arrival rate

[frien...@googlemail.com]

I just want to clarify what a arrival rate is as i am getting some
absurd result in my simulation runs.

for example i have
an arrival of first packet at 30sec
an arrival of sec packet at 50sec
an arrival of third packet at 58sec
an arrival of fourth packet at 60sec

what i did is : arrival rate of packets is 1 /(mean interarrival
time)
1/ ( ( (50-30) + (58 -30) + (60 - 58) ) / 3)

3 bacause number of intervals are three.
do i need to write 3 or the number of packets i.e., 4

if i wrong in the formula or anywhere please correct me...

thanks for any help

[John D Salt]

"frien...@googlemail.com" <frien...@googlemail.com> wrote in
news:21318174-a8f7-487d...@8g2000hse.googlegroups.com:

Lambda equals one over mu, mu equals one over lambda.

In the example above you have 4 arrivals in 60 sec.

The arrival rate (lambda) is 4/60 = one-fifteenth of an arrival per unit
time (the unit in this case being the second).

The expected inter-arrival time (mu) is 60/4 = 15 time units, or 15
seconds.

Note that some simulation packages/languages have a negexp function that
takes mu as the argument, others have lambda.

All the best,

John.

[xyz]

On May 27, 10:38 pm, John D Salt
<jdsalt_AT_gotadsl.co...@giganews.com> wrote:

Thanks for the reply
i think the arrival rate in the case is 4/ (60 - 30)
time the last arrival - time the first arrival...
thanks bye

[John D Salt]

xyz <lavanya...@gmail.com> wrote in
news:2b171d91-8180-4ff6...@2g2000hsn.googlegroups.com:

No. Lambda really does equal one over mu, not something else.

The arrival rate for the first half-minute is nil.
The arrival rate for the second half-minute is 8 per minute.
The arrival rate for the whole minute is 4 per minute.

You start timing when the timer starts, not when the first arrival
happens.

Some simulationists, when generating arrivals in a Poisson arrival
stream, of forcing an arrival at time zero. This is generally a mistake.

All the best,

John.