
BoostCommit : 
From: pbristow_at_[hidden]
Date: 20070916 11:24:38
Author: pbristow
Date: 20070916 11:24:38 EDT (Sun, 16 Sep 2007)
New Revision: 39323
URL: http://svn.boost.org/trac/boost/changeset/39323
Log:
Some improvements but more needed.
Text files modified:
sandbox/math_toolkit/libs/math/example/negative_binomial_example1.cpp  87 +++++++++++++++++++++
1 files changed, 47 insertions(+), 40 deletions()
Modified: sandbox/math_toolkit/libs/math/example/negative_binomial_example1.cpp
==============================================================================
 sandbox/math_toolkit/libs/math/example/negative_binomial_example1.cpp (original)
+++ sandbox/math_toolkit/libs/math/example/negative_binomial_example1.cpp 20070916 11:24:38 EDT (Sun, 16 Sep 2007)
@@ 13,7 +13,8 @@
/*`
First we need to #define macros to control the error and discrete handling policies.
For this simple example, we want to avoid throwing an exception (the default policy) and just return infinity.
+For this simple example, we want to avoid throwing
+an exception (the default policy) and just return infinity.
We want to treat the distribution as if it was continuous, so we choose a policy of real,
rather than the default policy integer_round_outwards.
*/
@@ 65,27 +66,27 @@
int all_houses = 30; // The number of houses on the estate.
cout <<"Selling candy bars  using the negative binomial distribution."
 << "\n""by Dr. Diane Evans,"
 "\n""Professor of Mathematics at RoseHulman Institute of Technology,"
 << "\n""see http://en.wikipedia.org/wiki/Negative_binomial_distribution""\n"
+ << "\nby Dr. Diane Evans,"
+ "\nProfessor of Mathematics at RoseHulman Institute of Technology,"
+ << "\nsee http://en.wikipedia.org/wiki/Negative_binomial_distribution\n"
<< endl;
cout << "Pat has a sales per house success rate of " << success_fraction
 << ".""\n""Therefore he would, on average, sell " << nb.success_fraction() * 100
+ << ".\nTherefore he would, on average, sell " << nb.success_fraction() * 100
<< " bars after trying 100 houses." << endl;
cout << "With a success rate of " << nb.success_fraction()
 << ", he might expect, on average,""\n"
+ << ", he might expect, on average,\n"
" to need to visit about " << success_fraction * all_houses
<< " houses in order to sell all " << nb.successes() << " candy bars. " << endl;
/*`
To finish on or before the 8th house, Pat must finish at the 5th, 6th, 7th or 8th house.
(Obviously he could not finish on fewer than 5 houses because he must sell 5 candy bars.
So the 5th house is the first that he could possibly finish on).
+So the 5th house is the first that he could possibly finish on).
The probability that he will finish on EXACTLY on any house
is the Probability Density Function (pdf).
*/
cout << "Probability that Pat finishes on the " << sales_quota << "th house is "
 << "f(5) = " << pdf(nb, 5  sales_quota) << endl;
+ << pdf(nb, 5  sales_quota) << endl;
cout << "Probability that Pat finishes on the 6th house is "
<< pdf(nb, 6  sales_quota) << endl;
cout << "Probability that Pat finishes on the 7th house is "
@@ 105,26 +106,26 @@
// Or using the negative binomial *cumulative* distribution function
// (cdf instead sum of the pdfs):
 cout << "\n""Probability of selling his quota of " << sales_quota
 << " candy bars""\n""on or before the " << 8 << "th house is "
+ cout << "\nProbability of selling his quota of " << sales_quota
+ << " candy bars\non or before the " << 8 << "th house is "
<< cdf(nb, 8  sales_quota) << endl;
 cout << "\n""Probability that Pat finishes exactly on the 10th house is "
+ cout << "\nProbability that Pat finishes exactly on the 10th house is "
<< pdf(nb, 10  sales_quota) << endl;
 cout << "\n""Probability of selling his quota of " << sales_quota
 << " candy bars""\n""on or before the " << 10 << "th house is "
+ cout << "\nProbability of selling his quota of " << sales_quota
+ << " candy bars\non or before the " << 10 << "th house is "
<< cdf(nb, 10  sales_quota) << endl;
cout << "Probability that Pat finishes on the 11th house is "
<< pdf(nb, 11  sales_quota) << endl;
 cout << "\n""Probability of selling his quota of " << sales_quota
 << " candy bars""\n""on or before the " << 11 << "th house is "
+ cout << "\nProbability of selling his quota of " << sales_quota
+ << " candy bars\non or before the " << 11 << "th house is "
<< cdf(nb, 11  sales_quota) << endl;
cout << "Probability that Pat finishes on the 12th house is "
<< pdf(nb, 12  sales_quota) << endl;
 cout << "\n""Probability of selling his quota of " << sales_quota
 << " candy bars""\n""on or before the " << 12 << "th house is "
+ cout << "\nProbability of selling his quota of " << sales_quota
+ << " candy bars\non or before the " << 12 << "th house is "
<< cdf(nb, 12  sales_quota) << endl;
/*`
@@ 132,13 +133,18 @@
even after visiting all the houses.
Calculate the probability that he would sell on the (nonexistent) lastplus1 house.
*/
 cout << "\n""Probability of selling his quota of " << sales_quota
 << " candy bars""\n""on or before the " << all_houses + 1 << "th house is "
 << cdf(nb, all_houses + 1  sales_quota) << endl;
 // So the risk of failing even at the 31th (nonexistent) house is 1  this probability.
 cout << "\n""Probability of failing to sell his quota of " << sales_quota
 << " candy bars""\n""even after visiting all " << all_houses << " houses is "
 << 1  cdf(nb, all_houses  sales_quota + 1) << endl;
+ cout << "\nProbability of selling his quota of " << sales_quota
+ << " candy bars\non or before the " << all_houses << "th house is "
+ << cdf(nb, all_houses  sales_quota) << endl;
+
+/*`So the risk of failing even at the 31th (nonexistent) house is 1  this probability,
+ ``1  cdf(nb, all_houses  sales_quota``
+But using this expression may cause serious inaccuracy,
+so it would be much better to use the complement of the cdf:
+*/
+ cout << "\nProbability of failing to sell his quota of " << sales_quota
+ << " candy bars\neven after visiting all " << all_houses << " houses is "
+ << cdf(complement(nb, all_houses  sales_quota)) << endl;
double p = cdf(nb, (8  sales_quota));
cout << "Probability of meeting sales quota on or before 8th house is "<< p << endl;
@@ 146,7 +152,7 @@
cout << "If the confidence of meeting sales quota is " << p
<< ", then the finishing house is " << quantile(nb, p) + sales_quota << endl;
/*`
Also try demanding absolute certainty that all 5 will be sold,
+Also demanding absolute certainty that all 5 will be sold,
which implies an infinite number of trials.
(Of course, there are only 30 houses on the estate,
so he can't even be certain of selling his quota).
@@ 182,7 +188,7 @@
cout << "If we demand a confidence of meeting sales quota of unity"
", then we can never be certain of selling 5 bars, so the finishing house is infinite!" << endl;
 cout << "\n""Probability (%) House for (last) " << sales_quota << " th sale." << endl;
+ cout << "\nProbability (%) House for (last) " << sales_quota << " th sale." << endl;
for (int i = (int)sales_quota; i < all_houses + 5; i++)
{
cout << left << setw(6) << cdf(nb, i  sales_quota) << " " << setw(3) << i<< endl;
@@ 192,7 +198,7 @@
catch(const std::exception& e)
{ // Since we have set an overflow policy of ignore_error,
// an exception should never be thrown.
 std::cout << "\n""Message from thrown exception was:\n " << e.what() << std::endl;
+ std::cout << "\nMessage from thrown exception was:\n " << e.what() << std::endl;
}
return 0;
} // int main()
@@ 203,7 +209,7 @@
Output is:
Example 1 using the Negative Binomial Distribution. ..\..\..\..\..\..\boostsandbox\math_toolkit\libs\math\example\negative_binomial_example1.cpp Sun Aug 12 22:28:11 2007
+Example 1 using the Negative Binomial Distribution. ..\..\..\..\..\..\boostsandbox\math_toolkit\libs\math\example\negative_binomial_example1.cpp Sun Sep 16 11:52:52 2007
Selling candy bars  using the negative binomial distribution.
by Dr. Diane Evans,
Professor of Mathematics at RoseHulman Institute of Technology,
@@ 212,7 +218,7 @@
Therefore he would, on average, sell 40 bars after trying 100 houses.
With a success rate of 0.4, he might expect, on average,
to need to visit about 12 houses in order to sell all 5 candy bars.
Probability that Pat finishes on the 5th house is f(5) = 0.10033
+Probability that Pat finishes on the 5th house is 0.01024
Probability that Pat finishes on the 6th house is 0.03072
Probability that Pat finishes on the 7th house is 0.055296
Probability that Pat finishes on the 8th house is 0.077414
@@ 230,9 +236,11 @@
Probability of selling his quota of 5 candy bars
on or before the 12th house is 0.56182
Probability of selling his quota of 5 candy bars
on or before the 31th house is 0.99897
+on or before the 30th house is 0.99849
+Probability of failing to sell his quota of 5 candy bars
+even after visiting all 30 houses is 0.0015101
Probability of failing to sell his quota of 5 candy bars
even after visiting all 30 houses is 0.0010314
+even after visiting all 30 houses is 0.0015101
Probability of meeting sales quota on or before 8th house is 0.17367
If the confidence of meeting sales quota is 0.17367, then the finishing house is 8
If the confidence of meeting sales quota is 1, then the finishing house is 1.#INF
@@ 251,23 +259,21 @@
If confidence of meeting quota is 0.99, then finishing house is 25
If confidence of meeting quota is 0.999, then finishing house is 32
If confidence of meeting quota is 1, then finishing house is 1.#INF
If we demand a confidence of meeting sales quota of unity,
then we can never be certain of selling 5 bars,
so the finishing house is infinite!
+If we demand a confidence of meeting sales quota of unity, then we can never be certain of selling 5 bars, so the finishing house is infinite!
Probability (%) House for (last) 5 th sale.
0.01024 5
0.04096 6
0.096256 7
+0.096256 7
0.17367 8
0.26657 9
0.3669 10
+0.3669 10
0.46723 11
0.56182 12
0.64696 13
0.72074 14
0.78272 15
0.83343 16
0.874 17
+0.874 17
0.90583 18
0.93039 19
0.94905 20
@@ 279,12 +285,13 @@
0.99337 26
0.99539 27
0.99681 28
0.9978 29
+0.9978 29
0.99849 30
0.99897 31
0.9993 32
+0.9993 32
0.99952 33
0.99968 34
+0.99968 34
+
*/
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