Question

Carloads of customers arrive at a single-server station in accordance with a Poisson process with rate 4 per hour. The service times are exponentially distributed with rate 20 per hour. If each carload contains either 1,2, or 3 customers with respective probabilities \(\frac{1}{4}, \frac{1}{2}\), and \(\frac{1}{4}\), compute the average customer delay in queue.

Short answer

The average customer delay in the queue is \(\frac{1}{16}\) hours.

Step-by-step solution

Calculate the average arrival rate and average service rate

The arrival rate is given as 4 per hour for the Poisson process, so the average arrival rate (λ) is: \(λ = 4\) The service times are given to be exponentially distributed with a rate of 20 per hour, so the average service rate (µ) is: \(µ = 20\)

Calculate the utilization of the server

The utilization of a server (ρ) can be calculated as the ratio between the average arrival rate and the average service rate for the system: \(ρ = \frac{λ}{µ}\) Substituting the values for λ and µ that we found in Step 1: \(ρ = \frac{4}{20} = \frac{1}{5}\)

Compute the service time distribution probabilities

The service time distribution probabilities are given as follows: \(P_1 = \frac{1}{4}\), \(P_2 = \frac{1}{2}\), and \(P_3 = \frac{1}{4}\) Let's calculate the mean of the service time distribution: \(E[S] = 1 \cdot P_1 + 2 \cdot P_2 + 3 \cdot P_3 = 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{2} + 3 \cdot \frac{1}{4} = \frac{1}{4} + 1 + \frac{3}{4} = 2\) The variance of the service time distribution can be found by: \(Var(S) = E[S^2] - (E[S])^2\) Now let's calculate the mean of the square service time distribution: \(E[S^2] = 1^2 \cdot P_1 + 2^2 \cdot P_2 + 3^2 \cdot P_3 = 1 \cdot \frac{1}{4} + 4 \cdot \frac{1}{2} + 9 \cdot \frac{1}{4} = \frac{1}{4} + 2 + \frac{9}{4} = \frac{14}{4} = \frac{7}{2}\) Now, we can compute the variance: \(Var(S) = E[S^2] - (E[S])^2 = \frac{7}{2} - 2^2 = \frac{7}{2} - 4 = \frac{1}{2}\)

Calculate average customer delay in queue

With all necessary parameters calculated, we can now compute the average customer delay in queue using the M/G/1 queue formula: \(W_q = \frac{ρ \cdot Var(S)}{2 \cdot (1-ρ) \cdot (E[S])}\) Substitute the values we calculated previously: \(W_q = \frac{\frac{1}{5} \cdot \frac{1}{2}}{2 \cdot (1-\frac{1}{5}) \cdot 2} = \frac{\frac{1}{10}}{\frac{8}{5}}\) Simplifying and calculating the value: \(W_q = \frac{1}{10} \cdot \frac{5}{8} = \frac{1}{16}\) Therefore, the average customer delay in the queue is \(\frac{1}{16}\) hours.