question on boost.timer

This is an example from boost.timer,my question is ,in the ouput of
the program,where is the last line of the output come from?

//output:

timer::elapsed_min() reports 0.001 seconds
timre::elapsed_max() reports 2.14748e+006 seconds,which is 59.6523
hours

verify progress_display(0) doesn't divide by zero

0% 10 20 30 40 50 60 70 80 90 100%
|----|----|----|----|----|----|----|----|----|----|
***************************************************

determine 1 second iteration count
8246574 iterations
burn about 1seconds

0% 10 20 30 40 50 60 70 80 90 100%
|----|----|----|----|----|----|----|----|----|----|
***************************************************

burn about 1 seconds again

0% 10 20 30 40 50 60 70 80 90 100%
|----|----|----|----|----|----|----|----|----|----|
***************************************************

Leading string 1 0% 10 20 30 40 50 60 70 80 90
100%
Leading string 2 |----|----|----|----|----|----|----|----|----|----|
Leading string 3 ***************************************************

t1 elapsed: 1.993
t2 elapsed:1.993
t3 elapsed:1.002
t4 elapsed:1.002
t5 elapsed:1.002
t1 and t2 should report the same time (very approximately 2 seconds).
t3,t4 and t5 should report about the same time,
The following elapsed time should be slightly greater than t1.
1.99 s

//end of output
1.99s ==>where is this come from?
Could anybody help?Thanks in advance.

example source[timer_test.cpp];

// timer, job_timer, and progress_display sample program
-------------------//

// Copyright Beman Dawes 1998. Distributed under the Boost
// Software License, Version 1.0. (See accompanying file
// LICENSE_1_0.txt or copy at http://www.boost.org/LICENSE_1_0.txt)

// See http://www.boost.org/libs/timer for documentation.

// Revision History
// 12 Jan 01 Cut time to 1.0 secs to speed regression tests (Beman
Dawes)
// 25 Sep 99 added elapsed_min() and elapsed_max() reporting
// 16 Jul 99 Second beta
// 6 Jul 99 Initial boost version

#include <boost/progress.hpp>
#include <iostream>
#include <climits>

using boost::timer;
using boost::progress_timer;
using boost::progress_display;
using std::cout;
using std::endl;

int main() {

  timer t0; // used only for elapsed_max() and elapsed_min()

  cout << "timer::elapased_min() reports " << t0.elapsed_min() << "
seconds\n";
  cout << "timer::elapased_max() reports " << t0.elapsed_max()
       << " seconds, which is " << t0.elapsed_max()/3600.0 << " hours
\n";

  cout << "\nverify progress_display(0) doesn't divide by zero" <<
endl;
  progress_display zero( 0 ); // verify 0 doesn't divide by zero
  ++zero;

  long loops;
  timer loop_timer;
  const double time = 1.0;

  cout << "\ndetermine " << time << " second iteration count" << endl;
  for ( loops = 0; loops < LONG_MAX
     && loop_timer.elapsed() < time; ++loops ) {}
  cout << loops << " iterations"<< endl;

  long i;
  bool time_waster; // defeat [some] optimizers by storing result here

  progress_timer pt;
  timer t1;
  timer t4;
  timer t5;

  cout << "\nburn about " << time << " seconds" << endl;
  progress_display pd( loops );
  for ( i = loops;i>0; i--; )
    { time_waster = loop_timer.elapsed() < time; ++pd; }

  timer t2( t1 );
  timer t3;
  t4 = t3;
  t5.restart();

  cout << "\nburn about " << time << " seconds again" << endl;
  pd.restart( loops );
  for ( i = loops;i>0; i--; )
    { time_waster = loop_timer.elapsed() < time; ++pd; }

  if ( time_waster ) cout << ' '; // using time_waster quiets
compiler warnings
  progress_display pd2( 50, cout, "\nLead string 1 ", "Lead string 2
", "Lead string 3 " );
  for ( ; pd2.count() < 50; ++pd2 ) {}

  cout << "\nt1 elapsed: " << t1.elapsed() << '\n';
  cout << "t2 elapsed: " << t2.elapsed() << '\n';
  cout << "t3 elapsed: " << t3.elapsed() << '\n';
  cout << "t4 elapsed: " << t4.elapsed() << '\n';
  cout << "t5 elapsed: " << t5.elapsed() << '\n';
  cout << "t1 and t2 should report the same times (very approximately
"
       << 2*time << " seconds).\n";
  cout << "t3, t4 and t5 should report about the same times,\n";
  cout << "and these should be about half the t1 and t2 times.\n";
  cout << "The following elapsed time should be slightly greater than
t1."
       << endl;
  return 0;
  } // main