Given a M/M/c queue (Poisson arrival, exponentially distributed service time, c servers). The queue is unlimited in length and operates by first-in-first-out. Each customer who arrives and needs to queue has a patience time T, with a uniform distribution over an interval of [a,b]. That means if the customer has been waiting in queue for T minutes, they will abandon the queue.
I would like to find the average waiting time and the rate of queue abandonment for this queue. I suppose the average waiting time would be similar to a typical M/M/c queue. However I don't know how to approach finding the queue abandonment rate. Is it as simple as figuring out how many customer has a patience time below the average waiting time? Thanks in advance.
For uniform distribution case, I don't know.
For exponential distribution case, the resutl is as follows:
Mandelbaum and Zeltyn 2004 introduce Erlang A model in the paper Service Engineering in Action: The Palm/Erlang-A Queue, with Applications to Call Centers
In the Erlang A model $\ M/M/s/k+M\quad$, each caller posses an exponentially distributed patience time with mean $\theta^{-1}$
Stationary probability distribution:
$$ \begin{aligned} & P_n=\left\{\begin{array}{l} \frac{1}{n}\left(\frac{\lambda}{\mu}\right)^n P_0, 0 \leq n \leq s . \\ \frac{\lambda^n}{s ! \prod_{j=1}^{n-s}(s \mu+j \theta) \mu^s}, s+1 \leq n \leq k . \end{array}\right. \\ & P_0=\left(\sum_{n=0}^s \frac{1}{n !}\left(\frac{\lambda}{\mu}\right)^n+\sum_{n=s+1}^k \frac{\lambda^n}{s ! \prod_{j=1}^{n-s}(s \mu+j \theta) \mu^s}\right)^{-1} . \\ & \end{aligned} $$
Average queue size:
$$ L_{n}=\sum_{n=0}^{k}np_{n}=(\sum_{n=0}^{s}\frac{1}{(n-1)!}(\frac{\lambda}{m})^{n}+\sum_{n=s+1}^{k}\frac{n\lambda^{n}}{s ! \prod_{j=1}^{n-s}(s \mu+j \theta) \mu^s})P_{0} $$
Average waiting queue size:
$$ L_q=\sum_{n=s+1}^{k}(n-s)P_{n}=\left ( \sum_{n=s+1}^{k}\frac{(n-s)\lambda^{n}}{s ! \prod_{j=1}^{n-s}(s \mu+j \theta) \mu^s} \right ) P_{0}. $$
System occupancy rate:
$$ P_B=P_k=\left(\frac{\lambda^k}{s ! \prod_{j=1}^{k-s}(s \mu+j \theta) \mu^m}\right) P_0 $$
Average number of customers rejected due to busy lines:
$$ N_B=P_B \lambda $$
Number of clients lost due to impatience to give up waiting:
$$ N_L=L_q \theta $$
Call Loss Ratio:
$$ P_L=\frac{N_B+N_L}{\lambda} =P_B+ \frac{N_L}{\lambda} $$