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13. Bandwidth meter

In this chapter I am going to develop a simple bandwidth meter using the following functions from libiptc:

Also the function gettimeofday will be used to measure elapsed time and the function getopt to get options from the command line.

I don't know really if the term bandwidth meter is well used here. I interpret bandwidth as a reference to a flow capacity; perhaps a better term could be flow meter.

Here is the bandwidth meter bw.c. It's well commented to be easy followed by everyone:

 * How to use libiptc- program #4
 * /usr/local/src/bw.c
 * By Leonardo Balliache - 04.09.2002
 * e-mail:
 * --WELL COMMENTED-- to be easy followed by everyone.

/* include files required */
#include <getopt.h>
#include <sys/errno.h>
#include <sys/time.h>
#include <stdio.h>
#include <fcntl.h>
#include <stdlib.h>
#include <string.h>
#include <dlfcn.h>
#include <time.h>
#include <unistd.h>
#include "libiptc/libiptc.h"
#include "iptables.h"

/* colors to differentiate chains measures */
#define RED     "\033[41m\033[37m"
#define GREEN   "\033[42m\033[30m"
#define ORANGE  "\033[43m\033[30m"
#define BLUE    "\033[44m\033[37m"
#define MAGENTA "\033[45m\033[37m"
#define CYAN    "\033[46m\033[30m"
#define WHITE   "\033[47m\033[30m"
#define BLACK   "\033[40m\033[37m"
#define RESET   "\033[00m"

/* maximum number of chains to be processed */

/* time between measures in seconds; adjust as you like */
#define SLEEPTIME 1

/* structure to count bytes per chain */
struct bwcnt  {
  int start;           /* the chain was initialized */
  u_int64_t icnt;      /* bytes through; previous measure */
  u_int64_t ocnt;      /* bytes through; current measure */
  double bw;           /* bandwitdh (flow) on this chain */

/* function to calculate differences of time in seconds.
 * micro-seconds precision.
double delta(struct timeval a, struct timeval b)
  if (a.tv_usec & b.tv_usec)  {
    a.tv_usec += 1000000;
  return a.tv_sec-b.tv_sec + (a.tv_usec-b.tv_usec)/1000000.0;

/* main function */
int main(int argc, char *argv[])
  int i, j, ok;
  double totbw;
  iptc_handle_t h;
  int c, act_bw = 0;
  const char *chain = NULL;
  const char *tablename = "filter";
  struct timeval ti, to;
  struct bwcnt bw[MAXUSERCHAINS];
  struct ipt_counters *counters;

  program_name = "bw";
  program_version = NETFILTER_VERSION;

 /* check options
  * we have 2 options: 
  *        -c = display current flow (each SLEEPTIME).
  *        -a = display average flow (from start); default option.
  while ((c = getopt (argc, argv, "ac")) != -1)
  switch (c)  {
  case 'a':
    act_bw = 0;
  case 'c':
    act_bw = 1;
  case '?':
    if (isprint(optopt))
      fprintf (stderr, "Unknown option `-%c'.\n", optopt);
      fprintf (stderr,"Unknown option character `\\x%x'.\n",optopt);

  /* initialize array of chains */
  memset(&bw, 0, MAXUSERCHAINS * sizeof(struct bwcnt));

  /* get time to start meter on variable ti */
  gettimeofday(&ti, NULL);

  /* fire meter loop */
  if ( act_bw )  
    printf("Displaying current flow values ...\n");
    printf("Displaying average flow values ...\n");

  /* forever loop; abort the program with ^C */
  while ( 1 )  {
    /* you have to initialize for each measure to be done */
    if ( !(h = iptc_init(tablename)) )  {
      printf("Error initializing: %s\n", iptc_strerror(errno));
    ok = 0;    /* we start a new loop */
    gettimeofday(&to, NULL);  /* have a time shoot */

    /* iterate through each chain of the table */
    for (chain = iptc_first_chain(&h), i = 0; 
         chain = iptc_next_chain(&h))  {
      if ( iptc_builtin(chain, h) )
        continue;    /* if it is a built-in chain, ignore it */

      /* ok, read the counters of this chain */
      if ( !(counters = iptc_read_counter(chain, 0, &h)) )  {
         printf("Error reading %s: %s\n", chain, iptc_strerror(errno));

      /* check that we do not have more chains than we can process */
      if ( i >= MAXUSERCHAINS )  {
         printf("Maximum of %d user chains exceeded!!\n", MAXUSERCHAINS);

      /* this chain counter has not been initialized; initialize it */
      if ( bw[i].start == 0 )  {
        bw[i].icnt = counters->bcnt;
        bw[i].start = 1;

      /* this chain has a previous measure; take the current one */
      else  {
        bw[i].ocnt = counters->bcnt;
        if ( bw[i].ocnt == bw[i].icnt )    /* no new bytes flowing? */
          bw[i].bw = 0;                    /* flow is zero */
         /* flow in this chain is:
          *   current bytes count (bw[i].octn)    *minus*
          *   previous bytes count (bw[i].icnt)   *divided by*
          *   128.0 to convert bytes to kbits     *and divided by*
          *   difference in times in seconds      *to get*
          *   flow in kbits/sec that is what we want.
          bw[i].bw = (bw[i].ocnt - bw[i].icnt) / (128.0 * delta(to, ti));

       /* do you want current flow of this chain? initialize previous 
        * bytes count to current bytes count; we get the flow in last 
        * SLEEPTIME elapsed time.
        if ( act_bw )
          bw[i].icnt = bw[i].ocnt;
        ok = 1;    /* ok, we have some measure to show */
      ++i;  /* next chain, please */

    /* we iterate and i == 0; we do not have user chains at all */
    if ( i == 0 )  {
       printf("No user chains to meter!!\n");

   /* do you want to measure current flow? initialize previous time 
    * to actual time; we get the time elapsed in last SLEEPTIME.
    if ( act_bw )
      ti = to;

    /* do we have something to show? ok, display it */
    if ( ok )  {
      totbw = 0;
      for ( j = 0; j < i; ++j )  { 
        totbw = totbw + bw[j].bw;   /* calculate total flow */
      printf("%s%6.1fk:%s ", col[7], totbw, col[8]);  /* display total */
      for ( j = 0; j < i; ++j )  {  /* display flow of each chain in color */
        printf("%s%6.1fk%s ", col[j], bw[j].bw, col[8]);
    sleep(SLEEPTIME);  /* rest a little; you go too fast */
  }             /* give us enough time in order to let some bytes flow */

  exit(0);  /* bye, we have our measures!! */

} /* main */

Write your program and compile as before:

bash# ./ipt-cc bw

Before using the meter we need to set our environment.

First, we have to have at least 2 PCs connected in a network. This is our diagram configuration:

+--------+ eth0       eth0 +--------+
| PC #1  +-----------------+ PC #2  |
+--------+                 +--------+
eth0=           eth0=

Second, we need to install a very nice and useful package called netcat written by Hobbit. This excellent package will help us to inject and receive a flow of bytes between 2 NICs. If you don't have the package in your system, download it from

The version that I use is 1.10-277. To install it follow these instructions:

bash# cp netcat-1.10.tar.gz /usr/local/src
bash# tar xzvf netcat-1.10.tar.gz
bash# cd netcat-1.10

My version requires to make a patch first; check yours if you have a file with a .dif extension and apply it too:

bash# patch -p0 -i netcat-1.10.dif

Next compile the package using make:

bash# make linux

Copy the binary nc to your user bin directory:

bash# cp nc /usr/bin

And also to the second PC in your network:

bash# scp nc

We are going to use netcat to "listen" to a flow of bytes from PC #2 and to "talk" from PC #1. Using tty1 to tty4 consoles on PC #2 let's start netcat to listen from this PC. Go to PC #2 and in tty1 type:

bash# nc -n -v -l -s -p 1001 >/dev/null

netcat must respond with:

listening on [] 1001 ...

This command started netcat to listen from address using port number 1001. Arguments are: -n = use numeric address identification; -v = verbose; -l = listen. All the flow that netcat receives in will be redirected to the "black hole" in /dev/null.

Repeat the command in tty2, tty3 and tty4; change to tty2 using ALT-F2 and after logging in write:

bash# nc -n -v -l -s -p 1002 >/dev/null

Now we are "listening" to the same address but port number 1002.

Go on now with tty3:

bash# nc -n -v -l -s -p 1003 >/dev/null

And tty4:

bash# nc -n -v -l -s -p 1004 >/dev/null

Now we are listening in PC #2, address in ports 1001, 1002, 1003 and 1004.

Come back to PC #1 and let's set the environment to allow iptables to help us to complete our tests:

On PC #1, type the into tty1 as follows:

bash# iptables -F
bash# iptables -X
bash# iptables -N chn_1
bash# iptables -N chn_2
bash# iptables -N chn_3
bash# iptables -N chn_4
bash# iptables -A chn_1 -j ACCEPT
bash# iptables -A chn_2 -j ACCEPT
bash# iptables -A chn_3 -j ACCEPT
bash# iptables -A chn_4 -j ACCEPT
bash# iptables -A OUTPUT -o eth0 -p tcp --dport 1001 -j chn_1
bash# iptables -A OUTPUT -o eth0 -p tcp --dport 1002 -j chn_2
bash# iptables -A OUTPUT -o eth0 -p tcp --dport 1003 -j chn_3
bash# iptables -A OUTPUT -o eth0 -p tcp --dport 1004 -j chn_4

These commands will:

Now start the bw meter using current values:

bash# ./bw -c

It must respond with:

Displaying current flow values ...
   0.0k:    0.0k    0.0k    0.0k    0.0k
   0.0k:    0.0k    0.0k    0.0k    0.0k
   0.0k:    0.0k    0.0k    0.0k    0.0k
   0.0k:    0.0k    0.0k    0.0k    0.0k

It informs that measures are current flows. Every line is a measure taken each SLEEPTIME lapse (1 second in our program). First column (in black) are total flow, next columns (in red, green, orange and blue) are flows in chains chn_1, chn_2, chn_3 and chn_4 respectively. Of course we do not have any flow now. However let bw run and continue reading.

Let's start now one of our byte flows; go to tty2 in PC #1 with ALT-F2 and after logging in, type:

bash# yes 000000000000000000 | nc -n -v -s -p 2001 1001

netcat responds with:

(UNKNOWN) [] 1000 (?) open

Now we have a flow of bytes from PC #1 to PC #2. yes generates a constant flow of zeroes; this flow is piped to netcat through address, port 2001 and sends it to PC #2, address, port 1001 (where PC #2 is listening).

Check now the display of bw in tty1:

7653.2k: 7653.2k    0.0k    0.0k    0.0k
7829.5k: 7829.5k    0.0k    0.0k    0.0k
7786.7k: 7786.7k    0.0k    0.0k    0.0k
7892.1k: 7982.1k    0.0k    0.0k    0.0k

Your mileage can vary depending of the physical characteristics of your system. In mine I have a flow of aproximately 7700 kbits/sec in the first chain chn_1 which corresponds to port number 1001 in PC #2.

Let's start now the second bytes flow; go to tty3 in PC #1 with ALT-F3 and after logging in, type:

bash# yes 000000000000000000 | nc -n -v -s -p 2002 1002

netcat responds with:

(UNKNOWN) [] 1002 (?) open

Now we have 2 flows of bytes from PC #1 to PC #2; one from to and another from to

Now check the display of bw in tty1:

7819.6k: 4144.2k 3675.4k    0.0k    0.0k
8090.5k: 3923.9k 4166.6k    0.0k    0.0k
7794.7k: 3920.8k 3873.9k    0.0k    0.0k
7988.3k: 3754.6k 4233.7k    0.0k    0.0k

Now we have 2 flows; each of them is more or less 50% of the total flow going out of the computer. The Linux kernel tries to balance the bandwidth available between the 2 channels of output.

To continue, start the 2 aditional flows through channels and

In tty4 type:

bash# yes 000000000000000000 | nc -n -v -s -p 2003 1003

In tty5 type:

bash# yes 000000000000000000 | nc -n -v -s -p 2004 1004

The display of bw in tty1 will be something like:

8120.6k: 1705.3k 2354.9k 1898.6k 2161.8k
7765.3k: 1634.2k 2560.2k 2011.4k 1559.5k
7911.9k: 1699.8k 2090.3k 1768.0k 2353.8k
8309.4k: 1734.5k 1999.7k 1999.9k 2575.3k

Total bandwidth is distributed between the 4 channels of flow.

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