This document discusses genetic algorithms and provides examples of calculating fitness values and estimating the number of generations needed for a population to reach a target fitness level.

1. 31
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5. )! "# $ % &
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1000 eval(1000) = 8
1001 eval(1001) = 9
1010 eval(1010) = 10
1011 eval(1011) = 11
1100 eval(1100) = 12
1101 eval(1101) = 13
1110 eval(1110) = 14
1111 eval(1111) = 15
0 # & 0'* & ( &)" #
S1 & % :
5.118/)15141312111098(SF 1
+ &)" S2=0*** , ( 0
&( - & #:
0000 eval(0000) = 0
0001 eval(0001) = 1
0010 eval(0010) = 2
0011 eval(0011) = 3
0100 eval(0100) = 4
0101 eval(0101) = 5
0110 eval(0110) = 6
0111 eval(0111) = 7
0 # & 0'* & ( &)" #
S2 & % :
5.38/)76543210(SF 2
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&( " 0'* & ( 0 (& % :
F=0+1+2+…+15=120 & 0'* & 7,5
2 , & 0'* & ( S1 53.33% % # & # 0'* & # ( 0 (& %,
=0.53 ( S2, 53.33% ' , =0,53.
4 & 0 % 1 & & : 1(S, t+k) = 1(S, t)*(1+ )k
t=0 ) ( 1(S1,0)=8, 0 0 (0 & ( : 1(S1, t+k) = 8*(1+ )k.
) =0.53 1(S, t+k)=16 ( S1, 0 "& & ' 0 (& ') ) ( :
16 = 8 *(1+0.53)k = 8 * 1.53k 2 = 1.53k
& ( *% ) ( :
log 2 = k * log (1.53)
k = (0.301/0.185) k= 1.63. +( 0 #, 0' 2 # ( ' &)" S1
0 "& & 0 (& '.
5) = -0.53 1(S1, t+k) <1 ) ( :
1 > 8 *(1-0.53)k = 8 * (0.47)k
& ( *% ) ( :
0 > log 8 + k * log (0.47) 0.903 +k (-0.327) < 0 (-0.327) k < -0.903 k > 2.73.
2 , 0' 3 # ( ' &)" S2 1 / & 0' 0 (& '.
A.
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2: .& &)" S 0 ( 0 & 0 % , 0, & 60% '#
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25% % 0' & ' ( ) 0 (& %
: 4 & 0 % 1 & & :
1(S, t+k) = 1(S, t)*(1+ )k
t=0 ) ( 1(S,0)=60, 0 0 (0 & ( : 1(S, t+k) = 60*(1+ )k.
=0.25 1(S, t+k)=100 ) ( :
100 = 60 *(1+0.25)k = 60 * 1.25k
0 ( Log *% ) ( :
Log 100 = Log 60 + k * Log (1.25) k=2.28, 0 ( & k=3.
2 , 0' 3 # &)" S 0 "& & 0 (& '.
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(2)… 1 / & 0' 0 (& ' ( S ) ( ' 10%
' 0' & ' ( ) 0 (& %
: 4 & 0 % 1 & & :
= -0.10 1(S, t+k) <1 ) ( :
1 > 60 *(1-0.10)k = 60 * 0.9k
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1 > Log 60 + k * Log (0.9) k=38.8, 0 ( & k=39.
2 , 0' 39 # &)" S 1 / & 0' 0 (& '.
A.
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6. 5. & "& #
/ " 1
"& # *( * # &( - & # 1=11101111, 2=00010100, 3=01000011
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1. # &( - & # , ( & &)" ;
2. 5 1 & ' " # &)" #
3. 60 & 0 ' 0 - & # &)" #, * & % &
0 ' * & % & # pc=0.85
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x2 & * & [0, 7]. 0 & & 1 # - "#
) & 0 % 3 bits (2
3
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-
-
(FITNESS)
101011
5 111000
111011
! 101000
111001
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(0 & % 0 ' # 0 "#.
# 0 0 & ( ' 0 '- ; "& 0 &" & #
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(
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( 2) (2/5) (0 & 0 ' 0 "# ' (, ) & 0 # 0 "
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( ) (3/15) (0 & ' ' / ' ( & 0' (
7 0 &" & # & 0 0 0 ).
! "# $ % &
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–
111 100
5 110 110
111 000
! 110 011
111 011
( 1) (2/7) .) - & 0'* & ( 0 (& % & 1 & &) & 0; '& )
- 0'* & ( % ( (# ( 0 (& % & 1 & &) & 0;

7. '! "# $ % &
( 2) (5/7) # ' # '# ' , &% / &) (Schema
Theorem), 0 ( & ) % & &)" S1=111*** & 2 "& ( ' / ',
* & % & 0 ' Pc=0.5 1 0 ' Pm=0.1;