Question

Show for the \(M / M / s\) system that the stationary queue size distribution \(\left\\{p_{n}, n=0,1,2, \ldots\right\\}\) is given by $$ \begin{aligned} &p_{0}=\left\\{\frac{(s \rho)^{s}}{s !(1-\rho)}+\sum_{i=0}^{s-1} \frac{(s \rho)^{t}}{i !}\right\\}^{-1} \\ &P_{n}= \begin{cases}P_{0} \frac{(s \rho)^{n}}{n !}, & 1 \leq n \leq s \\ p_{0} \rho^{n} \frac{s^{s}}{s !}, & s

Short answer

The stationary queue size distribution of an M/M/s system and the size of the queue without those being served can be calculated using the given formulas. These have been derived based on the understanding of an M/M/s system, service utilization ρ and using the concept of probabilities of different situations.

Step-by-step solution

Understanding the M/M/s system and the provided formulas

M/M/s system is a model for processes where arrivals follow a Poisson process, service times are exponentially distributed, and there are 's' servers. The formulas given for \(p_{n}\) are the probability of having 'n' customers in the system. It is divided into two sections: when \(n≤s\) and \(n ≠ s\). The given \(\rho\) formula is the server utilization, a measure of the average load on each server. The variable \(Q\) defined is the load on the system excluding those being served currently.

Proving the stationary queue size distribution

For \(n≤s\), as per the M/M/s system, the probability of 'n' customers in system \( p_n \), is expected to equal the probability that 'n' customers arrive at the server times the probability that all 's' servers are busy, which is mathematically given by \(P_{0} \frac{(s \rho)^{n}}{n !}\). Similarly, for \(n>s\), it is the probability that 'n' customers arrive times the probability that all 's' servers are busy and the extra \(n-s\) customers are waiting, which is mathematically given by \(p_{0} \rho^{n} \frac{s^{s}}{s !}\). For calculating \(p_0\), use the normalization condition on the probabilities. \(p_0\) should equal to 1 subtracted by the sum of probabilities for all other values of 'n'. Simplifying using the given equations, we get the provided equation for \(p_0\).

Proving the size of the queue

The quantity \(Q=\max(n-s,0)\) explains that only the customers exceeding the total servers \(s\) will be queued. The total probability that the queue size is 0, \(\gamma\), can be obtained by summing up all the situations where the number of customers does not exceed \(s\) (i.e., \(n≤s\)), which leads to given expression for \(γ\) after simplification. The expected queue length \(E(Q)\) is the average number of customers in the queue, excluding those being serviced currently. It can be obtained by taking the weighted sum of all possible queue lengths, excluding \(γ\), and then dividing it by probability of having a non-zero queue length \(1-γ\), which gives \(E(Q)=(1-γ)/(1-ρ)\) after simplification.