Answers.Body: <p>You can find the explanation by asking some question about the function. Suppose, the value of <span class="math-container">$\frac{E(y)-E(x)}{kT} &gt;&gt; 0$</span> is much more greater than zero. What does it mean? It means the value of <span class="math-container">$E(y)$</span> is much greater than <span class="math-container">$E(x)$</span> related to the <span class="math-container">$kT$</span> that is as a measure of temperature decreasing. Now, you want in this situation a probability which is near to zero. Hence, <span class="math-container">$e^{-\frac{E(y)-E(x)}{kT}}$</span> could be a good value for the probability of selection of worse solutions! </p>
  
  <p>Why <span class="math-container">$e$</span> instead of <span class="math-container">$2$</span> or other values greater than <span class="math-container">$1$</span>? Because it could be a good function in optimization problems as its derivative is more simple than others!</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2019-01-13T17:53:23.643Z
Answers.LastActivityDate: 2019-01-13T17:53:23.643Z
Answers.UserDisplayName: OmG
Body: <p>Why is the e-function used to decide whether to accept a worse solution or not? 
  To be more specific: Why was <span class="math-container">$e$</span> chosen as basis?</p>
  
  <p>The propability to accept a worse solution is described with:
  <span class="math-container">$p=e^{-\frac{E(y)-E(x)}{kT}}$</span></p>
  
  <p><span class="math-container">$E(y)$</span> is the energy from the old solution
  <span class="math-container">$E(x)$</span> is the energy from new solution <span class="math-container">$T$</span> is a constant temprature decreasing with a constant factor k in every iteration.</p>
  
Tags: <algorithm>
Title: Simulated Annealing: Why is e-function used as propability function to decide to accept a worse solution
UserDisplayName: Bryan McGill
ThreadText: # Simulated Annealing: Why is e-function used as propability function to decide to accept a worse solution
  
  (Asked by: username_0 on 2019-01-13)
  
  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.
  
  **Tags:** algorithm
  
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  ## 1 answers
  
  ### Answer by username_1 on 2019-01-13. Score: 0 | Accepted answer
  
  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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