pipantasi4

Answered

2022-07-08

A measurement of the traffic generated by the packet source indicates that the average traffic is $\lambda $ [packets/s] and maximum traffic is $\sigma $ [packets/s]. The classic exponential distribution is not appropriate for modeling such a source of traffic, because the exponential distribution contains only one parameter (and two parameters were measured).

To model such a source of motion, you can use the shifted exponential distribution which is described by two parameters ($\gamma ,\delta >0$):

$f(\tau )=\{\begin{array}{cc}0& \tau <d\\ \gamma {e}^{-\gamma (\tau -d)}& \tau \ge d\end{array}$

Random variable $\tau $ is the time interval between successive packets.

1. Draw a distribution graph (1). What is the relationship between the maximum traffic and $\delta $, for the distribution (#)?

2. Designate a mean value of the distribution (#).

3. Based on measurements of traffic sources $(\lambda ,\sigma )$ select firstly $\delta $ parameter for distribution (#), then $\gamma $ parameter for distribution (#).

To model such a source of motion, you can use the shifted exponential distribution which is described by two parameters ($\gamma ,\delta >0$):

$f(\tau )=\{\begin{array}{cc}0& \tau <d\\ \gamma {e}^{-\gamma (\tau -d)}& \tau \ge d\end{array}$

Random variable $\tau $ is the time interval between successive packets.

1. Draw a distribution graph (1). What is the relationship between the maximum traffic and $\delta $, for the distribution (#)?

2. Designate a mean value of the distribution (#).

3. Based on measurements of traffic sources $(\lambda ,\sigma )$ select firstly $\delta $ parameter for distribution (#), then $\gamma $ parameter for distribution (#).

Answer & Explanation

Alec Blake

Expert

2022-07-09Added 11 answers

Part 1 is simple. If $\delta $ is the minimum interarrival time for consecutive packets, where time is measured in seconds, then at most $1/\delta $ packets can arrive in one second. For instance, if $\delta =0.01$ seconds, meaning that we are assured that the time between packets is at least 0.01 seconds, then the most packets we can observe in one second is 1/0.01=100 packets, and this is $\sigma $. So

$\sigma \delta =1.$

For Part 3, use what you know from Parts 1 and 2 to establish a relationship between $\lambda $ and $\gamma $. The average traffic intensity $\lambda $ is simply the average you found in Part 2. Solve for $\gamma $ in terms of $\lambda $ and $\sigma $.

$\sigma \delta =1.$

For Part 3, use what you know from Parts 1 and 2 to establish a relationship between $\lambda $ and $\gamma $. The average traffic intensity $\lambda $ is simply the average you found in Part 2. Solve for $\gamma $ in terms of $\lambda $ and $\sigma $.

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