91Ó°ÊÓ

The exponential distribution is a continuous probability distribution that is often associated with the time between events in a Poisson process. When events follow a Poisson process, the times at which these events occur are random, and the duration between each event follows an exponential distribution.
In this particular exercise, the network log-ons follow a Poisson process with an average rate of three counts per minute. This means the time between each log-on is exponentially distributed. An important characteristic of the exponential distribution is that it is memoryless, which means the probability of an event occurring in the next instant is the same regardless of how much time has already passed.
The probability density function (PDF) of an exponential distribution is given by:

• \( f(x; \lambda) = \lambda e^{-\lambda x} \) for \( x \ge 0 \)

Here, \( \lambda \) is the rate parameter, which is equal to the mean number of events per time unit.

The mean time between counts in a Poisson process is a concept that can be easily derived from the exponential distribution model. Since we are dealing with a Poisson process at the rate \( \lambda = 3 \) counts per minute, the mean time between each log-on or count is calculated as the reciprocal of this rate.
Mathematically, the mean time \( \text{MTBC} \) between events is:

• \( \text{MTBC} = \frac{1}{\lambda} \)
• For our case, it equals \( \frac{1}{3} \) minutes or approximately 20 seconds.

Understanding MTBC is crucial for network management and predicting system loads, as it indicates the average time interval expected between consecutive events.

In statistical terms, the standard deviation (SD) provides a measure of the spread around the mean, indicating how varied or dispersed the values are from the average. When dealing with time gaps in a Poisson process, which are modeled by an exponential distribution, the standard deviation is particularly simple.
For an exponential distribution, the standard deviation is equal to the mean time between counts and is also calculated as the reciprocal of the rate parameter \( \lambda \).
In this exercise, as with the mean time between counts, the standard deviation is:

• \( \text{SD} = \frac{1}{\lambda} \)
• This value is \( \frac{1}{3} \) minutes, equating to about 20 seconds.

This relationship simplifies analysis and provides quick insights into the variability of the timing between log-ons in our network model.

The cumulative distribution function (CDF) of an exponential distribution plays a vital role in calculating the probability that an event occurs within a specific timeframe. For this exercise, it helps to determine the time by which there is a particular probability of at least one log-on occurring in the network.
The CDF for an exponential distribution is expressed as:

• \( F(x) = 1 - e^{-\lambda x} \)

Here, \( F(x) \) gives the probability that the time until the next event is less than or equal to \( x \) minutes.
To find when the probability of at least one log-on is 0.95, we solve:

• \( 1 - e^{-3x} = 0.95 \)
• Simplifying gives \( x = -\frac{\ln(0.05)}{3} \)
• By calculation, \( x \approx 1.00 \) minute.

Thus, it is very likely that at least one count will occur within the next minute of observation.