I struggle to understand the following queueing system (to avoid leaving out any important information, I will just copy the task):
"At a small river cars are brought from the left side to the right side of the river by a ferry. On average 15 cars by hour arrive according to a Poisson process. It takes the ferry an exponentially distributed time with a mean of 3 minutes to cross the river and return. The capacity of the ferry is equal to 2 cars. The ferry only takes off when there are two or more cars waiting.
- What is the fraction of time that the ferry is busy?
- Determine the distribution of the number of cars that are waiting on the ferry."
For the first question, I defined two cars as one job (since the ferry does not start and can also not handle more than two cars). That gives as the fraction of time the ferry is busy $\rho=3/8$.
I tried to work out question two using balance equations, but that did not work.
I know that the correct answers is $$p_0=\frac{1}{4}+\frac{3}{4} \Big( \frac{1}{2} \Big)^2 = \frac{7}{16}$$
$$p_1=\frac{3}{4}\Big( \frac{1}{2} \Big)^1 +\frac{3}{4} \Big( \frac{1}{2} \Big)^3 = \frac{15}{32}$$ $$p_n=\frac{3}{4}\Big( \frac{1}{2} \Big)^{n+2}, \qquad n\geq 2$$
Can somebody help me understanding this system?
The question is asked in a slightly confusing fashion. Let us denote by $k$, the system state, the number of cars currently in the system (i.e., either on the ferry or waiting for it). Let also the arrival rate $\lambda = 15$ cars per hour, and the ferry rate $\mu = 20$ trips per hour. If we let $r_k = P(\text{$k$ cars in the system})$, the equilibrium equations are as follows:
$$ \lambda r_0 = \mu r_2 \\ \lambda r_1 = \lambda r_0 + \mu r_3 \\ (\lambda+\mu) r_k = \lambda r_{k-1} + \mu r_{k+2}, \qquad k \geq 2 $$
The solution to this system is
$$ r_0 = \frac14 \\ r_k = \frac{3}{2^{k+2}}, \qquad k \geq 1 $$
Then, because there are $0$ cars waiting if $k = 0$ or $2$, and $1$ car waiting if $k = 1$ or $3$, then
$$ p_0 = r_0 + r_2 \\ p_1 = r_1 + r_3 \\ p_k = r_{k+2}, \qquad k \geq 2 $$