Why are departure rates of $M/M/c$ queues $n\mu$ but not $\mu$? I think this issue should have an answer somewhere but I could not find in any materials. In every textbook I read about $M/M/c$ queueing systems, it is always acknowledged from the beginning that the departure rates for that queue are defined as $\mu_n = n\mu$ if $n \le c$, and $\mu_n = c\mu$ if $n \ge c$, where $\mu$ is the mean service rate and $n$ is the number of customers in the system.
What is the intuitive meaning of this definition? And is there any strict proof for that too?
What if we have two different service rates $a$ and $b$? For example, there are $2$ servers, and currently there are $2$ customers with different service rates $a < b$ occupying the $2$ servers. How can we know the departure rate at each state $\mu_1, \mu_2, \mu_3, \dots$ in this case?
 A: If there are $k = \min\{n,c\}$ jobs, each of which is being processed at a rate of $\mu$, then the times until they are finished are independent exponential random variables: $\mathbf T_1, \mathbf T_2, \dots, \mathbf T_k \sim \operatorname{Exp}(\mu)$.
The time until the next one of them is finished is $\mathbf T =\min\{\mathbf T_1, \dots, \mathbf T_k\}$, and we can show that $\mathbf T \sim \operatorname{Exp}(k\mu)$, so $k\mu$ is the rate at which some job departs from the system.
To show this, we can characterize exponential variables by their CCDF: $\mathbf X \sim \operatorname{Exp}(\lambda)$ if and only if $\Pr[\mathbf X > x] = e^{-\lambda x}$ for all $x\ge 0$. We have $\mathbf T > t$ if and only if $\mathbf T_1, \dots, \mathbf T_k > t$, which happens with probability $\Pr[\mathbf T_i > t]^k = (e^{-\mu t})^k = e^{-(k\mu)t}$. Therefore $\mathbf T \sim \operatorname{Exp}(k\mu)$.
Similarly, is we have $k$ jobs being processed at rates of $\mu_1, \dots, \mu_k$, then we'll get $\mathbf T \sim \operatorname{Exp}(\mu_1 + \dots + \mu_k)$.

Actually, all of this logic is built into the setup, if we study queueing theory via continuous-time Markov chains. Say that state $S$ is a state with $k$ jobs being processed at a rate of $\mu$, and if one of them is done, we go to state $T$. Then we can draw $k$ arrows, each with a rate of $\mu$, from state $S$ to state $T$. Having $k$ arrows with a rate of $\mu$ is equivalent to having one arrow with a rate of $k\mu$.
