}^{\infty}\frac{3^x}{x!}$
$\Rightarrow \pi(0)e^3=1\iff \pi(0)=e^{-3}$
Hence we have that the proportion of time when no surfboard is $e^{-3}=4.9$%$ \quad\square $
(iv) We can now compute the busy period using the following $\mathbb E_{0}[T_{0,1}]=\mathbb E_{0}[T_1]+\mathbb E_{0}[B^{(1)}]\quad\Rightarrow \mathbb E_{0}[B^{(1)}]=E_{0}[T_{0,1}]-\mathbb E_{0}[T_1]=\frac{1}{q_{0}\pi(0)}-\frac{1}{1/10}=10e^3-10=190.9$ mins $\quad\square$
(v) As for this I am unsure how to tackle this I've had ideas of using Burke's Theorem but I have made no progress thus far :/

Comments
Found these questions tricky particularly unsure with parts (ii) and (v). Any help with either one or even alternate solutions to mine would be greatly appreciated :)
 A: This is a $M/M/\infty$ queue with arrival rate $\lambda = \frac1{10}$ and service rate $\mu=\frac1{30}$. We can model this as a birth-death process with birth rates uniformly $\lambda$ and death rates $\lambda_n=n\mu$, $n\geqslant 1$. The generator matrix of the corresponding continuous-time Markov chain would have superdiagonal elements equal to the birth rates, subdiagonal elements equal to the death rates, and diagonal elements such that each row sums to zero. Since
$$
\prod_{i=1}^n \frac{\mu_i}{\lambda_i} = \prod_{i=1}^n \frac{i\lambda}{\mu}=\left(\frac\lambda\mu\right)^nn!\stackrel{n\to\infty}\longrightarrow\infty,
$$
the process is recurrent. Moreover, the stationary distribution has Poisson distribution with parameter $\frac\lambda\mu$ (this can be verified by a straightforward computation). So the long-run proportion of time when no surfboard is in use is $e^{-\frac\lambda\mu}$.
The busy period has mean
$$
\frac1\lambda\sum_{i=1}^\infty \left(\frac\lambda\mu\right)^{i}\frac1{i!} = 10(e^{\frac13}-1).
$$
Guillemin, Fabrice; Simonian, Alain, Transient characteristics of an (M/M/\infty) system, Adv. Appl. Probab. 27, No. 3, 862-888 (1995). ZBL0829.60058.
Finally, the transient probabilities of a $M/M/\infty$ queue, as shown in Fundamentals of queueing theory (Gross, Donald; Harris, Carl M.; Shortle, John F.; Thompson, James M) are
$$
p_n(t)=\frac1{n!}\left((1-e^{-\mu t})\frac\lambda\mu\right)^n\exp\left(-(1-e^{-\mu t})\frac\lambda\mu\right),
$$
and so the probability that exactly two surfboards were returned in a $20$-minute window is
\begin{align}
\sum_{n=2}^\infty p_n(T)p_{n-2}(T+20),
\end{align}
which doesn't appear to have an easily expressed closed-form.
