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Here's an off the wall idea, though I'm a long ways from trying to code it. Anyway, I'll start babbling and let someone else refine this, if it is worth it.

Think of it visually, like a dark room, and each complaint is a a candle. The radience from each candle dwindles as per inverse square law. If two candles are near each other, then they would have a bright area between them. All you need to do is find the brightest place and go there.

How does this work in code? I'm not sure, but a matrix might be usefull. For every complaint, increment the value of nearby geographical points inversely proportional to the distance from the complaint.

When all complaints are plotted, look for the highest value in the matrix, send a tech there.

Remove up to X complaints within a radius of Y, and recompute. repeat as necessary.

The granularity of the matrix could be adjusted as needed. The "luminosity" of complaints will also need tweaking.

Update:
here's some code, but it doesn't quite do what I want. points on the grid next to a complaint are too "hot". Maybe it should be a more linear dropoff?

```my @complaints = (
{
'complaint_id' => 'a123',
'x'            => '45.2',
'y'            => '39.7'
},
{
'complaint_id' => 'b456',
'x'            => '46.5',
'y'            => '41.2'
},
{
'complaint_id' => 'c789',
'x'            => '11.9',
'y'            => '29.8'
},
{
'complaint_id' => 'd863',
'x'            => '95.3',
'y'            => '17.2'
},
{
'complaint_id' => 'e635',
'x'            => '65.5',
'y'            => '33.3'
},
) ;

my @highlight = (0.0,0.0);
my \$highvalue = 0.0;

foreach \$testpoint (@complaints) {
for (\$i = 0.0; \$i < 100.0; \$i += 1.0) {
for (\$j = 0.0; \$j < 100.0; \$j+= 1.0) {
\$dsqrd = (((\$testpoint->{'x'} - \$i)**2 + (\$testpoint->{'y'
+} - \$j)**2));
\$points[\$i][\$j] += \$dsqrd ? (1000 / (\$dsqrd * 4 * 3.14)) :
+ 0 ;
if (\$points[\$i][\$j] > \$highvalue) {
@highlight = (\$i,\$j);
\$highvalue = \$points[\$i][\$j];
}
}
}
}

print "\n\n \$highlight[0], \$highlight[1] has value \$highvalue\n";

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