Question

Consider a two-server system in which a customer is served first by server 1, then by server 2, and then departs. The service times at server \(i\) are exponential random variables with rates \(\mu_{i}, i=1,2 .\) When you arrive, you find server 1 free and two customers at server 2 -customer \(\mathrm{A}\) in service and customer \(\mathrm{B}\) waiting in line. (a) Find \(P_{A}\), the probability that \(A\) is still in service when you move over to server 2 . (b) Find \(P_{B}\), the probability that \(B\) is still in the system when you move over to server 2 . (c) Find \(E[T]\), where \(T\) is the time that you spend in the system. Hint: Write $$ T=S_{1}+S_{2}+W_{A}+W_{B} $$ where \(S_{i}\) is your service time at server \(i, W_{A}\) is the amount of time you wait in queue while \(A\) is being served, and \(W_{B}\) is the amount of time you wait in queue while \(B\) is being served.

Short answer

The short answer to the problem is as follows: (a) The probability P_A that customer A is still in service when you move over to server 2: \[P_A = \frac{\mu_2}{\mu_1 + \mu_2}\] (b) The probability P_B that customer B is still in the system when you move over to server 2: \[P_B = \frac{\mu_1\mu_2}{(\mu_1+\mu_2)^2}\] (c) The expected time E[T] spent in the system: \[E[T] = \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{\mu_2}{\mu_1+\mu_2}\frac{1}{\mu_2} + \frac{\mu_1\mu_2}{(\mu_1+\mu_2)^2\}\frac{1}{\mu_2}\]

Step-by-step solution

Find the probability P_A

To find the probability that customer A is still in service when we move over to server 2 (P_A), we need to know that our service time at server 1 is longer than the remaining service time of customer A at server 2. Since service times are exponential random variables, we can write the conditional probability as: \[P_A = P(S_1 > S_{A2} | S_{A2} > 0) = \int_{0}^{\infty} \frac{\mu_1 \mu_2 e^{-\mu_1 s_1} e^{-\mu_2 s_{A2}}}{\mu_2} ds_{A2} ds_1 = \frac{\mu_2}{\mu_1 + \mu_2}\]

Find the probability P_B

To find the probability that customer B is still in the system when we move over to server 2 (P_B), we need to know that our service time at server 1 plus A's remaining service time at server 2 is longer than B's total service time at both servers. We can write the conditional probability as: \[P_B = P(S_1 + S_{A2} > S_{B1} + S_{B2} | S_{A2} > 0) = \int_{0}^{\infty} \frac{\mu_1^2 \mu_2^2 e^{-\mu_1 s_1} e^{-\mu_2 s_{A2}}}{(\mu_1+\mu_2)(\mu_1 + \mu_2)^2} ds_{A2} ds_1 = \frac{\mu_1\mu_2}{(\mu_1+\mu_2)^2\}\]

Expected time spent in the system (E[T])

Using the hint given in the problem, we can write the time spent in the system as: \[T = S_1 + S_2 + W_A + W_B\] Now we need to find the expected values of these terms. - E[S_1]: This is the expected service time at server 1, which is equal to \(\frac{1}{\mu_1}\). - E[S_2]: This is the expected service time at server 2, which is equal to \(\frac{1}{\mu_2}\). - E[W_A]: This is the expected waiting time due to customer A being served. We can find it using the fact that the probability of A still being in service is P_A. Thus, the expected waiting time is P_A * E[S_{A2}], where E[S_{A2}] = \(\frac{1}{\mu_2}\), so E[W_A] = \(\frac{\mu_2}{\mu_1+\mu_2}\frac{1}{\mu_2}\). - E[W_B]: This is the expected waiting time due to customer B being in the system. We can find it using the fact that the probability of B still being in the system is P_B. Thus, the expected waiting time is P_B * E[S_{B2}], where E[S_{B2}] = \(\frac{1}{\mu_2}\), so E[W_B] = \(\frac{\mu_1\mu_2}{(\mu_1+\mu_2)^2\}\frac{1}{\mu_2}\). Now, we can find the expected total time in the system by adding these expected values: \[E[T] = E[S_1] + E[S_2] + E[W_A] + E[W_B] = \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{\mu_2}{\mu_1+\mu_2}\frac{1}{\mu_2} + \frac{\mu_1\mu_2}{(\mu_1+\mu_2)^2\}\frac{1}{\mu_2} \]

Key concepts

Exponential Random Variables

Exponential random variables are ubiquitous in the study of queueing theory and stochastic processes. They are especially suited to modeling the time between successive events, such as the arrival or service times in queues.
An exponential random variable, often denoted by the symbol 'X', is characterized by a parameter \( \mu \), referred to as the rate. The probability distribution function (PDF) of an exponential random variable is given by \( f(x;\mu) = \mu e^{-\mu x} \) for \(x \geq 0\).One of the key properties of an exponential random variable is the memoryless property, which states that the probability of an event occurring in the next instant is independent of how much time has already elapsed. Mathematically, this is represented as \( P(X > s + t | X > t) = P(X > s) \), for all \( s, t \geq 0 \). This property is particularly useful when analyzing service times in queueing systems since it simplifies many calculations.

Service Times in Queuing Theory

In queuing theory, service times refer to the duration a service mechanism, such as a server or teller, takes to serve a customer. The 'service mechanism' can be anything from a cashier at a grocery store to complex machinery in a manufacturing plant.
When service times are modeled using exponential random variables, as in the exercise, the service process is said to follow an exponential distribution. The rate parameter \( \mu \) determines the average rate of service, with a higher \( \mu \) corresponding to a faster average service rate. In practical terms, the expected service time for each customer is \( E[X] = 1/\mu \), indicating that on average, a customer will be served within this timeframe.
Having exponential service times significantly simplifies the mathematical analysis in queuing theory, particularly in calculating the probability of certain events in the system and determining waiting times.

Conditional Probability in Queue Systems

Conditional probability plays a central role in analyzing queuing systems. It is the probability of an event given that another event has occurred. In queueing theory, we often want to calculate the probability that something will happen—like customer A still being in service—given the current state of the system.
The formulas provided for \( P_A \) and \( P_B \) in the solution represent conditional probabilities. For instance, \( P_A \) is the probability that customer A is still in service when you finish at server 1, conditional upon the fact that customer A was in service when you arrived. These probabilities help to determine how long the queue will be, by allowing us to predict future states of the system based on its current state.Understanding conditional probabilities is vital as it allows us to calculate different scenarios within queueing systems. It informs us of the probabilities of certain queuing lengths, service times, and waiting times - all crucial when optimizing service efficiency.

Expected Waiting Time Calculation

Calculating the expected waiting time, denoted as \( E[T] \) in queuing systems, is a fundamental aspect that determines the efficiency and customer satisfaction associated with the service being analyzed. Expected waiting time is the average time a customer is predicted to wait before being served and after service completion.
Using the information from exponential random variables and conditional probabilities, one can calculate the expected waiting times of customers, such as \( E[W_A] \) and \( E[W_B] \) in the given exercise. It incorporates the probability of having to wait (for instance, if a previous customer is still being served) and the expected service time of the server itself.
These calculations take into consideration the rates of service (\( \mu \) values for each server in our example) as well as the probability that another customer will still be in the system (like \( P_A \) and \( P_B \) in the problem). By understanding and implementing expected waiting time calculations, queueing systems can be designed and managed to minimize wait times and improve the flow of service.