Question

In a certain system, a customer must first be served by server 1 and then by server \(2 .\) The service times at server \(i\) are exponential with rate \(\mu_{i}, i=1,2 .\) An arrival finding server 1 busy waits in line for that server. Upon completion of service at server 1 , a customer either enters service with server 2 if that server is free or else remains with server 1 (blocking any other customer from entering service) until server 2 is free. Customers depart the system after being served by server \(2 .\) Suppose that when you arrive there is one customer in the system and that customer is being served by server \(1 .\) What is the expected total time you spend in the system?

Short answer

The expected total time a customer spends in the system is given by the sum of the expected time spent waiting for server 1 and the expected time spent waiting for server 2. Thus, the total expected time spent in the system is \(1/\mu_1 + 1/\mu_2^2\).

Step-by-step solution

Expected time spent waiting for server 1

Since there is already a customer in the system being served by server 1, we need to find the expected time spent waiting for server 1 to finish serving that customer. Since the service time for server 1 is exponentially distributed with rate μ₁, the expected service time for server 1 is 1/μ₁. Therefore, the expected time spent waiting for the first server to finish serving the current customer is 1/μ₁.

Expected time spent waiting for server 2

The customer will enter service with server 2 after server 1 if server 2 is free. If server 2 is not free, the customer must wait until server 2 is free. The service times are independent, and each has its own exponential distribution. The customer will either wait 0 units of time or an amount equal to the remaining service time of server 2, assuming server 1 is now free. To compute the expected time spent waiting for server 2, we calculate the expected remaining time on server 2 given that a time 1/μ₁ has passed and server 2 is busy: \(E[remaining\_time] = \int_0^\infty x\cdot p(x) dx\) Considering that server 2's service time has an exponential distribution with rate μ₂, the probability density function is: \(p(x) = \mu_2 e^{-\mu_2 x}\) Hence, the expected remaining time is: \(E[remaining\_time] = \int_0^\infty x(\mu_2 e^{-\mu_2 x}) dx\) Integrating by parts gives: \(E[remaining\_time] = \left[-\frac{x e^{-\mu_2 x}}{\mu_2}\right]_0^\infty + \int_0^\infty \frac{e^{-\mu_2 x}}{\mu_2} dx\) \(E[remaining\_time] = \left[-\frac{x e^{-\mu_2 x}}{\mu_2}\right]_0^\infty - \left[\frac{ e^{-\mu_2 x}}{\mu_2^2}\right]_0^\infty\) \(E[remaining\_time] = 0 - 0 + \frac{1}{\mu_2^2}\) So the expected time spent waiting for server 2 is 1/μ₂².

Calculate the total expected time spent in the system

Now that we have the expected time spent waiting for server 1 (1/μ₁) and the expected time spent waiting for server 2 (1/μ₂²), we can find the total expected time spent in the system by adding these times together: Total expected time = Expected time waiting for server 1 + Expected time waiting for server 2 Total expected time = 1/μ₁ + 1/μ₂²

Key concepts

Exponential Service Time

Understanding the concept of exponential service time is crucial in queueing systems. Service time refers to the duration a service takes, such as how long a customer is being attended to by a server. In our context, the exponential service time implies that the time between events (customer service completions) follows an exponential distribution. This distribution is defined by its rate parameter, often denoted as \(\mu\), which also represents the average number of events occurring in a time unit.

The key property of this exponential service time is its memorylessness, which means the probability of an event occurring is the same, no matter how much time has already passed. This becomes particularly helpful in simplifying queueing problems because it implies that past events do not influence future events' probabilities. For example, no matter how long a customer has already been waiting, their expected remaining waiting time is still \(1/\mu\).

Queueing Theory

Queueing theory is the mathematical study of waiting lines or queues. This theory enables us to make predictions about queue lengths and waiting times, which are essential for designing efficient systems. It encompasses a range of concepts, from simple scenarios like a single queue with a single server to complex networks of multiple service points and interdependent queues.

In any system where resources such as servers have to handle random arrivals, determining the optimal allocation can be quite challenging. Queueing theory uses probability and statistics to analyze these random processes. The calculations often involve determining the expected waiting time and queue length, which heavily depend on service time distributions. Our exercise is an example of a queueing problem where customers have to wait for two service points consecutively, with different rates for each service point.

Probability Density Function

A probability density function (PDF) is a mathematical function that describes the likelihood of a random variable taking on a particular value. In the realm of continuous variables, the PDF is used to specify the probability of the variable falling within a certain range of values, rather than taking on any one value. The PDF for an exponentially distributed random variable has the form \(p(x) = \mu e^{-\mu x}\), where \(x\) is the time variable and \(\mu\) is the rate parameter.

In our exercise, service times are exponentially distributed with rate \(\mu_i\) for server \(i\). The provided PDF represents how likely it is for a service to take a given amount of time. When we compute expected time or probabilities, we use integration over the PDF to account for all possible service times.

Integration by Parts

The technique of integration by parts is a method used to integrate products of functions. It is derived from the product rule for differentiation and is given by the formula \(\int u dv=uv-\int v du\), where \(u\) and \(v\) are functions of a variable (usually \(x\)).

When faced with the expected service time calculation, integrating by parts comes in handy because it allows the decomposition of a complex integral into simpler parts. Often, one component of the integrand is chosen as \(u\) to be differentiated, and the other as \(dv\) to be integrated. By selecting these components strategically, we can often simplify the calculation significantly. For instance, when we apply integration by parts to the expected remaining time for server 2 in our exercise, we break down the integral into easier parts, which ultimately leads to a simple solution for the expected time, which is \(1/\mu_2^2\).