From: Jason Gardiner (gardiner@xxxxxxxxxx)
Date: Thu Apr 04 2002 - 13:54:06 GMT-3
Okay, this whole custom queueing thing gets me going on several levels.
First of all, queueing only kicks in once the buffers reach a saturation
point, AFAIK, which means that you're only talking about enforcing
distribution ratios at that point. Until saturation, the traffic
distribution can vary significantly, since queueing is not active.
And I don't understand why the average byte count makes a difference when
you're talking about percentages. Even if, say IPX, has a smaller average
packet size, the total ratio configured in the max byte count will make sure
that the bandwidth ratio stays consistent. The number of packets allowed
shouldn't make a significant difference.
Ths appears to be one of those areas where you can overanalyze yourself out
of some much needed time all too easily.
On Thursday 04 April 2002 11:09, glmorris48 wrote:
> Assuming your average byte counts for both IP and IPX were 1500, your
> philosophy would work fine. However, it is unlikely that this is going to
> be the case and you will find that your solution does not meet the the
> requirements of the problem.
>
> ----- Original Message -----
> From: "Todd Carswell" <acarswell@nc.rr.com>
> To: "Zhang, Stan" <stan.zhang@verizon.com>; "'Shaun Wakelen'"
> <Shaun.Wakelen@telindus.co.uk>; <steven.j.nelson@bt.com>;
> <ccielab@groupstudy.com>
> Sent: Thursday, April 04, 2002 7:52 AM
> Subject: Re: Custom Queueing (Again)
>
> > I don't use any formulas to calculate the byte-counts. I just think
> > about how many packets will fit in the queue when I start figuring out
> > the byte-counts. If I start monkeying around with a stinkin' formula on
> > the exam, I can easily get flustered and screw up my "plutification". To
> > keep it simple for myself, I think of it like this...
> >
> > The average packet is 1024 bytes, from what I've gathered thus far. I'd
> > like to fit more than 1 packet in the queue at any given time, so I make
>
> my
>
> > smallest queue at least 2048 bytes.
> >
> > Since I am a bit fond of nice round numbers, I would just bump that
> > number up to 3000 bytes to accomodate two packets of the maximum size of
> > 1500 bytes. From there, I'd figure out the rest of the queue sizes based
> > upon how much bandwidth I'd like to allocate.
> >
> > Example: 50% to IP, 25% IPX, 25% default
> >
> > queue-list 1 protocol ip 1
> > queue-list 1 protocol bridge 2
> > queue-list 1 protocol dlsw 3
> > queue-list 1 queue 1 byte-count 6000
> > queue-list 1 queue 2 byte-count 3000
> > queue-list 1 queue 3 byte-count 3000
> >
> >
> > Keep It Simple!!!
> >
> > Todd Carswell
> >
> >
> >
> >
> >
> > ----- Original Message -----
> > From: "Zhang, Stan" <stan.zhang@verizon.com>
> > To: "'Shaun Wakelen'" <Shaun.Wakelen@telindus.co.uk>;
> > <steven.j.nelson@bt.com>; <ccielab@groupstudy.com>
> > Sent: Thursday, April 04, 2002 9:14 AM
> > Subject: RE: Custom Queueing (Again)
> >
> > > Steve,
> > >
> > > There is another way of doing that. Instead of using division, you can
> >
> > use
> >
> > > multiplication to derive the common denominator. Either way works
> > > fine,
> >
> > and
> >
> > > both methods take about equal amount of time. I would have to concur
>
> with
>
> > > Shaun and say stick with what you know best, it'll always be there when
> >
> > you
> >
> > > need it. Best of the luck.
> > >
> > >
> > > Stan Zhang
> > >
> > >
> > >
> > > -----Original Message-----
> > > From: Shaun Wakelen [mailto:Shaun.Wakelen@telindus.co.uk]
> > > Sent: Thursday, April 04, 2002 7:45 AM
> > > To: steven.j.nelson@bt.com; ccielab@groupstudy.com
> > > Subject: RE: Custom Queueing (Again)
> > >
> > >
> > > Steve
> > >
> > > That is the formula I was shown and now use. I have seen other ways
>
> posted
>
> > > on here, which seem more complicated, but that may be down to the fact
> > > I have not done it that way. Stick with what you know best. If you know
> > > it gives the correct result, and you have to use it, then the points
> > > are in
> >
> > the
> >
> > > bag!
> > >
> > > Shaun
> > >
> > > -----Original Message-----
> > > From: steven.j.nelson@bt.com [mailto:steven.j.nelson@bt.com]
> > > Sent: 04 April 2002 13:26
> > > To: ccielab@groupstudy.com
> > > Subject: Custom Queueing (Again)
> > >
> > >
> > > All
> > >
> > > What is the simplest formula for working out bandwidth allocation when
> >
> > using
> >
> > > custom queueing.
> > >
> > > I have been using the following but I am looking for a quicker way.
> > >
> > > Step 1 For each queue, divide the percentage of bandwidth you want to
> > > allocate to the queue by the packet size, in bytes. For example, assume
> >
> > the
> >
> > > packet size for protocol A is 1086 bytes, protocol B is 291 bytes, and
> > > protocol C is 831 bytes. We want to allocate 20 percent for A, 60
>
> percent
>
> > > for B, and 20 percent for C. The ratios would be:
> > >
> > > 20/1086, 60/291, 20/831 or
> > >
> > > 0.01842, 0.20619, 0.02407
> > >
> > > Step 2 Normalize the numbers by dividing by the lowest number:
> > >
> > > 1, 11.2, 1.3
> > >
> > > The result is the ratio of the number of packets that must be sent so
>
> that
>
> > > the percentage of bandwidth that each protocol uses is approximately
> > > 20,
> >
> > 60,
> >
> > > and 20 percent.
> > >
> > > Step 3 A fraction in any of the ratio values means that an additional
> >
> > packet
> >
> > > will be sent. Round up the numbers to the next whole number to obtain
>
> the
>
> > > actual packet count.
> > >
> > > In this example, the actual ratio will be 1 packet, 12 packets, and 2
> > > packets.
> > >
> > > Step 4 Convert the packet number ratio into byte counts by multiplying
> >
> > each
> >
> > > packet count by the corresponding packet size.
> > > In this example, the number of packets sent is one 1086-byte packet,
> >
> > twelve
> >
> > > 291-byte packets, and two 831-byte packets or
> > > 1086, 3492, and 1662 bytes, respectively, from each queue. These are
> > > the byte counts you would specify in your custom
> > > queueing configuration.
> > >
> > > Step 5 To determine the bandwidth distribution this ratio represents,
> >
> > first
> >
> > > determine the total number of bytes sent after all three queues are
> > > serviced:
> > >
> > > (1 x 1086) + (12 x 291) +(2 x 831) = 1086 + 3492 + 1662 = 6240
> > >
> > > Step 6 Then determine the percentage of the total number of bytes sent
> >
> > from
> >
> > > each queue:
> > >
> > > 1086/6240, 3492/6240, 1662/6240 = 17.4, 56, and 26.6 percent
> > >
> > > As you can see, this is close to the desired ratio of 20/60/20.
> > >
> > >
> > >
> > > Steve Nelson
> > > Customer Engineer
> > > BT Ignite- National Solutions
> > > T: +44 (0)1422 338881 M: +44 (0)7811 944172
> > > e-mail: steven.j.nelson@bt.com
> > > pp HW A170, PO Box 200(HOM-NZ), London, N18 1ZF
> > >
> > > > British Telecommunications plc
> > > > Registered office: 81 Newgate Street London EC1A 7AJ
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