WSJT-X/boost/libs/math/example/find_root_example.cpp

// find_root_example.cpp

// Copyright Paul A. Bristow 2007, 2010.

// Use, modification and distribution are subject to 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)

// Example of using root finding.

// Note that this file contains Quickbook mark-up as well as code
// and comments, don't change any of the special comment mark-ups!

//[root_find1
/*`
First we need some includes to access the normal distribution
(and some std output of course).
*/

#include <boost/math/tools/roots.hpp> // root finding.

#include <boost/math/distributions/normal.hpp> // for normal_distribution
  using boost::math::normal; // typedef provides default type is double.

#include <iostream>
  using std::cout; using std::endl; using std::left; using std::showpoint; using std::noshowpoint;
#include <iomanip>
  using std::setw; using std::setprecision;
#include <limits>
  using std::numeric_limits;
#include <stdexcept>
  

//] //[/root_find1]

int main()
{
  cout << "Example: Normal distribution, root finding.";
  try
  {

//[root_find2

/*`A machine is set to pack 3 kg of ground beef per pack.
Over a long period of time it is found that the average packed was 3 kg
with a standard deviation of 0.1 kg.
Assuming the packing is normally distributed,
we can find the fraction (or %) of packages that weigh more than 3.1 kg.
*/

double mean = 3.; // kg
double standard_deviation = 0.1; // kg
normal packs(mean, standard_deviation);

double max_weight = 3.1; // kg
cout << "Percentage of packs > " << max_weight << " is "
<< cdf(complement(packs, max_weight)) << endl; // P(X > 3.1)

double under_weight = 2.9;
cout <<"fraction of packs <= " << under_weight << " with a mean of " << mean
  << " is " << cdf(complement(packs, under_weight)) << endl;
// fraction of packs <= 2.9 with a mean of 3 is 0.841345
// This is 0.84 - more than the target 0.95
// Want 95% to be over this weight, so what should we set the mean weight to be?
// KK StatCalc says:
double over_mean = 3.0664;
normal xpacks(over_mean, standard_deviation);
cout << "fraction of packs >= " << under_weight
<< " with a mean of " << xpacks.mean()
  << " is " << cdf(complement(xpacks, under_weight)) << endl;
// fraction of packs >= 2.9 with a mean of 3.06449 is 0.950005
double under_fraction = 0.05;  // so 95% are above the minimum weight mean - sd = 2.9
double low_limit = standard_deviation;
double offset = mean - low_limit - quantile(packs, under_fraction);
double nominal_mean = mean + offset;

normal nominal_packs(nominal_mean, standard_deviation);
cout << "Setting the packer to " << nominal_mean << " will mean that "
  << "fraction of packs >= " << under_weight
  << " is " << cdf(complement(nominal_packs, under_weight)) << endl;

/*`
Setting the packer to 3.06449 will mean that fraction of packs >= 2.9 is 0.95.

Setting the packer to 3.13263 will mean that fraction of packs >= 2.9 is 0.99,
but will more than double the mean loss from 0.0644 to 0.133.

Alternatively, we could invest in a better (more precise) packer with a lower standard deviation.

To estimate how much better (how much smaller standard deviation) it would have to be,
we need to get the 5% quantile to be located at the under_weight limit, 2.9
*/
double p = 0.05; // wanted p th quantile.
cout << "Quantile of " << p << " = " << quantile(packs, p)
  << ", mean = " << packs.mean() << ", sd = " << packs.standard_deviation() << endl; //
/*`
Quantile of 0.05 = 2.83551, mean = 3, sd = 0.1

With the current packer (mean = 3, sd = 0.1), the 5% quantile is at 2.8551 kg,
a little below our target of 2.9 kg.
So we know that the standard deviation is going to have to be smaller.

Let's start by guessing that it (now 0.1) needs to be halved, to a standard deviation of 0.05
*/
normal pack05(mean, 0.05);
cout << "Quantile of " << p << " = " << quantile(pack05, p)
  << ", mean = " << pack05.mean() << ", sd = " << pack05.standard_deviation() << endl;

cout <<"Fraction of packs >= " << under_weight << " with a mean of " << mean
  << " and standard deviation of " << pack05.standard_deviation()
  << " is " << cdf(complement(pack05, under_weight)) << endl;
//
/*`
Fraction of packs >= 2.9 with a mean of 3 and standard deviation of 0.05 is 0.9772

So 0.05 was quite a good guess, but we are a little over the 2.9 target,
so the standard deviation could be a tiny bit more. So we could do some
more guessing to get closer, say by increasing to 0.06
*/

normal pack06(mean, 0.06);
cout << "Quantile of " << p << " = " << quantile(pack06, p)
  << ", mean = " << pack06.mean() << ", sd = " << pack06.standard_deviation() << endl;

cout <<"Fraction of packs >= " << under_weight << " with a mean of " << mean
  << " and standard deviation of " << pack06.standard_deviation()
  << " is " << cdf(complement(pack06, under_weight)) << endl;
/*`
Fraction of packs >= 2.9 with a mean of 3 and standard deviation of 0.06 is 0.9522

Now we are getting really close, but to do the job properly,
we could use root finding method, for example the tools provided, and used elsewhere,
in the Math Toolkit, see __root_finding_without_derivatives.

But in this normal distribution case, we could be even smarter and make a direct calculation.
*/
//] [/root_find2]

  }
  catch(const std::exception& e)
  { // Always useful to include try & catch blocks because default policies
    // are to throw exceptions on arguments that cause errors like underflow, overflow.
    // Lacking try & catch blocks, the program will abort without a message below,
    // which may give some helpful clues as to the cause of the exception.
    std::cout <<
      "\n""Message from thrown exception was:\n   " << e.what() << std::endl;
  }
  return 0;
}  // int main()

/*
Output is:

//[root_find_output

Autorun "i:\boost-06-05-03-1300\libs\math\test\Math_test\debug\find_root_example.exe"
Example: Normal distribution, root finding.Percentage of packs > 3.1 is 0.158655
fraction of packs <= 2.9 with a mean of 3 is 0.841345
fraction of packs >= 2.9 with a mean of 3.0664 is 0.951944
Setting the packer to 3.06449 will mean that fraction of packs >= 2.9 is 0.95
Quantile of 0.05 = 2.83551, mean = 3, sd = 0.1
Quantile of 0.05 = 2.91776, mean = 3, sd = 0.05
Fraction of packs >= 2.9 with a mean of 3 and standard deviation of 0.05 is 0.97725
Quantile of 0.05 = 2.90131, mean = 3, sd = 0.06
Fraction of packs >= 2.9 with a mean of 3 and standard deviation of 0.06 is 0.95221

//] [/root_find_output]
*/