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296 lines
12 KiB
C++
296 lines
12 KiB
C++
// negative_binomial_example1.cpp
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// Copyright Paul A. Bristow 2007.
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// Use, modification and distribution are subject to the
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// Boost Software License, Version 1.0.
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// (See accompanying file LICENSE_1_0.txt
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// or copy at http://www.boost.org/LICENSE_1_0.txt)
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// Example 1 of using negative_binomial distribution.
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//[negative_binomial_eg1_1
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/*`
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First we need to #define macros to control the error and discrete handling policies.
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For this simple example, we want to avoid throwing an exception (the default policy) and just return infinity.
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We want to treat the distribution as if it was continuous, so we choose a policy of real,
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rather than the default policy integer_round_outwards.
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*/
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#define BOOST_MATH_OVERFLOW_ERROR_POLICY ignore_error
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#define BOOST_MATH_DISCRETE_QUANTILE_POLICY real
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/*`After that (vital to #include *after* #defines), we need some includes
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to provide easy access to the negative binomial distribution,
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and some std library iostream, of course.
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*/
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#include <boost/math/distributions/negative_binomial.hpp>
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// for negative_binomial_distribution
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using boost::math::negative_binomial; // typedef provides default type is double.
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using ::boost::math::cdf;
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using ::boost::math::pdf; // Probability mass of negative_binomial.
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using ::boost::math::quantile;
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#include <iostream>
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using std::cout;
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using std::endl;
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using std::noshowpoint;
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using std::fixed;
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using std::right;
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using std::left;
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#include <iomanip>
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using std::setprecision;
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using std::setw;
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#include <limits>
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using std::numeric_limits;
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int main()
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{
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cout << "Example 1 using the Negative Binomial Distribution.";
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#if defined(__FILE__) && defined(__TIMESTAMP__)
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cout << " " << __FILE__ << ' ' << __TIMESTAMP__ << "\n";
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#endif
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cout << endl;
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cout.precision(5);
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// None of the values calculated have a useful accuracy as great this, but
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// INF shows wrongly with < 5 !
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// https://connect.microsoft.com/VisualStudio/feedback/ViewFeedback.aspx?FeedbackID=240227
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try
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{
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double sales_quota = 5; // Pat's sales quota - successes (r).
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double success_fraction = 0.4; // success_fraction (p) - so failure_fraction is 0.6.
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negative_binomial nb(sales_quota, success_fraction); // double by default.
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int all_houses = 30; // The number of houses on the estate.
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cout <<"Selling candy bars - using the negative binomial distribution."
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<< "\n""by Dr. Diane Evans,"
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"\n""Professor of Mathematics at Rose-Hulman Institute of Technology,"
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<< "\n""see http://en.wikipedia.org/wiki/Negative_binomial_distribution""\n"
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<< endl;
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cout << "Pat has a sales per house success rate of " << success_fraction
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<< ".""\n""Therefore he would, on average, sell " << nb.success_fraction() * 100
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<< " bars after trying 100 houses." << endl;
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cout << "With a success rate of " << nb.success_fraction()
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<< ", he might expect, on average,""\n"
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" to need to visit about " << success_fraction * all_houses
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<< " houses in order to sell all " << nb.successes() << " candy bars. " << endl;
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/*`
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To finish on or before the 8th house, Pat must finish at the 5th, 6th, 7th or 8th house.
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(Obviously he could not finish on fewer than 5 houses because he must sell 5 candy bars.
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So the 5th house is the first that he could possibly finish on).
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The probability that he will finish on EXACTLY on any house
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is the Probability Density Function (pdf).
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*/
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cout << "Probability that Pat finishes on the " << sales_quota << "th house is "
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<< "f(5) = " << pdf(nb, nb.successes()) << endl;
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cout << "Probability that Pat finishes on the 6th house is "
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<< pdf(nb, 6 - sales_quota) << endl;
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cout << "Probability that Pat finishes on the 7th house is "
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<< pdf(nb, 7 - sales_quota) << endl;
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cout << "Probability that Pat finishes on the 8th house is "
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<< pdf(nb, 8 - sales_quota) << endl;
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// Sum of the probabilities for these houses is the Cumulative Distribution Function (cdf).
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cout << "Probability that Pat finishes on or before the 8th house is sum "
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"\n" << "pdf(sales_quota) + pdf(6) + pdf(7) + pdf(8) = "
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// Sum each of the mass/density probabilities for houses sales_quota = 5, 6, 7, & 8.
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<< pdf(nb, sales_quota - sales_quota) // 0
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+ pdf(nb, 6 - sales_quota) // 1
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+ pdf(nb, 7 - sales_quota) // 2
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+ pdf(nb, 8 - sales_quota) // 3
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<< endl;
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// Or using the negative binomial *cumulative* distribution function
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// (cdf instead sum of the pdfs):
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cout << "\n""Probability of selling his quota of " << sales_quota
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<< " candy bars""\n""on or before the " << 8 << "th house is "
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<< cdf(nb, 8 - sales_quota) << endl;
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cout << "\n""Probability that Pat finishes exactly on the 10th house is "
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<< pdf(nb, 10 - sales_quota) << endl;
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cout << "\n""Probability of selling his quota of " << sales_quota
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<< " candy bars""\n""on or before the " << 10 << "th house is "
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<< cdf(nb, 10 - sales_quota) << endl;
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cout << "Probability that Pat finishes on the 11th house is "
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<< pdf(nb, 11 - sales_quota) << endl;
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cout << "\n""Probability of selling his quota of " << sales_quota
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<< " candy bars""\n""on or before the " << 11 << "th house is "
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<< cdf(nb, 11 - sales_quota) << endl;
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cout << "Probability that Pat finishes on the 12th house is "
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<< pdf(nb, 12 - sales_quota) << endl;
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cout << "\n""Probability of selling his quota of " << sales_quota
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<< " candy bars""\n""on or before the " << 12 << "th house is "
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<< cdf(nb, 12 - sales_quota) << endl;
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/*`
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Finally consider the risk of Pat not selling his quota of 5 bars
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even after visiting all the houses.
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Calculate the probability that he would sell on the (non-existent) last-plus-1 house.
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*/
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cout << "\n""Probability of selling his quota of " << sales_quota
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<< " candy bars""\n""on or before the " << all_houses + 1 << "th house is "
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<< cdf(nb, all_houses + 1 - sales_quota) << endl;
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// So the risk of failing even at the 31th (non-existent) house is 1 - this probability.
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cout << "\n""Probability of failing to sell his quota of " << sales_quota
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<< " candy bars""\n""even after visiting all " << all_houses << " houses is "
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<< 1 - cdf(nb, all_houses - sales_quota + 1) << endl;
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double p = cdf(nb, (8 - sales_quota));
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cout << "Probability of meeting sales quota on or before 8th house is "<< p << endl;
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// Probability of meeting sales quota on or before 8th house is 0.174
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cout << "If the confidence of meeting sales quota is " << p
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<< ", then the finishing house is " << quantile(nb, p) + sales_quota << endl;
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/*`
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Also try demanding absolute certainty that all 5 will be sold,
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which implies an infinite number of trials.
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(Of course, there are only 30 houses on the estate,
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so he can't even be certain of selling his quota).
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*/
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cout << "If the confidence of meeting sales quota is " << 1.
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<< ", then the finishing house is " << quantile(nb, 1) + sales_quota << endl;
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// 1.#INF == infinity.
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cout << "If the confidence of meeting sales quota is " << 0.
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<< ", then the finishing house is " << quantile(nb, 0.) + sales_quota << endl;
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cout << "If the confidence of meeting sales quota is " << 0.5
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<< ", then the finishing house is " << quantile(nb, 0.5) + sales_quota << endl;
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cout << "If the confidence of meeting sales quota is " << 1 - 0.00151 // 30 th
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<< ", then the finishing house is " << quantile(nb, 1 - 0.00151) + sales_quota << endl;
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/*
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If the opposite is true, we don't want to assume any confidence, then this is tantamount
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to assuming that all the first sales_quota trials will be successful sales.
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*/
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cout << "If confidence of meeting quota is zero (we assume all houses are successful sales)"
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", then finishing house is " << sales_quota << endl;
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double ps[] = {0., 0.001, 0.01, 0.05, 0.1, 0.5, 0.9, 0.95, 0.99, 0.999, 1.};
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// Confidence as fraction = 1-alpha, as percent = 100 * (1-alpha[i]) %
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for (int i = 0; i < sizeof(ps)/sizeof(ps[0]); i++)
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{
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cout << "If confidence of meeting quota is " << ps[i]
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<< ", then finishing house is " << ceil(quantile(nb, ps[i])) + sales_quota
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<< endl;
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}
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cout << "If we demand a confidence of meeting sales quota of unity"
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", then we can never be certain of selling 5 bars, so the finishing house is infinite!" << endl;
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cout << "\n""Probability (%) House for (last) " << sales_quota << " th sale." << endl;
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for (int i = (int)sales_quota; i < all_houses + 5; i++)
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{
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cout << left << setw(6) << cdf(nb, i - sales_quota) << " " << setw(3) << i<< endl;
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}
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cout << endl;
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}
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catch(const std::exception& e)
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{ // Since we have set an overflow policy of ignore_error,
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// an exception should never be thrown.
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std::cout << "\n""Message from thrown exception was:\n " << e.what() << std::endl;
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}
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return 0;
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} // int main()
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//] [/ negative_binomial_eg1_1]
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/*
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Output is:
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Example 1 using the Negative Binomial Distribution. ..\..\..\..\..\..\boost-sandbox\math_toolkit\libs\math\example\negative_binomial_example1.cpp Sun Aug 12 22:28:11 2007
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Selling candy bars - using the negative binomial distribution.
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by Dr. Diane Evans,
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Professor of Mathematics at Rose-Hulman Institute of Technology,
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see http://en.wikipedia.org/wiki/Negative_binomial_distribution
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Pat has a sales per house success rate of 0.4.
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Therefore he would, on average, sell 40 bars after trying 100 houses.
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With a success rate of 0.4, he might expect, on average,
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to need to visit about 12 houses in order to sell all 5 candy bars.
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Probability that Pat finishes on the 5th house is f(5) = 0.10033
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Probability that Pat finishes on the 6th house is 0.03072
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Probability that Pat finishes on the 7th house is 0.055296
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Probability that Pat finishes on the 8th house is 0.077414
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Probability that Pat finishes on or before the 8th house is sum
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pdf(sales_quota) + pdf(6) + pdf(7) + pdf(8) = 0.17367
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Probability of selling his quota of 5 candy bars
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on or before the 8th house is 0.17367
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Probability that Pat finishes exactly on the 10th house is 0.10033
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Probability of selling his quota of 5 candy bars
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on or before the 10th house is 0.3669
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Probability that Pat finishes on the 11th house is 0.10033
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Probability of selling his quota of 5 candy bars
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on or before the 11th house is 0.46723
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Probability that Pat finishes on the 12th house is 0.094596
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Probability of selling his quota of 5 candy bars
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on or before the 12th house is 0.56182
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Probability of selling his quota of 5 candy bars
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on or before the 31th house is 0.99897
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Probability of failing to sell his quota of 5 candy bars
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even after visiting all 30 houses is 0.0010314
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Probability of meeting sales quota on or before 8th house is 0.17367
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If the confidence of meeting sales quota is 0.17367, then the finishing house is 8
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If the confidence of meeting sales quota is 1, then the finishing house is 1.#INF
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If the confidence of meeting sales quota is 0, then the finishing house is 5
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If the confidence of meeting sales quota is 0.5, then the finishing house is 11.337
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If the confidence of meeting sales quota is 0.99849, then the finishing house is 30
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If confidence of meeting quota is zero (we assume all houses are successful sales), then finishing house is 5
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If confidence of meeting quota is 0, then finishing house is 5
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If confidence of meeting quota is 0.001, then finishing house is 5
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If confidence of meeting quota is 0.01, then finishing house is 5
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If confidence of meeting quota is 0.05, then finishing house is 7
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If confidence of meeting quota is 0.1, then finishing house is 8
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If confidence of meeting quota is 0.5, then finishing house is 12
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If confidence of meeting quota is 0.9, then finishing house is 18
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If confidence of meeting quota is 0.95, then finishing house is 21
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If confidence of meeting quota is 0.99, then finishing house is 25
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If confidence of meeting quota is 0.999, then finishing house is 32
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If confidence of meeting quota is 1, then finishing house is 1.#INF
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If we demand a confidence of meeting sales quota of unity,
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then we can never be certain of selling 5 bars,
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so the finishing house is infinite!
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Probability (%) House for (last) 5 th sale.
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0.01024 5
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0.04096 6
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0.096256 7
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0.17367 8
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0.26657 9
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0.3669 10
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0.46723 11
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0.56182 12
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0.64696 13
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0.72074 14
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0.78272 15
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0.83343 16
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0.874 17
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0.90583 18
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0.93039 19
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0.94905 20
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0.96304 21
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0.97342 22
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0.98103 23
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0.98655 24
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0.99053 25
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0.99337 26
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0.99539 27
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0.99681 28
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0.9978 29
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0.99849 30
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0.99897 31
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0.9993 32
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0.99952 33
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0.99968 34
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*/
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