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copied!<p>Due to the use of <code>PIN</code> and <code>ceiling</code>, I don't see a easy way to invert the calculation. Assuming that <code>bid</code> has a fixed precision (I'd guess two decimals behind the dot) you can always use a binary search (as the functions are monotone).</p> <p>Edit: After thinking about it some more, I observed that, taking <code>x = bid*1.02 + 100</code>, we have that the final costs are between x+15 (exclusive) and x+70 (inclusive) (i.e. <code>x+15 < final cost < x+70</code>). Given the size of this range (<code>70-15=55</code>) and the fact that the special values (see note below) for <code>bid</code> are all apart more than this, you can take <code>x+15 = final cost</code> and <code>x+70 = final cost</code>, get the right cases/values of usage and added costs and simply solve that equation (which no longer has either <code>PIN</code> or <code>ceiling</code> in it).</p> <p>To illustrate, let the final cost be <code>222</code>. From <code>x+15 = 222</code> it follows that <code>bid = 107/1.02 = 104.90</code>. Then we have that the usage costs are given by <code>bid*0.1</code> and that the additional costs are <code>5</code>. In other words, we get <code>final cost = bid*0.1 + bid*0.02 + 5 + 100 + bid = bid*1.12 + 105</code> and therefore <code>bid = (222-105)/1.12 = 104.46</code>. As this value of <code>bid</code> means the right values for usage and additional costs were taken, we know that this is the solution.</p> <p>However, if we would have first looked at <code>x+70 = 222</code>, we would get the following. First we get that for this assumption that <code>bid = 52/1.02 = 50.98</code>. This means that usage costs are <code>10</code> and the additional costs are <code>5</code>. So we get <code>final costs = 10 + bid*0.02 + 5 + 100 + bid = bid*1.02 + 115</code> and therefore <code>bid = (222-115)/1.02 = 104.90</code>. But if <code>bid</code> is <code>104.90</code> then the usage costs are not <code>10</code> but <code>bid*0.1</code>, so this isn't the right solution.</p> <p>I hope I explained it clearly enough. If not, please let me know.</p> <p>N.B.: With special values I mean those for which the function defining the values of usage and added costs change. For example, for usage cost these values are <code>100</code> and <code>500</code>: below <code>100</code> you use <code>10</code>, above <code>500</code> you use <code>50</code> and in between you use <code>bid*0.1</code>.</p>

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