Question

The arrival process of customers to a service facility is Poisson with an average arrival rate of 4 customers per hour. The service facility has a single employee that serves the customers. This person takes on average 10 minutes to serve one customer (assume this time is exponentially distributed). (a) What average number of persons can be expected in the line? (b) What average number of persons can be expected in the system? (c) What is the average amount of time that the person can expect to spend in line? (d) On average how much time will it take for a person to be served, including waiting time? (e) On weekends, the arrival rate can be expected to increase to over 8 per hour. What effect will this have on the number in the waiting line? (f) What is the probability that a customer arrives, and nobody is front him/her in the line?

Answer 1

Using the M/M/1 queue model, we find the average number in the line to be 4, in the system to be 4.67, average time in line to be 1 hour, and total time in the system to be 70 minutes. An increase in arrival rate leads to a longer queue. The probability of arriving and finding no line is 33%.

Step-by-step explanation:

Queuing Theory Problem

For a service system with Poisson arrivals and exponential service times, known as an M/M/1 queue, we use the arrival rate (λ = 4 customers/hour) and the service rate (μ = 6 customers/hour, since one customer takes 10 minutes to serve) to answer the following:

• (a) Average number in the line (Lq) can be calculated using Lq = λ^2 / (μ * (μ - λ)). This gives Lq = 4^2 / (6 * (6 - 4)) = 4.
• (b) Average number in the system (Ls) is Lq + λ/μ. Using the values above, Ls = 4 + 4/6, giving us Ls = 4.67.
• (c) Average time in the line (Wq) can be calculated as Wq = Lq / λ. This results in Wq = 1 hour.
• (d) Average time in the system (Ws) includes service time, Ws = Wq + 1/μ, which gives Ws = 1 + 1/6 hours (or 70 minutes).
• (e) If the arrival rate increases to 8 customers/hour, the new Lq will be higher since Lq increases with λ. Specifically, Lq = 8^2 / (6 * (6 - 8)), demonstrating queue instability as λ now exceeds μ.
• (f) The probability of zero in the system (P0) is 1 - λ/μ. For the given rates, P0 = 1 - 4/6, resulting in a probability of 0.33, or 33%.

The calculations here illustrate key properties of M/M/1 queue systems and show that as arrival rates increase without a corresponding increase in service rates, the queue length and wait times will grow, potentially leading to an overburdened system.