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/
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