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### Re^4: Efficient Assignment of Many People To Many Locations?

 on Feb 26, 2005 at 06:19 UTC ( #434737=note: print w/replies, xml ) Need Help??

There is a 10% chance at any iteration of choosing a value in the top 10% (duh) when you are just selecting linearly numbers on a number line. So, the probability of choosing a number in the top 10% over 30 iterations is simply:

```printf("%.0f%%",100*(1-.9**30));

I don't see the magic of the number 30 here-- the cost/benefit of iterations to improved results appears to be linear to me. You get results that are twice as good by doing twice as many iterations.

It's easy to conceive of a non-linear results set where you are likely to come nowhere near the maximum value by picking some smallish number of random iterations.

Try, for instance, finding the maximum value of 1/10**N where N is an integer between 0 and 1,000,000. What is the percent error between your randomly selected max over thirty iterations and the actual max value? My money says it is 100%. Of course, evaluating the linear variance of an exponential function may be a bit of a stretch...

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Re^5: Efficient Assignment of Many People To Many Locations?
by ff (Hermit) on Feb 27, 2005 at 01:35 UTC
Concerning the number 30, you are correct, there's nothing particularly magical about it except that it's not too difficult to remember and it gives pretty good results for this general problem, i.e. if you want to be 95.8% sure that one of your randomly generated solutions is in the top 10% of all solutions, run 30 trials.

But to play with your equation, I submit the following:

```use strict;
use warnings;

# We'd like to see the likelihood that by running a given number
# of trials that one of these trials gives a result in some top X%
# of all the possible results.
#
# As an example, suppose we need to generate a solution in the top
# 10% of all possible solutions, i.e. we want to exceed a "clip
# level" of 0.90.  How many trials would we need to run to have an
# X% chance that one of those solutions out of all those trials is
# in that top 10%?
#
# Generating a simple little table shows us the percentage chance
# that one of those trials will be in that bracket.  For a 95.8%
# chance that one of our trials is in the top 10%, our little
# table shows us that we'd need to run 30 trials.
#
# Running 31 trials would boost our likelihood of being in that
# top 10% by about 0.4%.
#
# Changing the clip level to 0.95 and upping our upper bound to
# 65 shows that running 62 trials would give us that 95.8%
# confidence of one of our trials being in that set of solutions
# in the 95-100% range of best solutions.  By running only 30
# trials, we'd only be 78.5% sure that one of our solutions is
# superior to 95% of all possible solutions.

my \$want_soln_above_this_clip_level = 0.90;
my \$min_no_trials = 15;
my \$max_no_trials = 35;

my \$value;
my \$prev_value = 0;

foreach ( \$min_no_trials .. \$max_no_trials ) {
print "\$_\t";
#printf("%.1f%%",100*(1-.9**\$_));
\$value = 100*(1-\$want_soln_above_this_clip_level**\$_);
printf("%.1f%%", \$value);
print "\t";
printf("%.1f%%", \$value - \$prev_value);
print "\n";
\$prev_value = \$value;
}

__DATA__
15    79.4%    79.4%
16    81.5%    2.1%
17    83.3%    1.9%
18    85.0%    1.7%
19    86.5%    1.5%
20    87.8%    1.4%
21    89.1%    1.2%
22    90.2%    1.1%
23    91.1%    1.0%
24    92.0%    0.9%
25    92.8%    0.8%
26    93.5%    0.7%
27    94.2%    0.6%
28    94.8%    0.6%
29    95.3%    0.5%
30    95.8%    0.5%
31    96.2%    0.4%
32    96.6%    0.4%
33    96.9%    0.3%
34    97.2%    0.3%
35    97.5%    0.3%

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