# Answer of Tutorial 3 - selected), the probability of...

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Answer of Tutorial 3 Q1. (a) 010001010011, 110011001111 (b) 010011011111, 110001000011 (c) 010001000111, 110011011011 Q2. 4 3 Q3. 30 (excluding the original parents) Q4. (a) (1, 3, 6, 8, 4, 5, 7, 2, 9, 10); (9, 6, 4, 2, 3, 5, 1, 7, 10, 8) (b) (2, 3, 6, 8, 4, 5, 7, 9, 10, 1); (6, 8, 4, 2, 3, 5, 1, 7, 10, 9) (c) (1, 3, 6, 2, 4, 5, 7, 8, 9, 10); (9, 6, 4, 8, 3, 5, 1, 7, 10, 2) Q5. (a) Assuming a maximization problem, and using their ranks as fitness: Chromosome Objective Fitness according to rank Fitness to the closest integer 01001 4 3.5 7 10100 4 3.5 7 00000 0 1.5 3 01111 8 5 10 00000 0 1.5 3 (b) Section without replacement (assuming that the portion reduced by 1 when
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Unformatted text preview: selected), the probability of selecting “01001” twice is 145 7 (c) The offspring will be exactly the same as parents after crossover. [Note: the answer (b) depends on your answer in (a).] Q6. (a) 1 0 1 0 1 1 0 0 0 1 1 0 1 (occurrence frequency = 4) 1 0 1 0 0 1 0 0 0 1 1 0 1 1 0 1 0 0 0 0 0 0 1 1 0 1 1 0 1 0 0 0 1 0 0 1 1 0 1 (occurrence frequency = 2) 1 0 1 0 0 0 1 0 1 1 1 0 1 (occurrence frequency = 2) 1 0 1 0 0 0 1 0 1 1 0 0 1 (occurrence frequency = 2) probability = ½ (b) probability = 3/5 Q7. (a) 1/80 (b) 1/78 (c) 5/182 Q8. population size=14...
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## This note was uploaded on 04/10/2011 for the course EE 4047 taught by Professor Kitsangtsang during the Spring '09 term at City University of Hong Kong.

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