Queue M/M/2, proportion of unserved customers

Question

In my work, I managed to arrive at the following scenario: Consider that I have a queue with two servers whose arrival rate is a poisson process $\lambda = 2 $ customers per hour and the service time is exponential with an average of $E(T) = 1$ hour . In this queue, if a customer arrives and no one is waiting, the customer waits, otherwise the customer leaves. My interest is to know the proportion of customers that will not be served.

In a stable queuing system it is necessary that $\lambda > \mu$, but as my interest is in knowing the proportion of unserved customers, I believe that this restriction is not a problem, but as for this case, $\rho$ (occupancy rate) is equal to $1$, I'm kind of lost. Does anyone have any suggestions.

Answer 1

What you have is an $M/M/2/3$ queue with $1$ waiting capacity (the last $3$ refers to the maximum number of people in the system.) The queue occupancy probabilities are found in the general case $(M/M/c/K)$ using the following formulae, which give the probability $p_n$ of $n \, (0 \leq n \leq K)$ people in the system:

$$\begin{align} p_n &=& &{\lambda^n \over n!\mu^n}p_0 && 1 \leq n < c \\ &=& &{\lambda^n \over c^{n-c}c!\mu^n} p_0 && c \leq n \leq K \end{align}$$

$p_0$ can be derived from the condition that the probabilities must sum to 1.

The probability that the system is full at any time is:

$$p_c = {r^K \over c^{K-c}c!}p_0 $$

where $r = \lambda / \mu$. Since Poisson Arrivals See Time Averages (PASTA), the fraction of arrivals that will see $K$ people in the queue when they arrive and will leave as a consequence is equal to the probability that there are $K$ people in the queue in the steady state.

In your case $r = 2$ and $K = 3$, giving:

$$p_0 + \left(2 + {2^2\over 2} + {2^3 \over 2\cdot 2}\right)p_0 = 1 $$

resulting in $p_0 = 1/7$ and, after some minor calculation, $p_3 = 2/7$.

We can demonstrate the $M/M/c/K$ queue result with a little simulation. Here we make use of the fact that since all the interarrival times are distributed Exponentially, time is irrelevant; we have a continuous time Markov chain, and the transition probabilities are independent of time due to the memoryless property of the Exponential distribution.

lambda <- 2
mu <- 1

n_arrive <- 0
n_balk <- 0
n_servers_occupied <- 0
queue_length <- 0

for (i in 1:1000000) {
  p_arrival <- lambda  / (lambda + n_servers_occupied*mu)
  if (runif(1) <= p_arrival) {
    n_arrive <- n_arrive + 1
    if (n_servers_occupied == 2 & queue_length == 1) {
      n_balk <- n_balk + 1
    } else if (n_servers_occupied == 2) {
      queue_length = 1
    } else {
      n_servers_occupied <- n_servers_occupied + 1
    }
  } else {
    if (queue_length == 1) {
      queue_length <- 0
    } else {
      n_servers_occupied <- n_servers_occupied - 1
    }
  }
}

n_balk / n_arrive
[1] 0.2862546
> 2/7
[1] 0.2857143