You could get into some heavy-duty math here! &;-)
Oftentimes, packets aren't queued at all. A fast router that is
transmitting to a fast link may have no need to put packets in a queue, as
you mentioned below. The queuing delay is then zero, of course. With the
"show interface" command you can see how many frames are currently in the
queue.
Queuing becomes necessary when the outbound link is heavily used. So, one
way to think about this is to base your calculation on the utilization of
the outbound link. Basic queuing theory says that the number of packets in
a queue is utilization/(1-utilization).
Here's an example from my book, "Top-Down Network Design."
A router is supporting five applications, each offering packets at a rate
of 10 packets per second. The average length of the packets is 1024 bits.
The router needs to transmit this data over a 56-Kbps WAN circuit.
Load = 5 x 10 x 1024 = 51,200 bps
Utilization = 51,200/56,000 = 91.4 percent
Average number of packets in queue = (0.914)/(1-0.914) = 10.63 packets
Once you know the average depth of the queue (how many packets are in it),
you could also calculate the time a packet is in the queue. You would have
to continue to make some assumptions about packet size, of course.
With fancy queuing such as PQ, CQ, WFQ, etc, you could make sure that
"important" packets jump to the head of that queue. In this example, that
would be a good idea, since having 10 packets in the queue is a lot. A
better idea might be to upgrade the 56K link, but that wrecks the example. &;-)
So, that's the basics that hopefully give you some more food for thought.
Priscilla
At 04:03 PM 10/2/00, Laurent Lange wrote:
>Hi all,
>
>when calculating the delay for a certain packet size between 2 sites,
>I usually use the following formula:
>
>delay = access transmission+ backbone transmission+ access transmission
>
>where:
>backbone transmission = data provided by the network service provider (for
>example "on my frame relay network, a packet of 128 bytes is transmitted
>between two pops in maximum 100ms)
>
>Access transmission = serialization delay + transmission delay + queuing
>delay.
>
>serialization delay = packet size*8/access line speed
>(for example 128 bytes on a 128 kbps line = 8 ms)
>
>transmsission delay is the delay to transport the bits at copper wire
>speed (usually max 2ms depending on length of line)
>
>Queuing delay (in the router) = ??? that's my question.
>
>If there is no traffic queuing delay = 0 ?
>
>If line loaded at 20, 50, 80 % what is the queuing delay?
>What is the theoretical formula for queuing delay?
>What is the influence of QoS (CB-WFQ, PQ, CQ, Fragmentation)?
>
>I have never found quantitative information about this on cisco web site
>despite several attempts, only qualitative info describing how
>prioritization works.
>I would be grateful if anyone could provide me with this info, based on
>theory, statistical or empirical sources.
>
>thanks in advance
>Laurent
>
>
>
>_________________________________________________________________________
>Get Your Private, Free E-mail from MSN Hotmail at http://www.hotmail.com.
>
>Share information about yourself, create your own public profile at
>http://profiles.msn.com.
>
>**NOTE: New CCNA/CCDA List has been formed. For more information go to
>http://www.groupstudy.com/list/Associates.html
>_________________________________
>UPDATED Posting Guidelines: http://www.groupstudy.com/list/guide.html
>FAQ, list archives, and subscription info: http://www.groupstudy.com
>Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
________________________
Priscilla Oppenheimer
http://www.priscilla.com
**NOTE: New CCNA/CCDA List has been formed. For more information go to
http://www.groupstudy.com/list/Associates.html
_________________________________
UPDATED Posting Guidelines: http://www.groupstudy.com/list/guide.html
FAQ, list archives, and subscription info: http://www.groupstudy.com
Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
