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### Server Traffic

```
Date: 01/02/99 at 06:54:43
From: Yaeli
Subject: Poisson Statistics

A server is open for a year. N transactions occur each year. These are
randomly (evenly) spread throughout the year and the working hours of
each day.

Assume all transactions are during working hours (8.5 hours = 510
minutes) per day. Assume approximately 200 working days per year, which
is 510 * 200 = 102,000 minutes available per year.

Assume each transaction takes 2.5 minutes.

Now using the Poisson distribution, I should be able to work out the
probability that at any given moment there will be M simultaneous
transactions. So what's the probability of a particular minute having
M = 0, M = 1, M = 2, M = 3, ....

What is the formula?
```

```
Date: 01/02/99 at 07:51:39
From: Doctor Anthony
Subject: Re: Poisson Statistics

The mean number of transactions per minute is N/102000.

So  P(0) =  e^(-N/102000) = chance of no transactions in any minute

P(1) =  N/102000 e^(-N/102000) = chance of 1 transaction

P(2) = (N/102000)^2/2! e^(-N/102000) = chance of 2 transactions

P(3) = (N/102000)^3/3! e^(-N/102000) = chance of 3 transactions

and so on.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 01/04/99 at 18:50:56
From: Yaeli
Subject: Poisson distribution

Hi Doctor Anthony,

First, thanks for your quick reply. Could you elaborate on this a

The time per transaction is 2.5 minutes. Isn't that somehow relevant?
If I each transaction took 102,000 minutes, the mean number would be a
lot more.

```

```
Date: 01/05/99 at 11:32:54
From: Doctor Anthony
Subject: Re: Poisson distribution

The 2.5 minutes are irrelevant to the number of customers arriving per
minute. They become relevant to things like 'waiting times' if you
limit the number of available servers and you are then into 'queuing
theory' with the number of arrivals in any interval exceeding the
number of departures after the serving time of 2.5 minutes. The
question then is about the expected waiting times or queue lengths, but
realistically you would be better calculating these for a given number
of servers and using an actual Poisson mean number of arrivals per unit
time, rather than a general number in terms of N. But I emphasize again
the 'arrivals' are entirely independent of the serving time.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 01/05/99 at 18:25:08
From: Yaeli
Subject: Poisson distribution

Again, thanks for the reply. The thing is, I'm not interested in the
number of arrivals. I'm looking at the number of transactions in
progress at any moment of time.

For example, I want to know what percent of the time more than 4 more
transactions will be in progress at any moment in time, given N at
different values.

This allows me to calculate the number of phone lines needed for a
dial-up system (without giving those dialing up lots of "engaged
tones"), assuming uniform traffic throughout the working year.

Once again thank you very much.
```

```
Date: 01/06/99 at 05:37:19
From: Doctor Anthony
Subject: Re: Poisson distribution

Without going into the rather difficult subject of queuing theory you
could use the following ideas.

Take 2.5 as the unit of time. If m = mean number of calls per minute as
calculated by  N/102000, then the mean number of calls in 2.5 minutes
is 2.5m. We then require:

P(0) =     e^(-2.5m)
P(1) =  2.5m x e^(-2.5m)
P(2) = (2.5m)^2/2! x e^(-2.5m)
P(3) = (2.5m)^3/3! x e^(-2.5m)

and so on.

The probability that 1, 2, 3, or 4 people are calling in any 2.5-minute
interval is given by P(1) + P(2) + P(3) + P(4), and if this probability
is say 20% you might wish to install more lines. With more lines you
divide N by the number of lines to calculate the traffic for any one
line.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Probability

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