From: Simon Baxter (Simon.Baxter@xxxxxxxxxxxxxx)
Date: Tue Aug 22 2000 - 05:20:13 GMT-3
I had it wrong. The byte count is the only thing you really need. The
limit is (sort of) incoming to the queue - kind of loke (analogy) how many
people can be in the queue at once. The byte count is how much is taken
from the queue in each round of service.
ie queue-list 1 queue 1 byte-count 1086 limit 5
allow 5 entries to be queued up at once, and allow 1086 bytes to be taken at
each service of the queue.
The limit number should be something that's big enough to store all the
entries you need, but doens't effect the router. Basically, the defaults
are fine unless you're experiencing buffer shortages due to too many
messages in the queue.
hope this helps!
-----Original Message-----
From: STC Chen Duan [mailto:STCCDU@sbell.com.cn]
Sent: Tuesday, August 22, 2000 5:46 PM
To: Simon Baxter
Subject: RE: custom queueing and byte count
So how do you get the limit number?
> ----------
> From: Simon Baxter[SMTP:Simon.Baxter@au.logical.com]
> Reply To: Simon Baxter
> Sent: Tuesday, August 22, 2000 9:07 AM
> To: Joe Harris; ccielab@groupstudy.com
> Subject: RE: custom queueing and byte count
>
> So would you do the following :
>
> queue-list 1 protocol appletalk 1
> queue-list 1 protocol ip 2
> queue-list 1 protocol ipx 3
> queue-list 1 queue 1 byte-count 1086 limit 5
> queue-list 1 queue 2 byte-count 3260 limit 56
> queue-list 1 queue 3 byte-count 1080 limit 6
>
>
> would you need to both the queue depth and byte count?
>
> -----Original Message-----
> From: Joe Harris [mailto:JoeH@globaldatasys.com]
> Sent: Tuesday, August 22, 2000 7:21 AM
> To: ccielab@groupstudy.com
> Subject: RE: custom queueing and byte count
>
>
> John,
>
> I calculated the formula and I venture to say that you would get it
> correct.
> (Not that I am the proctor though). There is actually a simple (well maybe
> not so simple) formula to determine if you have the appropriate byte-count
> specified and whether or not you are allocating the appropriate percentage
> of the bandwidth. To determine the byte counts follow the steps below.
>
> 1. Produce a ratio of all frame sizes, dividing into the largest frame
> size. For example, assume that the frame size for protocol A is 1086
> bytes,
> for protocol B is 291 bytes, and for protocol C is 831 bytes. The ratios
> would be:
>
> 1086/1086 1086/291 1086/831
>
> 2. Now multiply the results by the percentage of bandwidth you want each
> protocol to have. For example, I will allocate 20 percent for A, 60
> percent
> for B, and 20 percent for C. This will give you:
>
> 1086/1086(0.2) 1086/291(0.6) 1086/291(0.2)
> or
> .2 2.239 0.261
>
> 3. Normalize the ratio by dividing each value by the smallest value, in
> our
> case:
>
> .2/.2 2.239/.2 0.261/.2
> or
> 1 11.2 1.3
>
> This is the ratio of the number of frames that must be sent out of each
> queue so that the percentage of bandwidth that each protocol uses is
> approximately in the ratio of 20, 60, and 20 percent.
>
> 4. Note that any fraction in any of the ratio values means that an
> additional frame will be sent. In the example above, the number of frames
> sent would be one 1086 byte frame, twelve 291 byte frames, and two 831
> byte
> frames or 1086, 3492, and 1662 bytes, respectively from each queue. These
> are the byte counts that you would specify in your custom queuing config.
> To determine the bandwidth distribution this 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
>
> 5. Then determine the percentage of the 6240 bytes that was 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 (but no
> cigar
> yet). The bandwidth allocation can even to tailored further from this
> point
> to represent a true 20:60:20. By multiplying the original ratio above of
> in
> step 3, 1:11.2:1.3 by an integer, and trying to get as close as possible
> to
> the real 20:60:20. For example we could multiply the ratio by 2, (1 x 2,
> 11.2 x 2, and 1.3 x 2) this would give us 2, 22.4, and 2.6. Now you would
> send two 1086 byte frames, twenty-three 291 byte frames, and three 831
> byte
> frames or 2172+6693+2493, for a total of 11358 bytes. The resulting ratio
> would be 19:59:22 percent. Which of course is much closer to the original
> 20:60:20.
>
>
> Now in your case you are using the default 1500 byte frame size, using
> four
> queues and attempting to allocate 25 percent of the bandwidth to each
> queue
> correct?
>
> Applying the formula above you get (you only need to run the formula on
> one
> queue because the others will be identical):
>
> Step 1: 1500/1500 = 1
> Step 2: 1500/1500(.25) = 0.25
> Step 3: 0.25/0.25 = 1
> Step 4: The number of frames sent would be one 1500 byte frame from
> each
> queue. Now determine the bandwidth distribution this represents.
> 1500+1500+1500+1500 = 6000
> Step 5: Now determine the percentage of the 6000 bytes that was sent
> from each queue:
> 1500/6000 1500/6000 1500/6000 1500/6000 = 25%,
> 25%,
> 25%, and 25%
>
> I would have to say you are correct. Also Marc is correct (run the
> formula). This one is not that bad because they asked you for 4 queues
> with
> 25 percent of the bandwidth so just about any number you use will work as
> long as they all are the same. Hope this helps.
>
> - Joe Harris
>
>
> -----Original Message-----
> From: John Conzone [mailto:jkconzone@home.com]
> Sent: Saturday, August 19, 2000 3:33 PM
> To: ccielab
> Subject: custom queueing and byte count
>
>
> I've been working on bootcamp lab three, and have a question on custom
> queueing.
> The requirement is to define 4 queues, each with %25 of bandwidth. The
> queue is to be placed on S0. My config matches Marcs except Marc specifies
> the byte count on each queue at 1000 bytes. I didn't specify the byte
> count
> on mine, leaving them at 1500. The other interface on the router is a
> token
> ring interface.
> Token ring is 4000 bytes or 8000 bytes and ethernet is 1500 bytes. Is
> he
> doing this to adjust for token ring packet size? If so, how and why?
> Fragments?
> If not, does anyone know why he is doing this?
> Also, if I got this on the lab and configured four equal length queues
> with the default size, anyone venture to guess if it would count?
> Thanks!
>
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