Malathi r k | 26 May 17:58 2016
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Directed acyclic graph for RPL

Hello all,

How a node can be removed from network without disconnecting the network, from DAG root? DAG root need to remove all routes from a particular node to all other nodes, as well as from DAG root?
What all APIs need to be called for this to be done.

Please help me,

Thanks
Malathi

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Malathi r k | 26 May 17:28 2016
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(no subject)

Hello all,

How a node can be removed from network without disconnecting the network, from DAG root? I need to remove all routes from a particular node to all other nodes, as well as from DAG root?
What all APIs need to be called for this to be done.

Please help me,

Thanks
Malathi



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Graeme Bragg | 26 May 13:34 2016
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Re: Trickle Algorithm

>> Result= ... = number that is in [0 , i_cur/2] )

That is not quite correct. It will return a number in the range of [i_cur/2 <= x < i_cur] as per the leading comment of the function. 

This is because it is limited by being i_cur/2 PLUS the random remainder. The upper limit comes from the fact that the remainder of a division by (i_cur/2) can only be in the range of [0 <= remainder <= ((i_cur/2) - 1)].


Rima, looking at the code quickly, this function is basically used when "doubling" the trickle timer interval to give a random factor to it.

Thanks,
Graeme

Sent from my iPad

On 26 May 2016, at 09:44, Spilios <spilios-Re5JQEeQqe8AvxtiuMwx3w@public.gmane.org> wrote:

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Hi,

You need to improve a bit your C skills then I am afraid. This is a simple math function that divides i_cur first / 2  ( see i_cur>>=1;) , then it ads on that a random number that is between 0 and i_cur/2  (see tt_rand() % i_cur) . This is a modulo function that returns the remaining portion of the division. Since they divide by i_cur (which is now half of the original) it will return a number between 0 and i_cur (that again is now half of the original i_cur).

 

So in fact it does that :

 

 

Result= (i_cur/2) + (remaining of division random_number/(i_cur/2) = number that is in [0 , i_cur/2] )

 

Kind regards,

 

Spilios

 

 

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From: rima shissah [mailto:rima.shissah-Re5JQEeQqe8AvxtiuMwx3w@public.gmane.org]
Sent: Thursday, May 26, 2016 3:37 AM
To: Contiki developer mailing list <contiki-developers-5NWGOfrQmneRv+LV9MX5uipxlwaOVQ5f@public.gmane.org>
Subject: [Contiki-developers] Trickle Algorithm

 

Hi All,

im new with contiki and i am very interested to improve RPL performances based on Trickle algorithm.

The trickle timer is defined in the trickle-timer.c file which is located in core -> lib -> trickle-timer.c

All the functions for setting the parameter values are defined here in this file. Among these function we find the functuon that return random time t

/* Returns a random time point t in [I/2 , I) */
static clock_time_t get_t(clock_time_t i_cur)
{
  i_cur >>= 1;

  PRINTF("trickle_timer get t: [%lu, %lu)\n", (unsigned long)i_cur, (unsigned long)(i_cur << 1));

  return i_cur + (tt_rand() % i_cur);
}

I really did not understand this piece of code? There is someone who can explain it to me?

Regards,

Rima

---------

Karima SHISSAH
Master Sécurité des Systèmes d'Information
+212632125644

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Narmadha Thangam | 26 May 07:58 2016
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reg cpu, memory and power consumption

Hi all,
I am working with telosb sensor to implement my algorithms.
can anyone suggest me how to measure CPU cycles, memory and power consumption measurements in contiki/cooja.
Thanks in advance.
regards,
Narmadha

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rima shissah | 26 May 03:36 2016
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Trickle Algorithm

Hi All,

im new with contiki and i am very interested to improve RPL performances based on Trickle algorithm.

The trickle timer is defined in the trickle-timer.c file which is located in core -> lib -> trickle-timer.c

All the functions for setting the parameter values are defined here in this file. Among these function we find the functuon that return random time t

/* Returns a random time point t in [I/2 , I) */
static clock_time_t get_t(clock_time_t i_cur)
{
  i_cur >>= 1;

  PRINTF("trickle_timer get t: [%lu, %lu)\n", (unsigned long)i_cur, (unsigned long)(i_cur << 1));

  return i_cur + (tt_rand() % i_cur);
}

I really did not understand this piece of code? There is someone who can explain it to me?

Regards,
Rima
---------
Karima SHISSAH
Master Sécurité des Systèmes d'Information
+212632125644
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rajasekaran | 25 May 18:16 2016
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Setting up JVM heap

Dear All,

I want to make rpl simulation for more than 5000 nodes. As you said, I need to increase JVM heap. But I dont know how to implement these commands for cooja simulator.

I dont know anything about java. Could you please help me.

-
Xms<size> set initial Java heap size -Xmx<size> set maximum Java heap size -Xss<size> set java thread stack size
Thank you,
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Emmanouil Palavras | 25 May 14:57 2016

DTLS-COAP simulation on Z1

Hello,

 

I am experimenting with DTLS COAP on Z1. On my Cooja simulation I have 3 motes: slip-radio (ID:1), dtls-coap-server (ID:2), dtls-coap-client (ID:3).

The server is based on 6lbr-demo and the client on er-example-client.

First of all, I made sure that both are working without DTLS and then I modified the code in order to add the DTLS functionality.

Due to the memory constraints of Z1, I have reduced any possible value from limiting the REST resources to 1 (device uptime) to reducing QUEUEBUF to 3 packets and all RPL related constants (such as MAX_NEIGHBORS, MAX ROUTES etc) to 2.

I have also installed mspgcc-4.7.0.

The problem is that I cannot find out why the handshake never gets completed.

So, when I press the button on mote 3 I get the following output:

 

00:26.148             ID:3        --Requesting /dev/uptime--

00:26.155             ID:3        11.033 DEBG /[00E:0 03]:5683:

00:26.162             ID:3        11.034 DEBG DTLSv12: initialize HASH_SHA256

00:26.193             ID:2        11.078 DEBG dtls_handle_message: PEER NOT FOUND

00:26.197             ID:2        11.078 DEBG      .[:

00:26.204             ID:2        11.078 DEBG got packet 12106 (0 bytes)

00:26.210             ID:2        11.079 DEBG receive header: (%zu bytes):

00:26.220             ID:2        00000000 16 FE FD 00 00 00 00 00  00 00 00 00 36

00:26.227             ID:2        11.080 DEBG receive unencrypted: (%zu bytes):

00:26.238             ID:2        00000000 01 00 00 2A 00 00 00 00  00 00 00 2A FE FD 00 00

00:26.249             ID:2        00000010 00 0B 73 36 30 69 A9 99  2E 1B CF C6 5C E8 65 27

00:26.261             ID:2        00000020 3A BC EB 00 48 2E E1 D4  06 62 C7 C0 F4 E6 00 00

00:26.266             ID:2        00000030 00 02 C0 A8 01 00

00:26.271             ID:2        11.083 DEBG DTLS_HT_CLIENT_HELLO

00:26.346             ID:2        11.087 DEBG create cookie: (%zu bytes): D34B75D781F8C8BCA42E3E72AAB4FECC

00:26.353             ID:2        11.088 DEBG compare with cookie: (%zu bytes):

00:26.358             ID:2        11.088 DEBG cookie len is 0!

00:26.365             ID:2        11.089 DEBG server hello verify was sent

00:32.334             ID:1        <at> <at> *5!RAz3:b <at> *5az3:d <at> *5Az;:,

00:39.233             ID:1        <at> <at> *5Az;:,

00:46.165             ID:3        21.034 INFO Connection timeout, removing peer

00:46.171             ID:3        21.034 DEBG send header: (%zu bytes):

00:46.180             ID:3        00000000 15 FE FD 00 00 00 00 00  00 00 01 00 02

00:46.187             ID:3        21.035 DEBG send unencrypted: (%zu bytes):

00:46.190             ID:3        00000000 02 00

00:46.200             ID:3        removed peer [fe80::c30c:0:0:2]:5683

 

I think it is worth mentioning that if I set REST_MAX_CHUNK_SIZE to 16 there is absolutely no response from the server (ID:2). In that case the output is like the above one without the ID:2.

I have also tried cf-client-1.1.0, with the same results.

 

Thank you for your time,

Manos.

 

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Monali Mavani | 25 May 13:02 2016
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energy consumption in cooja

Hi all

Energest module calculates energy consumption for each packet in terms of time.In this distance between two nodes does not have any effect.If I want to profile energy consumption with respect to distance and path loss,How can it be done?

Thanks

Monali mavani
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Torsten Zimmermann | 25 May 08:25 2016
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Cooja CFS Support for Wismote

Hi,

i'm having some problems running a Contiki application that uses CFS in
Cooja on the Wismote platform. I tried to add a file to the CFS via the
interface viewer and also using a CSC script such as:

...
mote = sim.getMoteWithID(1);&#xD;
fs = mote.getFilesystem();&#xD;
success = fs.insertFile(file);&#xD;
...

However, in the interface viewer you can't see the CFS interface and the
script fails. I tried to apply this approach
(https://sourceforge.net/p/contiki/mailman/message/30735346/) that works
for Z1 for the Wismote, unfortunately without success.

Anyone knows how to solve this?

Greetings,
Torsten

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Malathi r k | 25 May 06:16 2016
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Fwd:


Hello,

How can we find number of nodes in network and node information using API? I need API to find address and other information of all nodes participating in network ,when that API used in DAG root? Is ther any?

How can we discard some node or add some node to a network using API? I had used the function uip_ds6_route_lookup and uip_ds6_route_rm for removing some node. But , it did not. If we use list_remove(), can we completely discard node? Or else , any other way?

How the linked list here used to add nodes to network?

In which contiki source file,  the information regarding the above questions  located?

I need these information for an application.

Please help me to clear these.

Thanks
Malathi










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Adila Nordin | 24 May 17:00 2016
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Energy per probe

Hi all,

I'm currently looking into powertrace.
In a normal transmission, there'd be 2 or 3 probes before the packet is ACKed which gives approximately for 2 probes 0.462mJ per probe (just the energy for transmission and listening) and for 3 probes 0.380mJ

When I try to have retransmissions, it will try to send 5 probes (because the phase-lock might have changed so it will send more to get the ACK) - which gives 0.262mJ per probe

When it totally have no idea of the timing, no phase lock, it will keep sending a lot more. In my case, it tried to send 36 probes which gives 0.221mJ per probe

My question is, why does it take less energy when there are more probes for a packet?
Is this suppose to happen or I misunderstood some part of it?
If the extra energy is from the ACK for successful packet, why does 3 probes take less than 2 probes energy?


Thanks,
Adila
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