[Boost-commit] svn:boost r38818 - sandbox/math_toolkit/libs/math/example

From: pbristow_at_[hidden]
Date: 2007-08-21 09:16:01

Author: pbristow
Date: 2007-08-21 09:16:01 EDT (Tue, 21 Aug 2007)
New Revision: 38818
URL: http://svn.boost.org/trac/boost/changeset/38818

Log:
This replaced binomial_example_3 which can be removed.  TODO
Added:
   sandbox/math_toolkit/libs/math/example/binomial_example_NAG_C.cpp   (contents, props changed)

Added: sandbox/math_toolkit/libs/math/example/binomial_example_NAG_C.cpp
==============================================================================
--- (empty file)
+++ sandbox/math_toolkit/libs/math/example/binomial_example_NAG_C.cpp	2007-08-21 09:16:01 EDT (Tue, 21 Aug 2007)
@@ -0,0 +1,91 @@
+// Copyright Paul A. 2007
+// Copyright John Maddock 2007
+// 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)
+
+// Simple example of computing probabilitiesfor a binomial random variable.
+// Replication of source nag_binomial_dist (g01bjc).
+
+// Shows how to replace NAG C library calls by Boost Math Toolkit C++ calls.
+// Note that the default policy does not replicate the way that NAG
+// library calls handle 'bad' arguments, but you can define policies that do,
+// as well as other policies that may suit your application even better.
+// See the examples of changing default policies for details.
+
+#include <boost/math/distributions/binomial.hpp>
+
+#include <iostream>
+  using std::cout; using std::endl; using std::ios; using std::showpoint;
+#include <iomanip>
+  using std::fixed; using std::setw;
+
+int main()
+{
+  cout << "Using the binomial distribution to replicate a NAG library call." << endl;
+  using boost::math::binomial_distribution;
+
+  // This replicates the computation of the examples of using nag-binomial_dist
+  // using g01bjc in section g01 Somple Calculations on Statistical Data.
+  // http://www.nag.co.uk/numeric/cl/manual/pdf/G01/g01bjc.pdf
+  // Program results section 8.3 page 3.g01bjc.3
+    //8.2. Program Data
+    //g01bjc Example Program Data
+    //4 0.50 2 : n, p, k
+    //19 0.44 13
+    //100 0.75 67
+    //2000 0.33 700
+    //8.3. Program Results
+    //g01bjc Example Program Results
+    //n p k plek pgtk peqk
+    //4 0.500 2 0.68750 0.31250 0.37500
+    //19 0.440 13 0.99138 0.00862 0.01939
+    //100 0.750 67 0.04460 0.95540 0.01700
+    //2000 0.330 700 0.97251 0.02749 0.00312
+
+  cout.setf(ios::showpoint); // Trailing zeros to show significant decimal digits.
+  cout.precision(5); // Might calculate this from trials in distribution?
+  cout << fixed;
+  //  Binomial distribution.
+
+  // Note  that  cdf(dist, k) is equivalent to NAG library plek probability of <= k
+  // cdf(complement(dist, k)) is equivalent to NAG library pgtk probability of > k
+  //             pdf(dist, k) is equivalent to NAG library peqk probability of == k
+
+  cout << " n        p     k     plek     pgtk     peqk " << endl;
+  binomial_distribution<>my_dist(4, 0.5);
+  cout << setw(4) << (int)my_dist.trials() << "  " << my_dist.success_fraction()
+  << "   " << 2 << "  " << cdf(my_dist, 2) << "  "
+  << cdf(complement(my_dist, 2)) << "  " << pdf(my_dist, 2) << endl;
+
+  binomial_distribution<>two(19, 0.440);
+  cout << setw(4) << (int)two.trials() <<  "  "  << two.success_fraction()
+    << "  " << 13 << "  " << cdf(two, 13) << "  "
+    << cdf(complement(two, 13)) << "  " << pdf(two, 13) << endl;
+
+  binomial_distribution<>three(100, 0.750);
+  cout << setw(4) << (int)three.trials() << "  " << three.success_fraction()
+    << "  " << 67 << "  " << cdf(three, 67) << "  " << cdf(complement(three, 67))
+    << "  " << pdf(three, 67) << endl;
+  binomial_distribution<>four(2000, 0.330);
+  cout << setw(4) << (int)four.trials() <<  "  "  << four.success_fraction()
+  << " " << 700 << "  "
+    << cdf(four, 700) << "  " << cdf(complement(four, 700))
+    << "  " << pdf(four, 700) << endl;
+
+  return 0;
+} // int main()
+
+/*
+
+Example of using the binomial distribution to replicate a NAG library call.
+ n        p     k     plek     pgtk     peqk
+   4  0.50000   2  0.68750  0.31250  0.37500
+  19  0.44000  13  0.99138  0.00862  0.01939
+ 100  0.75000  67  0.04460  0.95540  0.01700
+2000  0.33000 700  0.97251  0.02749  0.00312
+
+
+ */
+