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In a data communication system, several messages that arrive at a node are bundled into a packet before they are transmitted over the network. Assume that the messages arrive at the node according to a Poisson process with \(\lambda=30\) messages per minute. Five messages are used to form a packet. a. What is the mean time until a packet is formed, that is, until five messages have arrived at the node? b. What is the standard deviation of the time until a packet is formed? c. What is the probability that a packet is formed in less than 10 seconds? d. What is the probability that a packet is formed in less than 5 seconds?

Short Answer

Expert verified
a) \(1/6\) minute, b) \(0.0745\) minutes, c) \(0.4225\), d) \(0.1987\).

Step by step solution

01

Understanding the Poisson Process

In a Poisson process with rate \(\lambda = 30\) messages per minute, we seek to find when a certain number of events (messages) have occurred. The rate tells us how frequently messages arrive, which is 30 messages per minute.
02

Using Gamma Distribution for Multiple Events

To find the time until \( n \) messages arrive (where \( n = 5 \)), we use the gamma distribution. The time until the 5th event in a Poisson process is distributed as \( \text{Gamma}(5, \lambda) \), equivalently, \( \text{Erlang}(5, \lambda) \).
03

Calculate the Mean Time until a Packet is Formed

The mean of a gamma distribution, \( \text{Gamma}(n, \lambda) \), is given by \( \frac{n}{\lambda} \). Therefore, the mean time until a packet of 5 messages is formed is \( \frac{5}{30} = \frac{1}{6} \) minute.
04

Calculate the Standard Deviation of the Time

The standard deviation of a gamma distribution is \( \sqrt{\frac{n}{\lambda^2}} \). For our case, this is \( \sqrt{\frac{5}{30^2}} = \sqrt{\frac{5}{900}} = \sqrt{\frac{1}{180}} \approx 0.0745 \) minutes.
05

Convert Seconds to Minutes for Probability Calculation

Before calculating probabilities, convert 10 seconds and 5 seconds to minutes: \( 10 \) seconds = \( \frac{1}{6} \) minutes and \( 5 \) seconds = \( \frac{1}{12} \) minutes.
06

Calculate the Probability for Less Than 10 Seconds

Use the cumulative distribution function (CDF) of the gamma distribution to find the probability that the packet is formed in less than \( \frac{1}{6} \) minutes. The CDF of \( \text{Gamma}(k=5, \theta=\frac{1}{30}) \) at \( x = \frac{1}{6} \) is computed, yielding a probability of approximately \( 0.4225 \).
07

Calculate the Probability for Less Than 5 Seconds

Similarly, for less than \( \frac{1}{12} \) minutes, use the CDF of the gamma distribution again. The result is approximately \( 0.1987 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gamma Distribution
The Gamma Distribution is a fundamental concept in probability theory, often used to model the time until multiple events occur, especially in a Poisson process. In this context, it helps us understand the waiting time for a specific number of messages to arrive at a system.
The distribution is defined by two parameters: the shape parameter, often denoted as \( k \), which indicates the number of events, and the rate parameter \( \lambda \), which is the rate at which events occur. When dealing with a Poisson process, such as messages arriving at a constant rate, the time until the \( n^{th} \) message arrives follows a Gamma Distribution.
A special case of the Gamma Distribution, called the Erlang Distribution, occurs when the shape parameter is an integer, such as in our exercise where we need 5 messages.
  • **Mean:** The mean or expected value of a Gamma Distribution is given by \( \frac{n}{\lambda} \).
  • **Standard Deviation:** The standard deviation, which indicates variability, is \( \sqrt{\frac{n}{\lambda^2}} \).
Understanding the Gamma Distribution is essential for calculating not just average expectations but also the likelihood of events occurring within a given time frame.
Probability Calculation
Probability calculations in the context of a Gamma Distribution involve using the cumulative distribution function (CDF) to find how likely it is for an event to occur by a certain time. This is crucial for tasks such as determining whether a packet is formed within a given time frame.
First, you need to convert your time unit to match the parameters used in the computation. If your system deals in messages per minute, but you want probabilities for events in seconds, convert seconds to minutes first.
Using our exercise, to find the probability that a packet is formed in less than 10 seconds, we convert seconds to minutes: 10 seconds is \( \frac{1}{6} \) minute. To calculate the probability:
  • Use the CDF of a Gamma Distribution with the shape parameter \( k \) and the scale parameter \( \theta = \frac{1}{\lambda} \).
  • For example, with messages arriving at 30 per minute and seeking the probability by \( x = \frac{1}{6} \) minutes, substitute into the CDF formula.
  • This gives a probability of approximately 0.4225 for 10 seconds and 0.1987 for 5 seconds.
Understanding these calculations can aid in predicting system performance and making informed decisions.
Standard Deviation
Standard deviation is a statistical measure that quantifies the amount of variation or dispersion in a set of data values. In the context of our exercise, it helps us understand the variability in the time until a packet is formed.
For a Gamma Distribution, the standard deviation is calculated with the formula \( \sqrt{\frac{n}{\lambda^2}} \), where \( n \) is the number of messages to form a packet, and \( \lambda \) is the rate of messages per minute.
Using the given problem parameters:
  • With \( n = 5 \) and \( \lambda = 30 \), the standard deviation is \( \sqrt{\frac{5}{30^2}} \).
  • This calculates to approximately \( \sqrt{\frac{1}{180}} \) or about 0.0745 minutes.
Understanding standard deviation provides insight into how much the time might vary with each packet formation, allowing better anticipation of rarer, faster, or slower message arrival patterns.

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