From: Ahmed Mamoor Amimi (mamoor@xxxxxxxx)
Date: Thu Apr 04 2002 - 14:43:21 GMT-3
What about if ur protocols have different MTUs that is ipx 1400 , ip 1300
and dlsw 1000. then i dont think ur formula will work.
-Mamoor
----- Original Message -----
From: Ferguson, Francis <Francis_Ferguson@icgcomm.com>
To: <ccielab@groupstudy.com>
Sent: Thursday, April 04, 2002 9:20 PM
Subject: RE: Custom Queueing (Again)
> here is a simple easy way to calculate bandwidth for custom queues
>
> Example create queues with 30% 20% 12% 18% 10% and 10%
>
> pick a byte number (any number),100 for instance and multiple each value
by
> the percentage required
> what you get 3000,2000,1200,1800,1000 and 1000 for the
>
> if you would like a particular queue size for one queue i.e.(30% = 4500)
> then the base byte number is (4500/30 = 150)
> now your queues look like this
>
> 4500 3000 1800 2700 1500 1500
>
>
>
> -----Original Message-----
> From: Todd Carswell [mailto:acarswell@nc.rr.com]
> Sent: Thursday, April 04, 2002 8:53 AM
> To: Zhang, Stan; 'Shaun Wakelen'; steven.j.nelson@bt.com;
> ccielab@groupstudy.com
> 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
> > > Registered in England no. 1800000.
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