# Simulated Annealing: Why is e-function used as propability function to decide to accept a worse solution

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!