erlang B implementation, calculating max users given acceptable GOS

erlangB.pl

#/usr/bin/env perl -w
# by jeremy warner, f14

use strict;

# determine number of channels to test
die "enter \# channels, then sectors\n" unless @ARGV == 2;
my ($chan, $sect) = @ARGV;

# program constants
my $au = 0.2; # erlangs
my $user = 1000; # max user
my $maxGos = 0.005;
my @atest=(0..$user);

# display contants
print "\nTraffic/user: \t\t$au\n";
print "Channels/sector: \t$chan\n";
print "# of Sectors:  \t\t$sect\n";
print "Max GOS:  \t\t$maxGos\n";

# factorial init
my %table; $table{0} = 1;

foreach my $a (@atest) {

    my $traffic = $a * $au;

    my $num = ( $traffic ** $chan ) / factorial($chan);

    my $denom = 0;
    for(my $i = 0; $i < $chan+1; $i++){
        $denom += ( $traffic ** $i ) / factorial($i);
    }

    my $result = ($num/$denom);
    if ($result > $maxGos)
    {
        print("\n\tresults:\n\n");
        my $endResults = "supported users:\t". ($a-1) ."\n";
        $endResults .= "supported traffic:\t". ($traffic-$au) . " [erlangs]\n";
        $endResults .= "traffic/sector:\t\t". ($traffic-$au)*$sect . " [erlangs]\n";
        die $endResults;
    }
    # user for debus
    #print("users = ". $traffic/$au .", gos = $result\n");
}


sub factorial {
    my $number = shift;
    my $factorial;

    if ( exists $table{$number} ) {
        return $table{$number}; # Input found in the look-up table, returning stored value.
    }
    else {
        $factorial = factorial($number - 1) * $number;
    }

    $table{$number} = $factorial;
    return $factorial;
}