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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 
little, please?

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.

Please explain!


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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