From: Sonja Koestler (skoestle@xxxxxxxxx)
Date: Wed Apr 17 2002 - 07:07:33 GMT-3
gerd,
a good explanation has been postetd lately to this list - look at this:
rgds, sonja
-----snip----------------------
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.
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com]On Behalf Of
Gerd Thuemmler
Sent: Mittwoch, 17. April 2002 08:47
To: ccielab@groupstudy.com
Subject: custom queueing, byte-count, size
Hi,
..i know that exist an formula to calculate the byte-counts for the queues,
but i have forgotten it .... =8-(
Gerd Thuemmler
Berlin, Germany
48 h to the Lab....
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