Why is the e-function used to decide whether to accept a worse solution or not? To be more specific: Why was $e$ chosen as basis?

The propability to accept a worse solution is described with: $p=e^{-\frac{E(y)-E(x)}{kT}}$

$E(y)$ is the energy from the old solution $E(x)$ is the energy from new solution $T$ is a constant temprature decreasing with a constant factor k in every iteration.


You can find the explanation by asking some question about the function. Suppose, the value of $\frac{E(y)-E(x)}{kT} >> 0$ is much more greater than zero. What does it mean? It means the value of $E(y)$ is much greater than $E(x)$ related to the $kT$ that is as a measure of temperature decreasing. Now, you want in this situation a probability which is near to zero. Hence, $e^{-\frac{E(y)-E(x)}{kT}}$ could be a good value for the probability of selection of worse solutions!

Why $e$ instead of $2$ or other values greater than $1$? Because it could be a good function in optimization problems as its derivative is more simple than others!


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