Hi Karim
Sorry for the chain emails, but I just realised that I mixed up x1 != x2
and x1 == x2 in my explanation. So please keep that in mind while
interpreting my explanation. You can mail me if you are still confused. You
can refer to https://youtube.com/playlist?list=PL8EC1756A7B1764F6 playlist
(Lectures 78 onwards) to get a deeper understanding of this topic.
Thanks
Setu
On Fri, Mar 10, 2023, 2:04 AM Setu Gupta <setu18...@iiitd.ac.in> wrote:
> Hi Karim
>
> Sorry I forgot to attach the picture.
>
> Thanks
> Setu
>
> On Fri, Mar 10, 2023, 2:02 AM Setu Gupta <setu18...@iiitd.ac.in> wrote:
>
>> Hi Karim
>>
>> Adding to what Hoa said, the topology looks something like the topology
>> I've drawn in the attached picture.
>>
>> The "Mesh" part of Mesh_XY specify connectivity of various routers. In
>> your case, since there are 4 CPUs, there are 4 routers. Moreover since the
>> number of rows specified by you is 2, you have a mesh which is 2 nodes wide
>> and 2 nodes high giving a total of 2x2 = 4 nodes as expected. Note that
>> mesh is simply a criss cross grid where each grid point represents a
>> node/router.
>>
>> Each node/router has 5 connections. 4 connection out of these five are
>> north, south, east and west. If a router is on the edge of the network, it
>> will not have one or more of these connections. That's why your output only
>> shows 3 connections for each router as all the four routers are on the edge
>> in your example. These four connections connect adjacent routers, i.e
>> neighbours, in a mesh topology. The fifth port is connected to what's
>> called a processing element which in your case is a CPU core. This fifth
>> port is called the local port as it forms a local connection between a
>> router and its processing element.
>>
>> The "XY" part in Mesh_XY specifies the routing algorithm which in your
>> case is the XY routing algorithm. The XY routing algorithm prioritises the
>> transfer along X first. Other routing algorithms may chose a different
>> strategy fpr routing. If a packet originates from source A and it is
>> destined to destination B, the first thing that the algorithm does is
>> computing the number of hops required in the X direction and the Y
>> direction.
>>
>> For example, if a packet originates from 1 and it is destined to 2, it
>> needs to hop 1 connection in the X direction and 1 connection in the Y
>> direction. So the traversal can be broken down into two steps: first
>> traverse along X, and then traverse along Y.
>>
>> Since we are following XY routing, the packet will first travel in the
>> west direction from 1, and it will reach 0. Then it will travel along the Y
>> direction to reach 2 from 0. As you can see, in this example, the traversal
>> first happened in the X direction and then in the Y direction.
>>
>> Hopefully you can see from this example that the routing table is
>> completed deterministic, i.e. XY is a fixed routing algorithm. Give a
>> source destination pair (A ,B) the path taken by packets will always be
>> exactly the same in the case of XY routing.
>>
>> If the current router's coordinates are (x1, y1) and the destination
>> coordinates are (x2, y2), the packet will go to the north or the south
>> router if and only if x1 != x2. The choice between north and south will
>> depend on the ordering between x1 and x2. If x1 = x2, the packet will then
>> travel either westwards or eastwards based on y1 and y2.
>>
>> Hopefully this clarifies your question. Please note that the topology
>> that you drew is not entirely correct. So that may be the source of
>> confusion.
>>
>> Thanks
>> Setu
>>
>>
>> On Fri, Mar 10, 2023, 1:20 AM Hoa Nguyen via gem5-users <
>> gem5-users@gem5.org> wrote:
>>
>>> Hi Karim,
>>>
>>> I'm not very familiar with the Mesh_XY topology, but here is my attempt
>>> to explain this.
>>>
>>> I think the layout of the topology looks like this,
>>>
>>> 2 (0,1) <-----> 3 (1,1)
>>> ^ ^
>>> | |
>>> v v
>>> 0 (0,0) <-----> 1 (1,0)
>>>
>>> It's straightforward to send a packet from one node to its
>>> adjacent node, such as sending a package from 2 to 3 consists of having the
>>> node 2 sending a local packet to its West side.
>>>
>>> When the source and the destination are more than 1 hop away, it
>>> involves sending a packet to immediate node. It showed in your example's
>>> line 5, there's a packet that is sent from 1 (1,0) to 2 (0,1).
>>> It can't be done in one hop. So, at first, the packet was sent from the
>>> local of node 1 to its West side. In the next step, node 0 receives the
>>> packet from its East side, and routes that packet to its North side. That
>>> explains the example's line 9.
>>>
>>> Hope this helps!
>>>
>>> Regards,
>>> Hoa Nguyen
>>>
>>> On Wed, Mar 8, 2023 at 11:25 PM Karim Soliman via gem5-users <
>>> gem5-users@gem5.org> wrote:
>>>
>>>> Hi, for my research I'm trying to implement a custom routing algorithm
>>>> in gem5/garnet.
>>>> Someone from the mailing list recommended that I should go for the
>>>> Mexh_XY routing algorithm to understand how it's working.
>>>>
>>>> The mesh topology i'm using contains only 4 cpus and 4 dirs and number
>>>> of mesh rows is 2, and the routing-algorithm option is set to 1
>>>>
>>>> here is the full simulation command i'm using
>>>>
>>>> *sudo build/NULL/gem5.debug --debug-flags=RubyNetwork
>>>> configs/example/garnet_synth_traffic.py --num-cpus=4 --num-dirs=4
>>>> --mesh-rows=2 --network=garnet --routing-algorithm=1 --topology=Mesh_XY
>>>> --synthetic=uniform_random --injectionrate=0.1*
>>>>
>>>> In the file mem/ruby/garnet/RoutingUnit.cc --> the function
>>>> outportComputeXY
>>>>
>>>> int RoutingUnit::outportComputeXY(RouteInfo route, int inport,
>>>> PortDirection inport_dirn)
>>>>
>>>> I'm trying to get some output during the simulation to understand the
>>>> routing table at each router and how the lookup table works.
>>>>
>>>> The following is my code. I'm using a file to get the data during the
>>>> simulation
>>>> So, the file will print the following:
>>>>
>>>> [RouterId] [DestinationRouter] [hops_x] [hops_y] [InportDirn] -->
>>>> [outport_dirn] [router_x, router_y] [des_x, dest_y].
>>>> and output the lookup table attached to the router.
>>>>
>>>> std::ofstream myfile;
>>>> // open the file in appending mode
>>>> myfile.open ("routing_output.txt", std::ios_base::app);
>>>> myfile << m_router->get_id() << "\t" << route.dest_router << " |
>>>> hops_x: " << x_hops << " | hops_y: " << y_hops << "\t";
>>>> myfile << inport_dirn << " --> " << outport_dirn << "\t" <<
>>>> m_outports_dirn2idx[outport_dirn] << "\t" << my_x << ", " << my_y;
>>>> myfile << " --> "<< dest_x << ", " << dest_y << std::endl;
>>>> // iterate over the outports table map and write the data to the file.
>>>> std::map<std::__cxx11::basic_string<char>, int>::iterator it;
>>>> for(it = m_outports_dirn2idx.begin(); it != m_outports_dirn2idx.end();
>>>> it++)
>>>> {
>>>> myfile << it->first << "\t" << it->second << std::endl;
>>>> }
>>>> myfile.close();
>>>>
>>>> Here are some samples of my output file and also the full txt file is
>>>> attached.
>>>>
>>>> [image: image.png]
>>>>
>>>> I don't really know how the packets are routed and also how the lookup
>>>> table works. also i can't imagine the connections between the nodes i tried
>>>> to draw a flow chart for each router
>>>> but some of the output data doesn't match the drawing -- like a packet
>>>> from router #2 will go east !!
>>>> [image: image.png]
>>>> Can anyone explain to me? or if there is any documentation I can check
>>>> to understand this.
>>>>
>>>> Best Regards,
>>>> *Eng. Karim Soliman*
>>>> Teaching Assistant
>>>> Computer Engineering Department
>>>> Pharos University in Alexandria (P.U.A)
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>>>>
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>>>
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