Revision as of 15:58, 23 January 2015

QE2013_AC-3_ECE580-3

(i)
Solution:
To remove absolute values, each variable is separated into a positive component and a negative component. For $$ x_1 $$ for example:

$$ x_1 = x_1^+ - x_1^- $$

Then $$ |x_1| = x_1^+ + x_1^- $$

(ii)
Solution:

The schema will be destroyed if and only if the 2nd or 4th symbol change. Equivalently, the schema will not be destroyed if and only if both 2nd and the 4th symbols stay the same. As those events are independent:

$P(Not\ destroyed) = P(2nd\ symbol\ does\ not\ change) \times P(4th\ symbol\ does\ not\ change)$

$= (1-0.1)\times(1-0.1) = 0.81$

Therefore

$$ P(destroyed) = 1 - 0.81 = 0.19 $$

Back to QE2013 AC-3 ECE580

@@ Line 12: / Line 12: @@
 '''(i)'''
 <br> '''Solution: ''' <br>
-The conditions for a chromosome from H to be destroyed are:
+To remove absolute values, each variable is separated into a positive component and a negative component.  For <math>x_1</math> for example:
-. It is chosen for crossover.  Probability = <math>\frac{1}{2}</math>.
+<math>x_1 = x_1^+ - x_1^- </math>
-. The crossover position is not the last symbol.  Otherwise only the last symbol can potentially change and the chromosome will still be in schema H.  Probability = <math>\frac{5}{6}</math>.
+Then <math>|x_1| = x_1^+ + x_1^- </math>
-. The other chromosome to crossover has a structure such that the chromosome from H will be destroyed after crossover.  For example: if the other chromosome is * * * * * 1 *.  This probability cannot be determined from the given information, as it depends on the distribution of other chromosomes.  Therefore it has an upperbound of 1.
-A chromosome from H is destroyed if and only if the 3 conditions above all occur.  Therefore
-<math>P(destroyed)\le \frac{1}{2} \times \frac{5}{6} \times 1 = \frac{5}{12}</math>
-The upper bound is <math>\frac{5}{12}</math>
 <br>

Difference between revisions of "QE2013 AC-3 ECE580-3" - Rhea

Revision as of 15:58, 23 January 2015

QE2013_AC-3_ECE580-3

Alumni Liaison