Queuing System and Simulation Question Taxis arrive randomly at rate $\alpha$ at a taxi stand with a maximum capacity of $N$ taxis. Passengers arrive randomly at rate $\beta(\alpha>\beta)$. Show that in the steady-state: 
a) $P$(No taxis or passengers present) = $(1-\frac{\beta}{\alpha})(\frac{\beta}{\alpha})^N$
b)$P$(Passengers queuing, but no taxis) = $(\frac{\beta}{\alpha})^{N+1}$
c) Expected number of passengers queuing is $(\frac{\beta}{\alpha})^{N+1}/(1-\frac{\beta}{\alpha})$. 
The hint was given below.
Hint: Use as a state $(n,r)$ where $n$ is the number of taxis present, and $r$ is the number of passengers present. 
In this question, I am confused on using the hint, particularly the $n$ and $r$ in solving the question. Can anyone provide me some guide on this question? 

Assuming that passengers depart immediately when a taxi arrives, the state space for this problem is $S=\{(n,r)\in \{0,1,\ldots\}\times\mathbb N_0: nr=0\}$. Let $\{X_m : m\geqslant 0\}$ be a continuous time Markov chain on $S$ with transition rates
$$
q_{(n,r),(n',r')} = \begin{cases}
\alpha,& n'=n+1, r'=r, n'<N\\
\alpha,& n'=n=0, r'=r-1, r>0\\
\beta,& n'=n, r'=r+1\\
\beta,& n'=n-1, n'>0, r=0.
\end{cases}
$$
We make use of the property that for the stationary distribution $\pi$, any subset $A\subset S$ satisfies
$$
\sum_{(n,r)\in A,(n',r')\notin A} q_{(n,r),(n',r')}\pi_{(n,r)} = \sum_{(n,r)\notin A, (n',r')\in A} q_{(n,r),(n',r')}\pi_{(n,r)}.
$$
This yields the equations $\beta\pi_{(0,r+1)} = \alpha\pi_r$, $r=0,1,2,\ldots$ and $\alpha\pi_{(n,0)} = \beta\pi_{(n+1,0)}$, $n=0,1,\ldots,N-1$. We thus have the recurrences
\begin{align}
\pi_{(0,r+1)} &= \left(\frac\beta\alpha\right)^r\pi_{(0,0)},\ r=0,1,\ldots\\
\pi_{(n+1,0)} &= \left(\frac\alpha\beta\right)^n\pi_{(0,0)},\ n=0,1,\ldots,N-1.
\end{align}
From $\sum_{(n,r)\in S}\pi_{(n,r)}=1$ we have 
$$
1 = \sum_{r=1}^\infty \left(\frac\beta\alpha\right)^r \pi_{(0,0)} + \sum_{n=0}^{N}\left(\frac\alpha\beta\right)^n \pi_{(0,0)} = \pi_{(0,0)}\left(\frac\beta{\alpha-\beta} + \frac{\alpha  \left(\frac{\alpha }{\beta }\right)^N-\beta }{\alpha -\beta }\right),
$$
so that
$$
\pi_{(0,0)} = \left(\frac\beta{\alpha-\beta} + \frac{\alpha  \left(\frac{\alpha }{\beta }\right)^N-\beta }{\alpha -\beta }\right)^{-1} = \frac{(\alpha -\beta ) \left(\frac{\alpha }{\beta }\right)^{-N}}{\alpha }.
$$
It follows that $$\pi_{(0,r)}= \left(\frac\beta\alpha\right)^r \frac{(\alpha -\beta ) \left(\frac{\alpha }{\beta }\right)^{-N}}{\alpha } = \frac{(\alpha -\beta ) \left(\frac{\alpha }{\beta }\right)^{-(N+r)}}{\alpha },\ r=0,1,\ldots$$ 
and
$$
\pi_{(n,0)} = \left(\frac\alpha\beta\right)^n\frac{(\alpha -\beta ) \left(\frac{\alpha }{\beta }\right)^{-N}}{\alpha } = \frac{(\alpha -\beta ) \left(\frac{\alpha }{\beta }\right)^{n-N}}{\alpha },\ n=0,1,\ldots,N-1.
$$
Hence for a) we have $\pi_{(0,0)} = \left(1-\frac\beta\alpha\right)\left(\frac\beta\alpha\right)^N$, for b) we have
$$
\sum_{r=1}^\infty \pi_{(0,r)} = \sum_{r=1}^\infty \frac{(\alpha -\beta ) \left(\frac{\alpha }{\beta }\right)^{-(N+r)}}{\alpha } = \frac{\beta  \left(\frac{\alpha }{\beta }\right)^{-N}}{\alpha } =\left(\frac\beta\alpha\right)^{N+1},
$$
and for c) we have
$$
\sum_{r=1}^\infty r\frac{(\alpha -\beta ) \left(\frac{\alpha }{\beta }\right)^{-(N+r)}}{\alpha } = \frac{\beta  \left(\frac{\alpha }{\beta }\right)^{-N}}{\alpha -\beta } = \frac{\left(\frac\beta\alpha\right)^{N-1}}{1-\frac\beta\alpha}.
$$
