Catch-22 claculating in PHP

[JasonGreen]

There doesn't seem to be a way around this. I want to be able to calculate values using variables but in such a way that the value continues to change - however, I cannot simply write out the number, I need a variable.

Here is the problem, look at my example below, if I were to use calculations based on define variable, I would end up using longer and longer calculations and that would be a horible mess.

Like this

print $serveryear . " " . $servercost . " " . ($servercost HIGH_DEPREC) . " " . ($servercost - $servercost HIGH_DEPREC) . "\n";
print ($serveryear + 1) . " " . ($servercost - $dodgecost HIGH_DEPREC) . " " . ($servercost - $servercost HIGH_DEPREC * HIGH_DEPREC) . " " . ETC . "\n";
It just grows and grows!!!!

At the same time I cannot define a variable for each and every calculation, that too is a mess that grows and grows.

DELL SERVER– Depreciation over 5-year period (Factor: .20)

YEAR Start Value Deprec Ending Value

2004 21297.00 4259.40 17037.60

2005 17037.60 3407.52 13630.08
2006
2007
2008

Any insights, comments, and suggestions will be greatly appreaciated.

Thanks

[NogDog]

You'll want to use some sort of loop. It's not clear to me what all the variables are ($dodgecost, for example?), but to emulate the results in your example output:

<style type="text/css">
td {
   text-align: right;
   padding: 0 0.5em;
}
</style>
<?php
$serveryear = 2004;
$servercost = 21297.00;
define('HIGH_DEPREC', 0.20);

$start = $servercost;
echo "<table>";
for($ix = 0; $ix < 5; $ix++)
{
   $deprec = $start * HIGH_DEPREC;
   $end = $start - $deprec;
   printf(
      "<tr><td>&#37;d</td><td>%.2f</td><td>%.2f</td><td>%.2f</td></tr>\n",
      $serveryear + $ix,
      $start,
      $deprec,
      $end
   );
   $start = $end;
}
echo "</table>";
?>