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Explain 'magic numbers' in the HFSC article

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Tom Eastep 2009-05-24 13:36:58 -07:00
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docs/traffic_shaping.xml
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@ -1367,6 +1367,73 @@ eth0 - 1000kbit hfsc</programlisting>
1:20 2 500kbit full 1
1:10:11 3 400kbit:53ms:1500b full 2
1:10:12 4 100kbit:30ms:1500b full 2</programlisting>
<para>The following sub-section offers some notes about the
article.</para>
<section>
<title>Where Did all of those Magic Numbers come from?</title>
<para>As you read the article, numbers seem to be introduced out of thin
air. I'll try to shed some light on those.</para>
<para>There is very clear development of these numbers:</para>
<itemizedlist>
<listitem>
<para>12ms to transfer a 1500b packet at 1000kbits/second.</para>
</listitem>
<listitem>
<para>100kbits per second with 1500b packets, requires 8 packets per
second.</para>
</listitem>
<listitem>
<para>A packet from class 1:12 must be sent every 120ms.</para>
</listitem>
<listitem>
<para>Total transmit delay can be no more than 132ms (120 +
12).</para>
</listitem>
</itemizedlist>
<para>We then learn that the queuing latency can be reduced to 30ms if
we use a two-part service curve whose first part is 400kbits/second.
Where did those come from?</para>
<itemizedlist>
<listitem>
<para>The latency is calculated from the rate. If it takes 12ms to
transmit a 1500 byte packet at 1000kbits/second, it takes 30ms to
transmit a 1500b at 400kbits/second.</para>
</listitem>
<listitem>
<para>For the slope of the first part of the service curve, in
theory we can pick any number between 100 and 500 (the parent
class's ceiling) with higher numbers providing lower latency.</para>
</listitem>
</itemizedlist>
<para>The final curious number is the latency for class 1:11 - 52.5ms.
It is a consequence of everything that has gone before.</para>
<para>To acheive 400kbits/second with 1500-byte packets, 33.33 packets
per second are required. So a packet from class 1:11 must be sent every
30 ms. As the article says, "...the maximum transmission delay of this
class increases from 30ms to a total of 52.5 ms.". So we are looking for
an additional 22.5 ms.</para>
<para>Assume that both class 1:11 and 1:12 transmit for 30 ms at
400kbits/second. That is a total of 800kbits/second for 30ms. So Class
1:11 is punished for the excess. How long is the punishment? The two
classes sent 24,000 bits in 30ms; they are only allowed 0.030 * 500,000
= 15,000. So they are 9,000 bits over their quota. The amount of time
required to transmit 9,000 bits at 400,000 bits/second is
22.5ms!.</para>
</section>
</section>
<section id="Downloads">