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Queuing delay

Unanswered Question
Jan 13th, 2014
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Hi


I'm trying to calculate the average dalay that an interface take to process the queue.

Is there any formula maybe based on bandwidth capacity of the interface?


Thanks

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Joseph W. Doherty Mon, 01/13/2014 - 08:57
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Posting


Average queuing delay would, as you note, depend on interface bandwidth, but it also depends on average amount of bits that have been queued.

aformisani Mon, 01/13/2014 - 11:56
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Hi Joseph thanks for you prompt reply.

Do you know if there is some document that can explain how to calculate an estimation time.

Supposing that I have 100Mb/s interface and 23000 total packets per seconds (in and out) and each packet have a size of 274 byte (2192 bit) what can be the delay.


Thanks

Joseph W. Doherty Mon, 01/13/2014 - 13:09
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In no event shall Author be liable for any damages whatsoever (including, without limitation, damages for loss of use, data or profit) arising out of the use or inability to use the posting's information even if Author has been advised of the possibility of such damage.


Posting


You would need to also know your arrival distribution.


But assuming a more or less random arrival rate, and (from your bandwidth consumption numbers) about 50%, I recall (assuming we can use a M/M/1 model) you would average less than 2 (or 1?) packet in the queue.  Assuming 1 packet packet in queue, 2K bits / 100 Mbps, average queuing delay of 2 ms.  (NB: there are some [free] queuing calculators that can be, but couldn't find a good one just now.)

aformisani Mon, 01/13/2014 - 23:56
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I joseph


I found the following formula:

1/(μ-λ)

where μ is the service rate (e.g., the number of packets per second the interface can sustain) and λ is the arrival rate (the average rate at which packets are arriving to be serviced).


The service rate μ = T/P (where T = the transmission rate of the facility and P = the average packet size).


Suppose to have 100Mbit/s connection (100.000.000 bit/s). From the captured traffic I will take the maximum packet size which is 274 bytes (2192 bit). So the service rate for a 100Mbit/s interface receiving those packets size is:

μ = 100.000.000/2192 = 45620,4 packets/sec

Right?


Considering that from the captured traffic the arrival rate on the interface is 23000 total packets/sec (in average) the delay in the interface directly connected to the server will be:


1/(45620-23000) = 0,000042 sec = 0,042 ms


I think I miss something or the formula is not correct.

What do you think

Thanks

Andrea

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