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Unformatted text preview: Chapter 13 Problem 13.11 a) b) 0) Tours Distance Tours Distance
123451 34 132451 32
123541 34 132541 26
1243—51 36 134—251 28
124531 31 135241 28
125341 30 142351 37
125431 25 143251 31 Optimal solution is 1254—31 (or the reverse tour 134—521)
Initial trial solution: 123451 Distance = 34 There are three possible subtour reversals that would improve upon this
solution: 123451 Distance = 34
Reverse 23 132451 Distance = 32
Reverse 234 143251 Distance = 31
Reverse 3—45 125431 Distance = 25 Choose 125431 as the next trial solution. There is no better subtour
reversals that would improve upon this solution. In fact, the solution 1254—31 is optimal.
Initial trial solution: 124351 Distance = 36 There are four possible subtour reversals that would improve upon this
solution: 12435—1 Distance = 36
Reverse 43 123451 Distance = 34
Reverse 35 124531 Distance = 31
Reverse 243 1—34—251 Distance = 28
Reverse 435 1—25341 Distance = 30 Choose 134251 as the next trial solution. l3~l d) There is only one possible subtour reversal that would improve upon this
solution: 1—3425—1
134521 Distance = 28
Reverse 25 Distance = 25 Choose 134521 as the next trial solution. There is no better subtour
reversals that would improve upon this solution. In fact, the solution 1345—2—1 is optimal.
Initial trial solution: 1423—5—1 Distance = 37 There are ﬁve possible subtour reversals that would improve upon this
solution: 142351 Distance = 37
Reverse 24 124—351 Distance = 36
Reverse 23 1432—51 Distance = 31
Reverse 35 142531 Distance = 28
Reverse 423 132451 Distance = 32
Reverse 235 1—4532—1 Distance = 34 Choose 142531 as the next trial solution. There is only one possible sub—tour reversal that would improve upon this
solution: 14—2531
145—231 Distance = 28 Reverse 25 Distance = 26 Choose 145231 as the next trial solution. There is only one possible subtour reversal that would improve upon this
solution: 14523—1
12—543—1 Distance = 26
Reverse 452 Distance = 25 Choose 134521 as the next trial solution. There is no better subtour
reversals that would improve upon this solution. In fact, the solution 12—543—1 is optimal. 132 Problem 13.12 a) b) Initial trial solution: 12456731 If the second reversal had been choosen, the next trial solution is 12354671. Then there is no better subtour reversals that would
improve upon this solution. Distance = 69 There are two possible subtour reversals that would improve upon this
solution: 12—456731 Distance = 69
Reverse 56 12465731 Distance = 66
Reverse 24567 17—564231 Distance = 68 Choose 12—465731 as the next trial solution. There is only one possible subtour reversal that would improve upon this
solution: Distance = 66
Distance = 63 L2465131 Reverse 57 124675—31 Choose 12467531 as the next trial solution. This is an optimal solution. Problem 13.13 a) b) Initial trial solution: 1234561 Tours Distance Tours Distance
1234561 64 126345—1 69
1234651 59 1523461 56
1236451 67 1524361 61
124465—1 64 1—62—3451 63
1—2—5~436—1 67 1632—451 66 Optimal solution is 15—2—3461 (or the reverse tour 16—4—3521)
Distance = 64 There are two possible subtour reversals that would improve upon this
solution: 12345—61 Distance = 64 l3~3 Reverse 56 1234651 Distance = 59
Reverse 2345 1543261 Distance = 63 Choose 12346—51 as the next trial solution. There is no better subtour
reversals that would improve upon this solution. c) Initial trial solution: 12—54361 Distance = 67 There are two possible subtour reversals that would improve upon this
solution: 125—4—361 Distance = 67
Reverse 2—5 1524361 Distance = 61
Reverse 543 l234—5~6l Distance = 64 Choose 152—4361 as the next trial solution. There is no better subtour
reversals that would improve upon this solution. (3s— I mw‘r WM, A; AD) 25, (,1) oar/f 37f?
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<5 BL 23 5 Sa1esman Prob1ems: % cities: 8.2 '5‘ Trave1in
Number 0 13—45—2—1 0
0
0
0
0 4 86406 3 45047 Best So1ution 12—4—351 2 80563 1.111111 so1ution:
r
25 0 111111 = Tabu Search A1gorithm: Initia1 tria1
Iteration
Best Distance 012345 8 Sa1esman Prob1ems: % Cities: Ml ‘51 Trave1in
Number 0 8 7190. .30
12 2..1 17256 376131
_ . . . . _ _ . . _ _
23567 765322
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(x C 132—S.txt 1—32—5—6—4—7—8—1 (c) Initia1 tria1 so1ution: Distance
111.0
104.0
101.0
99.0
98.0
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n. _ _ _ _ _ _ _
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T33388833 _ _ _ _ _ _ _ _ 11111111 Iteration 01234567 1—837—65—4—2—1 Best So1ution 98.0 Best Distance /3‘/0 10
10
35
29
36
25
35
16
14 16
34
24
27
14
23
10
14 18
23
44
34
37
25
32
1o
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31
28
19
11
10
32
23
35 21
23
13
12
10
25
14
25 21
29
18
10
12
11
37
27
36 21
11
10
13
19
34
24
29 13—26.txt
15 25
26
11
18
23
28
44
34
35 my
10
13
26
21
29
21
31
23
16
10 0
13
25
15
21 9
19
18 8
15 H.241
Trave1ing Sa1esman Prob1
Number of Cities:
I
Tabu Search A1gorithm: (a) llllllll
_______
0076660
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92....2
n__5555_
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1__4433_
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.13888899
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1111111 Iteration 0123456 Best Soiution = 19—8—10—23—45—7—6—1 121.0 Best Distance 1..
122.60
___871
943:0;
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________
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_______.
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________
11111111 Iteration (L) 01234567 167—5—432—1089—1 Best So1ution 121.0 Best Distance (c) 1....5.
02222.3
S____7_
0000.2
111116.
a.____0
1888991
F_____.
T999888 __._._
111111 Iteration 012345 1981023—457—61 Best So1ution 121.0 Best Distance 1’3 ~ 121 Problem 13.31
Zc = 30, T = 2.
(a) For maximization problem:
Zn = 29, x = (Zn  Zc)/T = 0.5, Prob{acceptance} = exp(x) = 0.607.
Zn = 34, Zn > Zc. Prob{acceptance} = 1
Zn = 31, Zn > Zc. Prob{acceptance} = 1
Zn = 24. x = (Zn  Zc)/T = 3, Prob{acceptance} = exp(x) = 0.05
(b) For minimization problem:
Zn = 29, Zn < Zc. Prob{acceptance} = 1
Zn = 34, x = (Zc  Zn)/T = 2, Prob{acceptance} = exp(x) = 0.135.
Zn = 31, x = (Zc  Zn)/T = 0.5, Prob{acceptance} = exp(x) = 0.607
Zn = 24. Zn < Zc. Prob{acceptance} = 1
Problem 132 Because of the randomness in the algorithm, its output will vary. Problem 13.33 (a) Initial trial solution: 123451 Zc = 34 T1 = 0.2*Zc = 6.8
0000003332 Subtour begins in slot 2
0333306666 Subtour begins in slot 3
0666709999 Subtour begins in slot 4 The random number is 0.09656: choose a subtour that begins in slot 2 Be beginning in slot 2, the subtour needs to end somewhere between slots 3 and 4. 0.00000.4999 Subtour begins in slot 3
0500009999 Subtour begins in slot 4 The random number is 0.96657: choose a sub—tour that ends in slot 4
Reverse 234: new solution: 14—32—5—1 Zn = 31
Since Zn < Zc, we accept 143251 as new trial solution. (b) Because of the randomness in the algorithm, its output will vary. /3~/3 Problem 13.34
Because of the randomness in the algorithm, its output will vary.
Problem 13.35
Because of the randomness in the algorithm, its output will vary.
Problem 13.36
Because of the randomness in the algorithm, its output will vary. Problem 13.37 (a) * ( 3+ﬁtl3‘7‘ly’ F0317“) ‘
 ﬁnal Mln‘rmum , ’0 F maX/mnm ,
or M xxija’ fa“) : )F’J‘lzi‘r 60 >0‘ M x":(o, fwup) : {4, ~‘)/\; :vévfu’
EMl rh«ﬁliiXMIfiw : €1.20 7 0) pull tum/«mum
(b) X:3I,f'(3) : (73 >0 1 ,Qomll mimwm 3 l V r
W fl)”: xéox +€iaux+ln (c) x =15.5, f(x) = Zc = 3558.9, T1 = 0.2*Zc = 671.775
L = 0, U = 31, sigma= (UL)/6 = 5.167 The random number we obtain from table 20.3 is 0.09656. From Appendix 5, P{standard normal S 1.315} 5 0.09656
So N(0, 5.167) = 1.315*5.167 = 6.79. x = 15.5 +N(0, 5.167)=15.5 — 6.79 = 8.71
Zn = f(x) = 4047.6
Since Zn > Zc, we will accept x = 8.71 as the next trial solution. ((1) Because of the randomness in the algorithm, its output will vary. Problem 1338
The nonconvex program in Sec 12.10 is Max 0 5’X56X$+ 154;)(3' 37 “(2 flax s.t. 0 S x S 5
(a) x = 2.5, f(x) = Zc = 3.5156, T1 = 0.2*Zc = 0.7031
L = 0, U = 5, sigma = (UL)/6 = 0.8333 The random number we obtain from table 20.3 is 0.09656. From Appendix 5, P{standard normal S 1.315} 5 0.09656
So N(0, 0.8333) = 1.315*0.8333 = 1.0958 x = 2.5 + N(0, 5.167) = 2.5 — 1.0958 = 1.4042
Zn = f(x) = 1.5782
Since (Zn — Zc)/T = 7.2488, the probability of accepting x = 1.4042 as the next trial solution is
Prob{acceptance} = exp(7.2488) = 0.00071 From table 20.3, the next random number is 0.96657 > 0.00071.
We reject x = 1.4042 as the next trial solution. (b) Because of the randomness in the algorithm, its output will vary. /3~/§ (b) Because of the randomness in the algorithm, its output will vary. Problem 13.39 Because of the randomness in the algorithm, its output will vary. Problem 13.3—10 Because of the randomness in the algorithm, its output will vary. Problem 13.4—1
(a) P1: 010011 and
P2: 100101
Only the last digit agree, then children then become
C1: xxxxxl and
C2: xxxxxl,
Where x indicates that this particular digit is not known yet. Random numbers are used
to identify these unknown digits and if random number is
0.00000  0.49999 x = 0, and
0.50000 — 0.99999 x = 1.
Starting from the front of the top row of Table 20.3, the ﬁrst 10 random number are:
0.09656, 0.96657, 0.64842, 0.49222, 0.49506, 0.10145, 0.48455, 0.23505, 0.90430,
0.04180. The corresponding digits are: 0, 1,1 0, 0, 0, 0, 0, 1, 0. Then children then
become:
C1: 011001 and
C2: 000101 Next, we consider the possibility of mutations. The probability of a mutation in any gene
is set at 0.1, and let random numbers 0.000000.99999 correspond to a mutation and
0.100000.99999 correspond to no mutation. Use Table 20.3 to obtain the next 12 random
numbers and the 8th and the 11th one gives mutation. Therefore, the ﬁnal conclusion is
that the two children are C1: 011001 and C2: 010111
(b) P1: 000010 and
P2: 001101 Only the last digit agree, then children then become
C1: 00xxxx and
C2: 00xxxx,
Where x indicates that this particular digit is not known yet. Random numbers are used
to identify these unknown digits and if random number is
0.00000 — 0.49999 x = 0, and
0.50000 — 0.99999 x = 1.
Starting from the front of the top row of Table 20.3, the ﬁrst 8 random number
correspond to digits: 0, 1, 1 0, 0, 0, 0, 0. Then children then become:
C1: 000110 and
C2: 000000 Next, we consider the possibility of mutations. The probability of a mutation in any gene
is set at 0.1, and let random numbers 0.00000—0.99999 correspond to a mutation and
0.10000—0.99999 correspond to no mutation. Use Table 20.3 to obtain the next 12 random
numbers and the 2nd and the 10th one gives mutation. Therefore, the ﬁnal conclusion is
that the two children are /3—/7 C1: 001001 and C2: 000011
(c) P1: 100000 and
P2: 101000 Only the last digit agree, then children then become
C 1 : 1 0x000 and
C2: 10x000,
Where x indicates that this particular digit is not known yet. Random numbers are used
to identify these unknown digits and if random number is
0.00000  0.49999 x = 0, and
0.50000 — 0.99999 x = 1.
Starting from the front of the top row of Table 20.3, the ﬁrst 2 random number
correspond to digits: 0, 1. Then children then become:
C1: 100000 and
C2: 101000 Next, we consider the possibility of mutations. The probability of a mutation in any gene
is set at 0.1, and let random numbers 0.000000.99999 correspond to a mutation and
0.100000.99999 correspond to no mutation. Use Table 20.3 to obtain the next 12 random
numbers and the 8th one gives mutation. Therefore, the ﬁnal conclusion is that the two
children are C1: 100000 and C2: 111000 Problem 13.42
(a) P1: 1234765—8—1 and
P2: 15—3678241 Start from city 1. Possible links: 12, 18, 15, 14. Random Number: 0.09656 Choose 12
0.96657 No mutation Start from city 2. Current Tour: 12 Possible links: 23, 28, 24. Random Number: 0.64842 Choose 28
0.49222 No mutation Start from city 8. Current Tour: 128 Possible links: 8—5, 87 Random Number: 0.49506 Choose 85
0.10145 No mutation /3/o“ Start from city 5. Current Tour: 1285 Possible links: 56, 53 Random Number: 0.48455 Choose 56
0.23505 No mutation Start from city 6. Current Tour: 12856
Possible links: 67, 67, 63
Random Number: 0.90430 Choose 63
0.04180 Mutation
Reject 63 and consider all other possible linksz64, 67
Random Number: 0.24712 Choose 64 Start from city 4. Current Tour: 128564 Possible links: 43, 47 Random Number: 0.55799 Choose 47
0.60857 No mutation Start from city 7. Current Tour: 1285647
Possible links: 71 Therefore, C1 = l2856471 (b) Pl: 164738251 and
P2: l2536847l Start from city 1. Possible links: 16, 15, 1—2, 17. Random Number: 0.09656 Choose 16
0.96657 No mutation Start from city 6. Current Tour: 1—6 Possible links: 64, 63, 68. Random Number: 0.64842 Choose 63
0.49222 No mutation Start from city 3. Current Tour: 163 Possible links: 37, 38, 35 Random Number: 0.49506 Choose 38
0.10145 No mutation Start from city 8. Current Tour: 1—6—38 Possible links: 82, 84 Random Number: 0.48455 Choose 8—2
0.23505 No mutation /3'/7 Start from city 2. Current Tour: 16382
Possible links: 2—8, 25, 25
Random Number: 0.90430 Choose 25
0.04180 Mutation
Reject 25 and consider all other possible links:24, 27
Random Number: 0.24712 Choose 24 Start from city 4. Current Tour: 163824 Possible links: 4—7 Random Number: 0.55799 Choose 47
0.60857 No mutation Start from city 7. Current Tour: 1638—247
Possible links: 71 Therefore, C1 = 16382471 (c) ) P1: 157462381 and
P2: 137256841 Start from city 1. Possible links: 15, 18, l—3, 14. Random Number: 0.09656 Choose 15
0.96657 No mutation Start from city 5. Current Tour: 15 Possible links: 57, 52, 56. Random Number: 0.64842 Choose 52
0.49222 No mutation Start from city 2. Current Tour: 152 Possible links: 2—6, 2—3, 27 Random Number: 0.49506 Choose 23
0.10145 No mutation Start from city 3. Current Tour: 1—523 Possible links: 38, 37 Random Number: 0.48455 Choose 38
0.23505 No mutation Start from city 8. Current Tour: 15238
Possible links: 86, 8—4
Random Number: 0.90430 Choose 84
0.041 80 Mutation
Reject 84 and consider all other possible 1inks:86, 87
Random Number: 0.24712 Choose 86 N'W Start from city 6. Current Tour: 152386 Possible links: 64 Random Number: 0.55799 Choose 64
0.60857 No mutation Start from city 4. Current Tour: 1523864
Possible links: 71 Therefore, C1 = 15238647—1
Problem 13.43 (a) Because of the randomness in the algorithm, its output will vary.
(b) Because of the randomness in the algorithm, its output will vary. /2~>I I3.46“4[
(a) 13—4—4(a).txt Integer Non1inear Programming: Object Function Max f(x) = 1 XA3  60 XAZ + 900
subject to
X1 <= 31
and
X1 >= 0.
Iter.  Best $01ution  Fitness
1 I (0)  900.0
Iteration 1
Popuiation:
Member I Popu1ation I So1ution
1 ICOOOOO) ICO)
2 (00001) (1)
3 (00100) (4)
4 (00110) (6)
5 l(01010) (10)
6 {(01110) (14)
7 (10111) (23)
8 (11010) (26)
9 (11100) (28)
10 (11101) (29)
Chi1dren:
Member  Parents  Chiidren 
5 (01010) ( [1][1]010 ) (26)
3 (00100) I( 01100 ) (12)
4 (00110) ( 0011[1] ) (7)
6 1(01110) I( 00110 ) IC6)
2 ICOOOOl) IC [1100[0]0) (16)
8 l(11010) I( 11000 ) (24) So1ution Fitness
0 .0
4 .0 OI—‘O 

I
I 1044.0
I—4100.0

I
I
I I homo —81l6.0 —18673.0
22084.0
24l88.0
—25171.0  Fitness
—22084.0
6012.0
—1697.0
1044.0
I10364.0
—19836.0 (b) Because of the randomness in the aigorithm, its output wi11 vary. 821 Problem 13.45 Because of the randomness in the algorithm, its output will vary. Problem 13.46 Because of the randomness in the algorithm, its output will vary. Problem 13.47 (a) Because of the randomness in the algorithm, its output will vary.
(b) Because of the randomness in the algorithm, its output will vary. Problem 13.48
(a) Genetic Algorithm:
Iteration  Best Solution I Fitness  1 I12—5—431 124.0 I Iteration 1: Member I Population I Fitness   Member  Children  Fitness 
l I 145321 I 33.0   10 I 134521 I 24.0
2  125431 I 24.0 I  3  132—451  310
3 I 132451  31.0 I I 5  123451 I 330
4 I 145321 I 33.0   1 I 123541  33 0
5  123451  33.0 I I 2  152431 I 280
6 I 134251 I 28.0   8 I 143251 I 310
7  152341  31.0 I 8 I 142351  37.0  9 I 152341 I 31.0  10  152431  28.0 I (b) Because of the randomness in the algorithm, its output will vary. Problem 13.49 Because of the randomness in the algorithm, its output will vary. /3 23 Problem 13 .4 l 0 Because of the randomness in the algorithm, its output will vary. Problem 13.51 See the solution for Problem 13.2—5(a) for the output from the basic tabu search
algorithm. Because of the randomness in the basic simulated annealing and genetic
algorithms, their outputs will vary. Problem 13.52 See the solution for Problem 13.26(a) for the output from the basic tabu search algorithm. Because of the randomness in the basic simulated annealing and genetic
algorithms, their outputs will vary. 13—l¥ ...
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 Fall '10
 ChoiJin
 Evolution, Randomness, random number, Statistical randomness

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