Question

Suppose that a customer of the \(M / M / 1\) system spends the amount of time \(x>0\) waiting in queue before entering service. (a) Show that, conditional on the preceding, the number of other customers that were in the system when the customer arrived is distributed as \(1+P\), where \(P\) is a Poisson random variable with mean \(\lambda\). (b) Let \(W_{Q}^{*}\) denote the amount of time that an \(M / M / 1\) customer spends in queue. As a by-product of your analysis in part (a), show that $$ P\left[W_{\mathrm{Q}}^{*} \leqslant x\right]=\left\\{\begin{array}{ll} 1-\frac{\lambda}{\mu} & \text { if } x=0 \\ 1-\frac{\lambda}{\mu}+\frac{\lambda}{\mu}\left(1-e^{-(\mu-\lambda) x}\right) & \text { if } x>0 \end{array}\right. $$

Short answer

The distribution of the number of customers in the system when a customer arrives is 1+P, where P is a Poisson random variable with mean λx, i.e., Q(n) = \( e^{-\lambda x} \frac{(\lambda x)^{n-1}}{(n-1)!} \). Furthermore, the probability distribution of the waiting time in the queue is: \( P\left[W_{Q}^{*} \leqslant x\right]=\left\\{\begin{array}{ll} 1-\frac{\lambda}{\mu} & \text { if } x=0 \\\ 1-\frac{\lambda}{\mu}+\frac{\lambda}{\mu}\left(1-e^{-(\mu-\lambda) x}\right) & \text { if } x>0 \end{array}\right. \)

Step-by-step solution

Identify Poisson process of arrivals

Since the system operates as an M/M/1 queue, we know that arrivals follow a Poisson process with rate λ.

Derive arrival distribution of previous customers

Conditional on the time spent in the queue by a customer, we can calculate the probability of a certain number of arrivals during that time. If there are n arrivals during that time, then the distribution of the number of arrivals (P(n)) will follow a Poisson distribution with mean λx: P(n) = \( e^{-\lambda x} \frac{(\lambda x)^n}{n!} \)

Shift the distribution by 1

Since we are given that the customer spends a positive amount of time (x>0) in the queue, there should be at least one customer in the system when this customer arrives. Therefore, we need to shift the distribution by 1 to account for this additional customer. The distribution of the number of customers in the system (Q) is Q(n) = P(n-1) = \( e^{-\lambda x} \frac{(\lambda x)^{n-1}}{(n-1)!} \) #Part b#: Deriving the probability distribution of time spent in queue

Obtain complementary cumulative distribution for interarrival time

Recall the system operates using exponential distributions for arrival and service rates with mean interarrival time 1/λ and mean service time 1/μ. The complementary cumulative distribution function (CCDF) of the interarrival time T between customers can be written as \( P(T > x) = e^{-\lambda x} \).

Calculate conditional CCDF for time in queue

If the service time S is less than the interarrival time T, the customer will spend no time in queue (x=0). Otherwise, the customer will spend the excess service time in the queue (x = S-T). We can express the conditional CCDF for the time spent in the queue as \( P(W^{*}_Q > x | S > T) = P(T > S - x | T < S) = \frac{ P(T > S-x)^2}{\mu - \lambda} \).

Convert the conditional CCDF to CDF for time in queue

The cumulative distribution function (CDF) for the time spent in the queue can be calculated as the complement of the conditional CCDF: \( P(W^{*}_Q \leq x) = 1 - P(W^{*}_Q > x | S > T) \).

Obtain the final probability distribution for time in queue

Plugging the values obtained in the previous steps, we find the probability distribution for the time spent in the queue: \( P\left[W_{\mathrm{Q}}^{*} \leqslant x\right]=\left\\{\begin{array}{ll} 1-\frac{\lambda}{\mu} & \text { if } x=0 \\\ 1-\frac{\lambda}{\mu}+\frac{\lambda}{\mu}\left(1-e^{-(\mu-\lambda) x}\right) & \text { if } x>0 \end{array}\right. \)