Question

There are two servers available to process \(n\) jobs. Initially, each server begins work on a job. Whenever a server completes work on a job, that job leaves the system and the server begins processing a new job (provided there are still jobs waiting to be processed). Let \(T\) denote the time until all jobs have been processed. If the time that it takes server \(i\) to process a job is exponentially distributed with rate \(\mu_{i}, i=1,2\), find \(E[T]\) and \(\operatorname{Var}(T)\)

Short answer

In summary, to find the expected time \(E[T]\) and variance \(\operatorname{Var}(T)\) for processing \(n\) jobs by two servers with exponentially distributed processing time, follow these steps: 1. Determine the distribution of the minimum processing time \(Z\) considering the processing time for each server is exponentially distributed with rates \(\mu_1\) and \(\mu_2\). 2. Find the cumulative distribution function (CDF) of \(Z\) and express terms using CDFs of \(X\) and \(Y\). 3. Find the probability density function (PDF) of \(Z\) and derive \(E[T]\) by summing up the expected minimum time spent on each job: \(E[T] = n \times \int_0^\infty z f_Z(z) dz\). 4. Compute the second moment, \(E[Z^2]\), and the variance, \(\operatorname{Var}(Z) = E[Z^2] - (E[Z])^2\), for the processing time of a single job. Finally, find the variance for \(n\) jobs: \(\operatorname{Var}(T) = n \times \operatorname{Var}(Z)\).

Step-by-step solution

Determine the distribution of the minimum processing time

To analyze this problem, let us first consider the time it takes for the servers to process a job collectively. Suppose server 1 takes time X and server 2 takes time Y to process a job. We are interested in the minimum time, Z = min(X, Y), because the first job to be completed is replaced by the next job in the queue. Since the processing time for each server is exponentially distributed with rates \(\mu_1\) and \(\mu_2\), their probability density functions (PDF) are given by: \(f_X(x) = \mu_1 e^{-\mu_1 x}\) (1) \(f_Y(y) = \mu_2 e^{-\mu_2 y}\) (2) We now need to find the PDF of Z. To do this, we will first find the CDF of Z: \(F_Z(z) = \mathbb{P}(Z \leq z)\) (3) This is the probability that the minimum of X and Y, the processing times of the two servers, is less than or equal to z. For this to happen, either X or Y has to be less than or equal to z. The probability of this happening is: \(F_Z(z) = 1 - [ \mathbb{P}(X > z) \times \mathbb{P}(Y > z)]\) (4) Next, we will express terms inside the bracket in terms of the CDFs of X and Y.

Express terms using CDFs of X and Y

Using the complementary probability rule, we can represent \(\mathbb{P}(X > z)\) and \(\mathbb{P}(Y > z)\) as \(1 - F_X(z)\) and \(1 - F_Y(z)\), respectively. From equations (1) and (2), we have: \(F_X(z) = 1 - e^{-\mu_1 z}\) \(F_Y(z) = 1 - e^{-\mu_2 z}\) Now, substitute these expressions into equation (4), we get: \(F_Z(z) = 1 - [(1 - e^{-\mu_1 z})(1 - e^{-\mu_2 z})]\). Simplify the expression: \(F_Z(z) = 1 - (1 - e^{-\mu_1 z} - e^{-\mu_2 z} + e^{-(\mu_1 + \mu_2) z})\). So, we get: \(F_Z(z) = e^{-\mu_1 z} + e^{-\mu_2 z} - e^{-(\mu_1 + \mu_2) z}\)

Find the PDF of Z and derive E[T]

Next, we will find the PDF of Z, f_Z(z), by differentiating the CDF F_Z(z) with respect to z: \(f_Z(z) = \frac{\partial}{\partial z} F_Z(z) = \mu_1 e^{-\mu_1 z} + \mu_2 e^{-\mu_2 z} - (\mu_1 + \mu_2) e^{-(\mu_1 + \mu_2) z}\). To find the expected time E[T] to process n jobs, we will sum up the expected minimum time spent on each job: \(E[T] = n \times E[Z]\). Using the PDF of Z, we can find the expected minimum time spent on each job as: \(E[Z] = \int_0^\infty z f_Z(z) dz\). Finally, substitute this expression for E[Z] into the equation for E[T]: \(E[T] = n \times \int_0^\infty z (\mu_1 e^{-\mu_1 z} + \mu_2 e^{-\mu_2 z} - (\mu_1 + \mu_2) e^{-(\mu_1 + \mu_2) z}) dz\). To get E[T], simply compute the integral and multiply by n.

Compute Var(T)

To find the variance, \(\operatorname{Var}(T)\), we need to compute the second moment, \(E[Z^2]\), of the distribution of Z: \(E[Z^2] = \int_0^\infty z^2 f_Z(z) dz\). Then, we have: \(\operatorname{Var}(Z) = E[Z^2] - (E[Z])^2\). Since the jobs are processed independently, the variance for n jobs is the sum of the variances for each individual job: \(\operatorname{Var}(T) = n \times \operatorname{Var}(Z)\).

Key concepts

Exponential Distribution

The exponential distribution is a continuous probability distribution that is often used to model the time between independent events that happen at a constant average rate. For example, it can describe the time a server takes to process a job in an operations research scenario.

Characterized by a single parameter, usually denoted as \( \lambda \) or \( \mu \) (the rate of occurrences), it has the unique property of being memoryless, meaning the probability of an event occurring in the next unit of time is the same, regardless of how much time has already elapsed.

The probability density function (PDF) of the exponential distribution is defined as:
\[ f(x) = \lambda e^{-\lambda x} \]
for x \( \geq 0 \), and 0 elsewhere. The rate, \( \lambda \), is the average number of events per unit time, and \( e \) is the base of the natural logarithm. This distribution plays a key role in modeling processing times in operations research.

Expected Value (E[T]) of Processing Time

The expected value, or mean, of a random variable is a crucial concept in probability and statistics, representing the long-term average outcome of a random event if it were repeated many times. In the context of processing times, if \( T \) represents the time taken by a server to process a job, the expected value \( E[T] \) denotes the average processing time for a job.

For example, when the processing time \( T \) follows an exponential distribution with rate \( \mu \), the expected value is given by the reciprocal of the rate:
\[ E[T] = \frac{1}{\mu} \]
This value is intuitive: if a server processes jobs at an average rate of \( \mu \) jobs per unit time, then on average, it takes \( 1/\mu \) units of time to process one job. In operations research, calculating the expected processing time is essential for predicting system performance and planning.

Variance of Processing Time (Var(T))

Variance is a statistical measure of dispersion that quantifies the spread of a set of data points or the variability of a random variable around its mean. In an operations research context, understanding the variance in processing time \( \operatorname{Var}(T) \) helps to assess the consistency and reliability of system performance.

For a processing time \( T \) that is exponentially distributed with rate \( \mu \), the variance is given by the square of the reciprocal of the rate:
\[ \operatorname{Var}(T) = \frac{1}{\mu^2} \]
This formula highlights that the higher the rate \( \mu \), the lower the variance, indicating more consistent processing times. Conversely, a lower rate suggests more variability in processing times, which could lead to less predictable system performance.

Probability Density Function (PDF)

The probability density function (PDF) is a fundamental concept in probability theory, describing the relative likelihood for a random variable to take on a particular value. In essence, the PDF for a continuous random variable provides a curve where the probability of the variable falling within a particular range can be found by the area under the curve between two points.

The PDF of the exponential distribution can be expressed as:
\[ f(x) = \mu e^{-\mu x} \]
This function quantifies the probability of a server finishing a job at any given time \( x \) and is characterized by the rate parameter \( \mu \). In operations research, the PDF can be used to derive other important measures, such as expected values and variances, and to gain insights into process efficiency and timelines.

Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) is another vital concept in the study of probability distributions. It describes the probability that a random variable \( X \) will have a value less than or equal to \( x \), capturing the cumulative effect of probabilities up to a certain point.

For the exponential distribution, the CDF is given by:
\[ F(x) = 1 - e^{-\mu x} \]
for \( x \geq 0 \), where \( \mu \) is the rate parameter. The CDF climbs steeply at first and levels off as it approaches 1, reflecting the fact that the likelihood of the server completing the processing of a job increases as time passes. In operations research, analyzing the CDF is particularly useful when scheduling jobs and predicting system throughput, as it directly relates to wait times and service levels.