In queueing-theory, M/M/1/K, if in steady state, is the mean arrival rate equal to mean departure rate(not service rate)? I am trying to calculate the Waiting time of a box that contains two queueing systems(qs) in serial. The arrivals are on left, they enter qs1,  exit , then enter qs2, then exit qs2 i.e exits the whole system. They are in steady state i.e arrival rate is lower than service rate for both qs1 and qs2.
Can I assume that mean arrival rate in qs1 is equal to mean departure rate at qs1?
(I ask this so that I can know the arrival rate of qs2)
 A: From your title I assume that both queues are $M/M/1/K$ queues.
No, that is not the case because an arrival at the first queue can find $K$ jobs at the first queue and therefore not be allowed to enter. So, certainly the number of arrivals to the first queue is larger than the number of departures from the first queue.
However, we can still say something about the departure rate at the first queue. Let $\pi_n$ be the steady-state probability of having $n$ jobs at the first queue and let $\lambda$ be the arrival rate to the first queue. By PASTA, an arriving job finds $K$ jobs in the first queue with probability $\pi_K$. If an arriving job finds a full system, then he is blocked (or lost). So, the arrival rate of jobs that enter the first queue is $\lambda (1 - \pi_K)$ and therefore the rate at which jobs leave the first queue is also $\lambda (1 - \pi_K)$. 
I also would like to note that two $M/M/1/K$ queues in series are always stable for any positive arrival rate and service rate, since the number of jobs in each queue can never exceed $K$.
