From: Ahmed Mamoor Amimi (mamoor@xxxxxxxx)
Date: Thu Apr 04 2002 - 13:16:34 GMT-3
how about if the packet size of ipx is 1400. then it will consume :
1400x3=4200
That gives 4200 - 3000 = 1200 extra.
U have to calculate through the MTU size of the protcol. so it would take as
minimum extra bytes as possilbe.
-Mamoor
----- Original Message -----
From: Todd Carswell <acarswell@nc.rr.com>
To: <steven.j.nelson@bt.com>; <stan.zhang@verizon.com>;
<Shaun.Wakelen@telindus.co.uk>; <ccielab@groupstudy.com>
Sent: Thursday, April 04, 2002 9:01 PM
Subject: Re: Custom Queueing (Again)
> OOPS, I screwed up my custom queue example! Here it is again...
>
> Example: 50% to IP, 25% IPX, 25% default
>
> queue-list 1 protocol ip 1
> queue-list 1 protocol ipx 2
> queue-list 1 default 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
>
> You see?! Nice and easy. I would not suggest trying to find any lowest
> common denominators or normalizing anything. Keep it simple.
>
> Todd
>
>
>
> ----- Original Message -----
> From: <steven.j.nelson@bt.com>
> To: <stan.zhang@verizon.com>; <Shaun.Wakelen@telindus.co.uk>;
> <ccielab@groupstudy.com>
> Sent: Thursday, April 04, 2002 9:24 AM
> Subject: RE: Custom Queueing (Again)
>
>
> > Thanks all,
> >
> > Better the devil you know then for me !!
> >
> > Steve
> >
> > -----Original Message-----
> > From: Zhang, Stan [mailto:stan.zhang@verizon.com]
> > Sent: 04 April 2002 15:14
> > To: 'Shaun Wakelen'; Nelson,SJ,Steven,IVNH25 C; ccielab@groupstudy.com
> > 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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