How to fix RX problem in Sender Based Orchestra Scheduling with CCA? - simulation

I am realizing a scenario of DAG root and DAG node in Contiki NG environment with Sender Based Orchestra while using UDP between nodes.But I have a problem of lack of RX in my Cooja Simulation after enabling Clear Channel Assesment(CCA) in my project-conf.h file. What should I change in my files to solve this problem?
I use one node for DAG root and one node for DAG node(while using the files under Contiki NG namely udp-server.c and udp-client.c). After CCA I have had an error like !dl-miss TXbeforeTX 2800 2120 in the mote output of COOJA. As I get it there was a problem in timing. Therefore, I have changed the CCA offset(default is 1800 us) to 1000 us and apparently it erased the warning. But this approach did not work on lack of RX in especially sender based unicast and EB slotframes. While seeing TX for mentioned slotframes, RX could not be observed. I could see in the mote output that the node tries to send a packet but it is never acknowledged. Also after sometime I have observed that the nodes were leaving the network.
I expect to have a RX slots for EB and Sender based unicast slotframes as well but actual result is I am getting RX just for Broadcast slotframe.
What should I change in my configuration files?

Related

CANopenNode: Master node sends continually a message

I am trying to use CANopenNode into a STM32L476 device by using libohiboard as HAL library. In the network, I have: (i) my board that operates as a master and (ii) a commercial node. At startup, the node sends HB message and SYNC message. When my board use
CO_NMT_sendCommand(CO->NMT,CO_NMT_ENTER_OPERATIONAL, 0x0A);
the master starts to send continually the same message without stopping!
With logic analyzer I see this:
Where Channel 0 is the TX pins of the microcontroller, and Channel 1 is the RX pin.
I can't understand why the message returns into RX pin immediately! I checked the microcontroller configuration and the loopback mode is OFF.
Thanks
Looks like normal CAN operation - all messages are immediately echoed back while they are sent or else bus arbitration wouldn't work. The only difference is the ACK bit which you can see are set on the rx line but not on tx. This bit is filled in by the other CAN node on the bus.
The reason why your node keeps sending the same message doesn't seem related to this.
I don't know how it works on your controller but usually you have to pay attention to send NMT_start_command only when your slave node doesn't return any heartbeat or if the heartbeat value is different than the mode expected (pre operational or operational as an example)
If the slave doesn't return anything there might be multiple reasons:
nothing activated so you have first to set a time using the right SDO
the slave use nodeguarding instead of heartbeat so you have to query first the slave with a message ID: 0x700 + Node ID, DLC: 0
Please let me know if it is not clear or doesn't help

Omnetpp application sends multiple streams

Let's say I have a car with different sensors: several cameras, LIDAR and so on, the data from this sensors are going to be send to some host over 5G network (omnetpp + inet + simu5g). For video it is like 5000 packets 1400 bytes each, for lidar 7500 packets 1240 bytes and so on. Each flow is encoded in UDP packets.
So in omnetpp module in handleMessage method I have two sentTo calls, each is scheduled "as soon as possible", i.e., with no delay - that corresponds to the idea of multiple parallel streaming. How does omnetpp handle situations, when it needs to send two different packets at the same time from the same module to the same module (some client, which receives sensor data streams)? Does it create some inner buffer on the sender or receiver side, therefore allowing really only one packet sending per handleMessage call or is it wrong? I want to optimize data transmission and play with packet sizes and maybe with sending intervals, so I want to know, how omnetpp handles multiple streaming at the same time, because if it actually buffers, maybe than it makes sense to form a single package from multiple streams, each such package will consist of a certain amount of data from each stream.
There is some confusion here that needs to be clarified first:
OMNeT++ is a discrete event simulator framework. An OMNeT++ model contains modules that communicate with each other, using OMNeT++ API calls like sendTo() and handleMessage(). Any call of the sendTo() method just queues the provided message into the future event queue (an internal, time ordered queue). So if you send more than one packet in a single handleMessage() method, they will be queued in that order. The packets will be delivered one by one to the requested destination modules when the requested simulation time is reached. So you can send as many packets as you wish and those packets will be delivered one by one to the destination's handleMessage() method. But beware! Even if the different packets will be delivered one by one sequentially in the program's logic, they can still be delivered simultaneously considering the simulation time. There are two time concepts here: real-time that describes the execution order of the code and simulation-time which describes the time passes from the point of the simulated system. That's why, while OMNeT++ is a single threaded application that runs each events sequentially it still can simulate infinite number of parallel running systems.
BUT:
You are not modeling directly with OMNeT++ modules, but rather using INET Framework which is a model directly created to simulate internet protocols and networks. INET's core entity is a node which is something that has network interface(s) (and queues belonging to them). Transmission between nodes are properly modeled and only a single packet can travel on an ethernet line at a time. Other packets must queue in the network interface queue and wait for an opportunity to be delivered from there.
This is actually the core of the problem for Time Sensitive Networks: given a lot of pre-defined data streams in a network, how the various packets interfere and affect each other and how they change the delay and jitter statistics of various streams at the destination, Plus, how you can configure the source and network gate scheduling to achieve some desired upper bounds on those statistics.
The INET master branch (to be released as INET 4.4) contains a lot TSN code, so I highly recommend to try to use it if you want to model in vehicle networks.
If you are not interested in the in-vehicle communication, bit rather want to stream some data over 5G, then TSN is not your interest, but you should NOT start to multiplex/demultiplex data streams at application level. The communication layers below your UDP application will fragment/defragment and queue the packets exactly how it is done in the real world. You will not gain anything by doing mux/demux at application layer.

How exactly do socket receives work at a lower level (eg. socket.recv(1024))?

I've read many stack overflow questions similar to this, but I don't think any of the answers really satisfied my curiosity. I have an example below which I would like to get some clarification.
Suppose the client is blocking on socket.recv(1024):
socket.recv(1024)
print("Received")
Also, suppose I have a server sending 600 bytes to the client. Let us assume that these 600 bytes are broken into 4 small packets (of 150 bytes each) and sent over the network. Now suppose the packets reach the client at different timings with a difference of 0.0001 seconds (eg. one packet arrives at 12.00.0001pm and another packet arrives at 12.00.0002pm, and so on..).
How does socket.recv(1024) decide when to return execution to the program and allow the print() function to execute? Does it return execution immediately after receiving the 1st packet of 150 bytes? Or does it wait for some arbitrary amount of time (eg. 1 second, for which by then all packets would have arrived)? If so, how long is this "arbitrary amount of time"? Who determines it?
Well, that will depend on many things, including the OS and the speed of the network interface. For a 100 gigabit interface, the 100us is "forever," but for a 10 mbit interface, you can't even transmit the packets that fast. So I won't pay too much attention to the exact timing you specified.
Back in the day when TCP was being designed, networks were slow and CPUs were weak. Among the flags in the TCP header is the "Push" flag to signal that the payload should be immediately delivered to the application. So if we hop into the Waybak
machine the answer would have been something like it depends on whether or not the PSH flag is set in the packets. However, there is generally no user space API to control whether or not the flag is set. Generally what would happen is that for a single write that gets broken into several packets, the final packet would have the PSH flag set. So the answer for a slow network and weakling CPU might be that if it was a single write, the application would likely receive the 600 bytes. You might then think that using four separate writes would result in four separate reads of 150 bytes, but after the introduction of Nagle's algorithm the data from the second to fourth writes might well be sent in a single packet unless Nagle's algorithm was disabled with the TCP_NODELAY socket option, since Nagle's algorithm will wait for the ACK of the first packet before sending anything less than a full frame.
If we return from our trip in the Waybak machine to the modern age where 100 Gigabit interfaces and 24 core machines are common, our problems are very different and you will have a hard time finding an explicit check for the PSH flag being set in the Linux kernel. What is driving the design of the receive side is that networks are getting way faster while the packet size/MTU has been largely fixed and CPU speed is flatlining but cores are abundant. Reducing per packet overhead (including hardware interrupts) and distributing the packets efficiently across multiple cores is imperative. At the same time it is imperative to get the data from that 100+ Gigabit firehose up to the application ASAP. One hundred microseconds of data on such a nic is a considerable amount of data to be holding onto for no reason.
I think one of the reasons that there are so many questions of the form "What the heck does receive do?" is that it can be difficult to wrap your head around what is a thoroughly asynchronous process, wheres the send side has a more familiar control flow where it is much easier to trace the flow of packets to the NIC and where we are in full control of when a packet will be sent. On the receive side packets just arrive when they want to.
Let's assume that a TCP connection has been set up and is idle, there is no missing or unacknowledged data, the reader is blocked on recv, and the reader is running a fresh version of the Linux kernel. And then a writer writes 150 bytes to the socket and the 150 bytes gets transmitted in a single packet. On arrival at the NIC, the packet will be copied by DMA into a ring buffer, and, if interrupts are enabled, it will raise a hardware interrupt to let the driver know there is fresh data in the ring buffer. The driver, which desires to return from the hardware interrupt in as few cycles as possible, disables hardware interrupts, starts a soft IRQ poll loop if necessary, and returns from the interrupt. Incoming data from the NIC will now be processed in the poll loop until there is no more data to be read from the NIC, at which point it will re-enable the hardware interrupt. The general purpose of this design is to reduce the hardware interrupt rate from a high speed NIC.
Now here is where things get a little weird, especially if you have been looking at nice clean diagrams of the OSI model where higher levels of the stack fit cleanly on top of each other. Oh no, my friend, the real world is far more complicated than that. That NIC that you might have been thinking of as a straightforward layer 2 device, for example, knows how to direct packets from the same TCP flow to the same CPU/ring buffer. It also knows how to coalesce adjacent TCP packets into larger packets (although this capability is not used by Linux and is instead done in software). If you have ever looked at a network capture and seen a jumbo frame and scratched your head because you sure thought the MTU was 1500, this is because this processing is at such a low level it occurs before netfilter can get its hands on the packet. This packet coalescing is part of a capability known as receive offloading, and in particular lets assume that your NIC/driver has generic receive offload (GRO) enabled (which is not the only possible flavor of receive offloading), the purpose of which is to reduce the per packet overhead from your firehose NIC by reducing the number of packets that flow through the system.
So what happens next is that the poll loop keeps pulling packets off of the ring buffer (as long as more data is coming in) and handing it off to GRO to consolidate if it can, and then it gets handed off to the protocol layer. As best I know, the Linux TCP/IP stack is just trying to get the data up to the application as quickly as it can, so I think your question boils down to "Will GRO do any consolidation on my 4 packets, and are there any knobs I can turn that affect this?"
Well, the first thing you can do is disable any form of receive offloading (e.g. via ethtool), which I think should get you 4 reads of 150 bytes for 4 packets arriving like this in order, but I'm prepared to be told I have overlooked another reason why the Linux TCP/IP stack won't send such data straight to the application if the application is blocked on a read as in your example.
The other knob you have if GRO is enabled is GRO_FLUSH_TIMEOUT which is a per NIC timeout in nanoseconds which can be (and I think defaults to) 0. If it is 0, I think your packets may get consolidated (there are many details here including the value of MAX_GRO_SKBS) if they arrive while the soft IRQ poll loop for the NIC is still active, which in turn depends on many things unrelated to your four packets in your TCP flow. If non-zero, they may get consolidated if they arrive within GRO_FLUSH_TIMEOUT nanoseconds, though to be honest I don't know if this interval could span more than one instantiation of a poll loop for the NIC.
There is a nice writeup on the Linux kernel receive side here which can help guide you through the implementation.
A normal blocking receive on a TCP connection returns as soon as there is at least one byte to return to the caller. If the caller would like to receive more bytes, they can simply call the receive function again.

Driver: Reception overrun: Messages are lost

The above two pictures are my measurement and simulation setups respectively. The Replay block plays a blf file of length 6 minutes containing total of 2,413,161 CAN frames from two CAN channels.
The above picture explains the bench setup. Canoe reads the blf file and transmits the CAN frames on two CAN channels. Microcontroller (MuC) receives the CAN frames converts them in to Ethernet IPV4 UDP packet and transmit again to the Canoe.
When i run this configuration, i am getting below errors.
1. System - CAN driver: Reception overrun - messages are lost
2. System CAN X : Message with ID = XXX could not be sent. Driver error 11 in TransmitCANFrame, "XL_ERR_QUEUE_IS_FULL"
3. System Warning: replay loading delay(s)
System ReplayBlock 1(blf_file.blf): 15 times, 7347.46 ms total
I assumes this was due to Canoe performance issue or CAN driver issue. So I did the below steps.
1. Modified the CANCaseXL Receive latency->Very fast under Vector hardware Config.
2. Increased the Transmit queue settings->32768 (maximum) under Vector hardware config -> Global settings.
3. I disabled all except one logging block (blf) [As you can see in the measurement setup].
But i still experience the same errors. What could be the problem? Is there any other ways to resolve this?
You need to have termination on both CANs (120 Ohm).
Such error indicates lack of termination.

How to find the right sending speed with UDP?

I want to transfer a large number of messages. The messages do not need to be reliable. UDP comes to mind as a protocol choice.
Latency is important as well. I do not want to suffer from TCP head-of-line blocking.
I'm concerned I might overload the network when I just start sending messages at maximum speed (e.g. while (messagesRemaining != 0) Send(...);). If I send more than some middle-box can relay then, I think, large numbers of messages might be dropped. Some messages being dropped is fine but most of them should arrive.
How can I address this issue? How can I find out how fast I can send? I want to maintain reasonable packet loss (a few percent) and otherwise maximize bandwidth.
Whether you will overload the network or not depends on what is between the sender and the receiver hosts. The iperf utility has a UDP option that could help you determine the max rate you could send for a certain level of acceptable packet loss.
That said, from personal experience:
If it's local Gigabit network and client/server on same subnet, I highly doubt you would lose any packets. I've done tests before with iperf in this type of environment and never lost any packets; iperf is going to be one of the fastest and most efficient ways to put UDP packets on to the wire from a PC and we still never lost packets. We were even running the packets through an intel Atom-based Linux host with bridged ports setup while doing tcpdump at the same time and still never lost packets (note that even cheapo switches would perform as good or better than a bridge setup in a PC). The only way we were ever able to get packets to be lost was when we used FPGA/ASIC test devices that could put packets onto the wire at true line-speed for long periods of time. Even at that, the test setup was only losing packets when the packets were less than around 500 bytes.
If you aren't on a local network though (i.e. going over internet or routers) you will just need to do some testing with iperf to see what is acceptable for your environment. Problem is though that what rate can be sustained one day isn't guaranteed to be the same the next day. UDP doesn't have any sort of congestion/flow control algorithms like TCP does so you will have to figure out on your own how fast you can send.