-.. _platform:
-
.. raw:: html
- <object id="TOC" data="graphical-toc.svg" width="100%" type="image/svg+xml"></object>
+ <object id="TOC" data="graphical-toc.svg" type="image/svg+xml"></object>
<script>
window.onload=function() { // Wait for the SVG to be loaded before changing it
var elem=document.querySelector("#TOC").contentDocument.getElementById("PlatformBox")
Of course, this is only one possible way to model these things. YMMV ;)
+.. _howto_parallel_links:
+
+Modeling parallel links
+***********************
+
+Most HPC topologies, such as fat-trees, allow parallel links (a
+router A and a router B can be connected by more than one link).
+You might be tempted to model this configuration as follows :
+
+.. code-block:: xml
+
+ <router id="routerA"/>
+ <router id="routerB"/>
+
+ <link id="link1" bandwidth="10GBps" latency="2us"/>
+ <link id="link2" bandwidth="10GBps" latency="2us"/>
+
+ <route src="routerA" dst="routerB">
+ <link_ctn id="link1"/>
+ </route>
+ <route src="routerA" dst="routerB">
+ <link_ctn id="link2"/>
+ </route>
+
+But that will not work, since SimGrid doesn't allow several routes for
+a single `{src ; dst}` pair. Instead, what you should do is :
+
+ - Use a single route with both links (so both will be traversed
+ each time a message is exchanged between router A and B)
+
+ - Double the bandwidth of one link, to model the total bandwidth of
+ both links used in parallel. This will make sure no combined
+ communications between router A and B use more than the bandwidth
+ of two links
+
+ - Assign the other link a `FATPIPE` sharing policy, which will allow
+ several communications to use the full bandwidth of this link without
+ having to share it. This will model the fact that individual
+ communications can use at most this link's bandwidth
+
+ - Set the latency of one of the links to 0, so that latency is only
+ accounted for once (since both link are traversed by each message)
+
+So the final platform for our example becomes :
+
+.. code-block:: xml
+
+ <router id="routerA"/>
+ <router id="routerB"/>
+
+ <!-- This link limits the total bandwidth of all parallel communications -->
+ <link id="link1" bandwidth="20GBps" latency="2us"/>
+
+ <!-- This link only limits the bandwidth of individual communications -->
+ <link id="link2" bandwidth="10GBps" latency="0us" sharing_policy="FATPIPE"/>
+
+ <!-- Each message traverses both links -->
+ <route src="routerA" dst="routerB">
+ <link_ctn id="link1"/>
+ <link_ctn id="link2"/>
+ </route>
+
.. _understanding_lv08
+
Understanding the default TCP model
-*****************************
+***********************************
When simulating a data transfer between two hosts, you may be surprised
by the obtained simulation time. Lets consider the following platform:
.. code-block:: xml
- <host id="A" speed="1Gf"/>
- <host id="B" speed="1Gf"/>
+ <host id="A" speed="1Gf" />
+ <host id="B" speed="1Gf" />
- <link id="link1" latency="10ms" bandwidth="1Mbps"/>
+ <link id="link1" latency="10ms" bandwidth="1Mbps" />
- <route src="A" dst="B>
- <link_ctn id="link1/>
+ <route src="A" dst="B">
+ <link_ctn id="link1" />
</route>
If host `A` sends `100kB` (a hundred kilobytes) to host `B`, one could expect
that this communication would take `0.81` seconds to complete according to a
simple latency-plus-size-divided-by-bandwidth model (0.01 + 8e5/1e6 = 0.81).
However, the default TCP model of SimGrid is a bit more complex than that. It
-accounts for three phenomena that directly impacts the simulation time even
+accounts for three phenomena that directly impact the simulation time even
on such a simple example:
+
- The size of a message at the application level (i.e., 100kB in this
- example) is not the size of what will actually be transferred over the
- network. To mimic the fact that TCP and IP header are added to each packet of
+ example) is not the size that will actually be transferred over the
+ network. To mimic the fact that TCP and IP headers are added to each packet of
the original payload, the TCP model of SimGrid empirically considers that
- `only 97% of the nominal bandwidth` are available. In other works, the
+ `only 97% of the nominal bandwidth` are available. In other words, the
size of your message is increased by a few percents, whatever this size be.
- In the real world, the TCP protocol is not able to fully exploit the
bandwidth of a link from the emission of the first packet. To reflect this
`slow start` phenomenon, the latency declared in the platform file is
multiplied by `a factor of 13.01`. Here again, this is an empirically
- determined value that may not correspond to every TCP implementation on
+ determined value that may not correspond to every TCP implementations on
every networks. It can be tuned when more realistic simulated times for
short messages are needed though.
- When data is transferred from A to B, some TCP ACK messages travel in the
opposite direction. To reflect the impact of this `cross-traffic`, SimGrid
simulates a flow from B to A that represents an additional bandwidth
- consumption of `0.05`. The bandwidth share allocated to the flow from A to
- B is then the available bandwidth (i.e., 97% of the nominal one) divided by
- 1.05 (i.e., the total consumption). This feature, activated by default, can be
- disabled by adding the `--cfg=network/crosstraffic:0` flag to command line.
+ consumption of `0.05`. The route from B to A is implicitly declared in the
+ platform file and uses the same link `link1` as if the two hosts were
+ connected through a communication bus. The bandwidth share allocated to the
+ flow from A to B is then the available bandwidth of `link1` (i.e., 97% of
+ the nominal bandwidth of 1Mb/s) divided by 1.05 (i.e., the total consumption).
+ This feature, activated by default, can be disabled by adding the
+ `--cfg=network/crosstraffic:0` flag to command line.
-As a consequence, the simulated time of a transfer of 100kB from A to B is not
-0.81 seconds but
+As a consequence, the time to transfer 100kB from A to B as simulated by the
+default TCP model of SimGrid is not 0.81 seconds but
.. code-block:: python
- 0.01 * 13.01 + 800000 / ((0.97 * 1e6) / 1.05) = 0.996079 seconds
+ 0.01 * 13.01 + 800000 / ((0.97 * 1e6) / 1.05) = 0.996079 seconds.
