![]() |
Home | Libraries | People | FAQ | More |
Imagine you have a critical component that you know will fail in 1 in
N "uses" (for some suitable definition of "use").
You may want to schedule routine replacement of the component so that
its chance of failure between routine replacements is less than P%. If
the failures follow a binomial distribution (each time the component
is "used" it either fails or does not) then the static member
function binomial_distibution<>::find_maximum_number_of_trials
can be used to estimate the maximum number of "uses" of that
component for some acceptable risk level alpha.
The example program binomial_sample_sizes.cpp demonstrates its usage. It centres on a routine that prints out a table of maximum sample sizes for various probability thresholds:
void find_max_sample_size( double p, // success ratio. unsigned successes) // Total number of observed successes permitted. {
The routine then declares a table of probability thresholds: these are the maximum acceptable probability that successes or fewer events will be observed. In our example, successes will be always zero, since we want no component failures, but in other situations non-zero values may well make sense.
double alpha[] = { 0.5, 0.25, 0.1, 0.05, 0.01, 0.001, 0.0001, 0.00001 };
Much of the rest of the program is pretty-printing, the important part is in the calculation of maximum number of permitted trials for each value of alpha:
for(unsigned i = 0; i < sizeof(alpha)/sizeof(alpha[0]); ++i) { // Confidence value: cout << fixed << setprecision(3) << setw(10) << right << 100 * (1-alpha[i]); // calculate trials: double t = binomial::find_maximum_number_of_trials( successes, p, alpha[i]); t = floor(t); // Print Trials: cout << fixed << setprecision(5) << setw(15) << right << t << endl; }
Note that since we're calculating the maximum number of trials permitted,
we'll err on the safe side and take the floor of the result. Had we been
calculating the minimum number of trials required
to observe a certain number of successes using
find_minimum_number_of_trials
we would have taken the ceiling instead.
We'll finish off by looking at some sample output, firstly for a 1 in 1000 chance of component failure with each use:
________________________
Maximum Number of Trials
________________________
Success ratio = 0.001
Maximum Number of "successes" permitted = 0
____________________________
Confidence Max Number
Value (%) Of Trials
____________________________
50.000 692
75.000 287
90.000 105
95.000 51
99.000 10
99.900 0
99.990 0
99.999 0
So 51 "uses" of the component would yield a 95% chance that no component failures would be observed.
Compare that with a 1 in 1 million chance of component failure:
________________________
Maximum Number of Trials
________________________
Success ratio = 0.0000010
Maximum Number of "successes" permitted = 0
____________________________
Confidence Max Number
Value (%) Of Trials
____________________________
50.000 693146
75.000 287681
90.000 105360
95.000 51293
99.000 10050
99.900 1000
99.990 100
99.999 10
In this case, even 1000 uses of the component would still yield a less than 1 in 1000 chance of observing a component failure (i.e. a 99.9% chance of no failure).