# Cost scaling and optimal strategy for synthesis

I'm trying to determine an optimal strategy to minimize synthesis cost in Before the Echo (previously known as Sequence). Does anyone know how synthesis cost scales with success rate?

For a brief recap, synthesis is the name of crafting in this game. The player can synthesise an item by paying a certain amount of experience points. The player can freely choose the success rate of each synthesis, up to `95%`, but the cost increases non-linearly with the success rate. Furthermore, each item has a level, which as far as I can tell only affects the base synthesis cost.

So, a synthesis trial has probability `p` to succeed and `1 - p` to fail, thus it can be modelled by a Bernoulli distributed random variable `X ~ B(p)`. Now, we can safely assume that all trials are independent, i.e. that the success rate of synthesis doesn't depend from the previous failures. Hence if, for the sake of simplicity, we suppose to always try synthesis with the same success rate, we can model repeated trials with a Bernoulli process. In particular, the success rate after `n` trials is `1 - (1 - p)^n`.

For example, a synthesis trial of `Key the Fifth` with `p = 0.5`, i.e. with `50%` success rate, costs `825 xp`, while one with `p = 0.75` costs `1708 xp`. Therefore two consecutive trials with `p = 0.5` cost less than one with `p = 0.75`, despite having the same success rate overall.

• Added the tag :) – Ave Nov 27 '15 at 18:51

With a bit of work I figured that the formula for the cost of synthesising a level `L` object with probability `0 <= x < 1` is
``````cost = floor(L^2 * tan(x * pi/2) + 8 * L)
In case you're wondering, to find it I first wrote down the costs of synthesising a few items of different levels with various success rates, sampled every `5%`. Then — using Sage, an open-source computer algebra system — I plotted the data points for the highest level object (which had the highest "resolution") and then tried to fit them with a couple of likely functions. Finally, I checked my solution against my other data sets.