Hill13 - Chapter 13 Problem 13.1-1 a b 0 Tours Distance...

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Unformatted text preview: Chapter 13 Problem 13.1-1 a) b) 0) Tours Distance Tours Distance 1-2-3-4-5-1 34 1-3-2-4-5-1 32 1-2-3-5-4-1 34 1-3-2-5-4-1 26 1-2-4-3—5-1 36 1-3-4—2-5-1 28 1-2-4-5-3-1 31 1-3-5-2-4-1 28 1-2-5-3-4-1 30 1-4-2-3-5-1 37 1-2-5-4-3-1 25 1-4-3-2-5-1 31 Optimal solution is 1-2-5-4—3-1 (or the reverse tour 1-3-4—5-2-1) Initial trial solution: 1-2-3-4-5-1 Distance = 34 There are three possible sub-tour reversals that would improve upon this solution: 1-2-3-4-5-1 Distance = 34 Reverse 2-3 1-3-2-4-5-1 Distance = 32 Reverse 2-3-4 1-4-3-2-5-1 Distance = 31 Reverse 3—4-5 1-2-5-4-3-1 Distance = 25 Choose 1-2-5-4-3-1 as the next trial solution. There is no better sub-tour reversals that would improve upon this solution. In fact, the solution 1-2-5-4—3-1 is optimal. Initial trial solution: 1-2-4-3-5-1 Distance = 36 There are four possible sub-tour reversals that would improve upon this solution: 1-2-4-3-5—1 Distance = 36 Reverse 4-3 1-2-3-4-5-1 Distance = 34 Reverse 3-5 1-2-4-5-3-1 Distance = 31 Reverse 2-4-3 1—3-4—2-5-1 Distance = 28 Reverse 4-3-5 1—2-5-3-4-1 Distance = 30 Choose 1-3-4-2-5-1 as the next trial solution. l3~l d) There is only one possible sub-tour reversal that would improve upon this solution: 1—3-4-2-5—1 1-3-4-5-2-1 Distance = 28 Reverse 2-5 Distance = 25 Choose 1-3-4-5-2-1 as the next trial solution. There is no better sub-tour reversals that would improve upon this solution. In fact, the solution 1-3-4-5—2—1 is optimal. Initial trial solution: 1-4-2-3—5—1 Distance = 37 There are five possible sub-tour reversals that would improve upon this solution: 1-4-2-3-5-1 Distance = 37 Reverse 2-4 1-2-4—3-5-1 Distance = 36 Reverse 2-3 1-4-3-2—5-1 Distance = 31 Reverse 3-5 1-4-2-5-3-1 Distance = 28 Reverse 4-2-3 1-3-2-4-5-1 Distance = 32 Reverse 2-3-5 1—4-5-3-2—1 Distance = 34 Choose 1-4-2-5-3-1 as the next trial solution. There is only one possible sub—tour reversal that would improve upon this solution: 1-4—2-5-3-1 1-4-5—2-3-1 Distance = 28 Reverse 2-5 Distance = 26 Choose 1-4-5-2-3-1 as the next trial solution. There is only one possible sub-tour reversal that would improve upon this solution: 1-4-5-2-3—1 1-2—5-4-3—1 Distance = 26 Reverse 4-5-2 Distance = 25 Choose 1-3-4-5-2-1 as the next trial solution. There is no better sub-tour reversals that would improve upon this solution. In fact, the solution 1-2—5-4-3—1 is optimal. 13-2 Problem 13.1-2 a) b) Initial trial solution: 1-2-4-5-6-7-3-1 If the second reversal had been choosen, the next trial solution is 1-2-3-5-4-6-7-1. Then there is no better sub-tour reversals that would improve upon this solution. Distance = 69 There are two possible sub-tour reversals that would improve upon this solution: 1-2—4-5-6-7-3-1 Distance = 69 Reverse 5-6 1-2-4-6-5-7-3-1 Distance = 66 Reverse 2-4-5-6-7 1-7—5-6-4-2-3-1 Distance = 68 Choose 1-2—4-6-5-7-3-1 as the next trial solution. There is only one possible sub-tour reversal that would improve upon this solution: Distance = 66 Distance = 63 L2465131 Reverse 5-7 1-2-4-6-7-5—3-1 Choose 1-2-4-6-7-5-3-1 as the next trial solution. This is an optimal solution. Problem 13.1-3 a) b) Initial trial solution: 1-2-3-4-5-6-1 Tours Distance Tours Distance 1-2-3-4-5-6-1 64 1-2-6-3-4-5—1 69 1-2-3-4-6-5-1 59 1-5-2-3-4-6-1 56 1-2-3-6-4-5-1 67 1-5-2-4-3-6-1 61 1-2-4-4-6-5—1 64 1—6-2—3-4-5-1 63 1—2—5~4-3-6—1 67 1-6-3-2—4-5-1 66 Optimal solution is 1-5—2—3-4-6-1 (or the reverse tour 1-6—4—3-5-2-1) Distance = 64 There are two possible sub-tour reversals that would improve upon this solution: 1-2-3-4-5—6-1 Distance = 64 l3~3 Reverse 5-6 1-2-3-4-6-5-1 Distance = 59 Reverse 2-3-4-5 1-5-4-3-2-6-1 Distance = 63 Choose 1-2-3-4-6—5-1 as the next trial solution. There is no better sub-tour reversals that would improve upon this solution. c) Initial trial solution: 1-2—5-4-3-6-1 Distance = 67 There are two possible sub-tour reversals that would improve upon this solution: 1-2-5—4—3-6-1 Distance = 67 Reverse 2—5 1-5-2-4-3-6-1 Distance = 61 Reverse 5-4-3 l-2-3-4—5~6-l Distance = 64 Choose 1-5-2—4-3-6-1 as the next trial solution. There is no better sub-tour reversals that would improve upon this solution. (3-s— I mw‘r WM, A; AD) 25, (,1) oar/f 37f? fefito I ' ‘ ~ r (A 1W6 1 Cori AL AP m5 CD 27; CE AB 275 AD [$4] e li/Imum ( CD 270 l BE | 345 Cuwwt ML: Ag gE CD C, Tall/x 1777‘ 1 CE ' C \ ITMPOV) 2 r/l‘mummw" ’w‘fl Yew/UL w W1 M DE 09 015 TQLM WK " CE, DE cum/«T (Mk 2 m3 BE (3/ DE 130/1360” 33 Min/MMM IJWAQ Jaw/L, 6x404 We” “’7‘ Ac BE 75’ 6’ "WW 13-6 B\7"3. HWMWM M57 0/), A3,, gc, out 3,4 ITEM/6M I: mrnimmm 15ml WV’A 4d% wide out ET DE /22 WM )6?" 0A, Ab, BE, cE , 57‘, mm Imf ET' Iflwfhn 2 m7nhnqu 19¢»17sfiarmf Mi WK W7C <5 BL 23 5 Sa1esman Prob1ems: % cities: 8.2 '5‘ Trave1in Number 0 1-3—4-5—2—1 0 0 0 0 0 4 86406 3 45047 Best So1ution 1-2—4—3-5-1 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 t 88444 t 444883 5 . . . _ . S . . . . . . 7 _ .988903 ..I 31673 :I 652714 . . 1 1 L s .. .. y .. 1 L 1 1 1 1 .. .. 1 441725 . 11.37613 . U . . . . _ _ 2 U . . . . . . . 8 .0 27.2356 . :0 8876537. . a 9 a I 1 a. y 4 a 1 I v y y .. u. 3 T 338844 _ T 3344488 _ 6 . .70509. . _ . . . _ 5 . . . . . . . 7 . .111. . 11.31.67 _ 4465271 . r0 6 IIIIIIIIII _ IllllulIlll . 7 5 . . 3 4 5 .011058. . . .221. 11. e 8 e 7. C . C . 1 n 1 1 n 1. . 3000 00 . 2.000 000 8 t . . .00 . . = 3 t . . .00 . . = . S370 . .23 _ $037. . .073 4 .4101080 7 10008900 n 8 110098000 n .11 1.1 2 . D1119911 O _ D11199111 O 6 .1 4 ..I _ IIIIIIIIII t . Illllllll t 5 U 7 U _ .l . 1| 4. o ,b 0 53011799 . S . S 1.1. 1.21 3 1.111111. 5 11111111 . n. . . . . . . ..L . n. . _ . . . . . t 2 000822222 5 2 0388800332 5 . .1. . . . . _ . e . .1. . _ . . . . _ e 1 t7744555 B 1 t83333823 B U. . . . . . . U. . . . . . . . 40340 .. 16655466 .. 144777744 1 1.12 n O. _ . . . . . n O. . . . . . . _ 0 55566647 0 577466655 .. .1 . . . . . . . .1 . . . . . . _ . m t 14477774 ..L 166645566 h U a. . _ _ . _ . 0 U a. _ . _ _ _ . . 0 t 1 .13283333 . .| .155554477 . .7 .1 O P. . . . . _ . oo 0 P. . . . . _ . . oo .1 “0| 5 T2338888 9 S T22222288 9 . . . . . . _ . . . . . . . . g 1 1.111111 = .| 11111111. .| a a A .1 IIIIIIIIII e .1 IIIIIIIIII e r C r C IIIIIIIIIIIIII h t n t n 1233456700 C a a r .I n t .I n t a a O S a O S e ..l ..l .1. ..I .1 ..l S t t D t t D ..l a ..I a U n ml t n r t b T. e S T. e S a t e t e T \«W 10123456 B m 101234567 B (x C 13-2—S.txt 1—3-2—5—6—4—7—8—1 (c) Initia1 tria1 so1ution: Distance 111.0 104.0 101.0 99.0 98.0 106.0 11111111 n. _ _ _ _ _ _ _ 082222222 .1. _ _ _ _ _ _ _ t75554445 u. _ _ _ _ _ _ _ 146445666 0. _ _ _ _ . _ _ 564666554 _ _ _ _ _ _ _ . 157777777 3. _ _ _ _ _ _ _ .128833388 P. _ _ _ _ _ _ _ T33388833 _ _ _ _ _ _ _ _ 11111111 Iteration 01234567 1—8-3-7—6-5—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 16 19 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—2-6.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 11..._1 _.6777. 92....2 n__5555_ 083____4 1__4433_ t74____3 U__3344_ 165....5 O_.2222_ 556____7 __0000_ 14711116 a_______ .13888899 P_______ T2999988 _______ 1111111 Iteration 0123456 Best Soiution = 1-9—8—10—2-3—4-5—7—6—1 121.0 Best Distance 1.. 122.60 ___871 943:0; yy1919 I11111111 ________ 22999888 _______. 00888990 11_____1 n__00006_ 0881111_2 i._._._7_ t992222_3 U______5_ 1666433_4 0......4_ S777344.5 ______3_ 1555555_7 a____._2_ 1433777_6 P______0. T34466619 ________ 11111111 Iteration (L) 01234567 1-6-7—5—4-3-2—10-8-9—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 1-9-8-10-2-3—4-5-7—6-1 Best So1ution 121.0 Best Distance 1’3 ~ 121 Problem 13.3-1 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 13-2 Because of the randomness in the algorithm, its output will vary. Problem 13.3-3 (a) Initial trial solution: 1-2-3-4-5-1 Zc = 34 T1 = 0.2*Zc = 6.8 00000-03332 Sub-tour begins in slot 2 03333-06666 Sub-tour begins in slot 3 06667-09999 Sub-tour begins in slot 4 The random number is 0.09656: choose a sub-tour that begins in slot 2 Be beginning in slot 2, the sub-tour needs to end somewhere between slots 3 and 4. 0.0000-0.4999 Sub-tour begins in slot 3 05000-09999 Sub-tour begins in slot 4 The random number is 0.96657: choose a sub—tour that ends in slot 4 Reverse 2-3-4: new solution: 1-4—3-2—5—1 Zn = 31 Since Zn < Zc, we accept 1-4-3-2-5-1 as new trial solution. (b) Because of the randomness in the algorithm, its output will vary. /3~/3 Problem 13.3-4 Because of the randomness in the algorithm, its output will vary. Problem 13.3-5 Because of the randomness in the algorithm, its output will vary. Problem 13.3-6 Because of the randomness in the algorithm, its output will vary. Problem 13.3-7 (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)”: x-éox +€iaux+ln (c) x =15.5, f(x) = Zc = 3558.9, T1 = 0.2*Zc = 671.775 L = 0, U = 31, sigma= (U-L)/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 133-8 The nonconvex program in Sec 12.10 is Max 0 5’X5-6X$+ 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 = (U-L)/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 13.3-9 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 0.00000-0.99999 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 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 0.00000-0.99999 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 8th one gives mutation. Therefore, the final conclusion is that the two children are C1: 100000 and C2: 111000 Problem 13.4-2 (a) P1: 1-2-3-4-7-6-5—8—1 and P2: 1-5—3-6-7-8-2-4-1 Start from city 1. Possible links: 1-2, 1-8, 1-5, 1-4. Random Number: 0.09656 Choose 1-2 0.96657 No mutation Start from city 2. Current Tour: 1-2 Possible links: 2-3, 2-8, 2-4. Random Number: 0.64842 Choose 2-8 0.49222 No mutation Start from city 8. Current Tour: 1-2-8 Possible links: 8—5, 8-7 Random Number: 0.49506 Choose 8-5 0.10145 No mutation /3-/o“ Start from city 5. Current Tour: 1-2-8-5 Possible links: 5-6, 5-3 Random Number: 0.48455 Choose 5-6 0.23505 No mutation Start from city 6. Current Tour: 1-2-8-5-6 Possible links: 6-7, 6-7, 6-3 Random Number: 0.90430 Choose 6-3 0.04180 Mutation Reject 6-3 and consider all other possible linksz6-4, 6-7 Random Number: 0.24712 Choose 6-4 Start from city 4. Current Tour: 1-2-8-5-6-4 Possible links: 4-3, 4-7 Random Number: 0.55799 Choose 4-7 0.60857 No mutation Start from city 7. Current Tour: 1-2-8-5-6-4-7 Possible links: 7-1 Therefore, C1 = l-2-8-5-6-4-7-1 (b) Pl: 1-6-4-7-3-8-2-5-1 and P2: l-2-5-3-6-8-4-7-l Start from city 1. Possible links: 1-6, 1-5, 1—2, 1-7. Random Number: 0.09656 Choose 1-6 0.96657 No mutation Start from city 6. Current Tour: 1—6 Possible links: 6-4, 6-3, 6-8. Random Number: 0.64842 Choose 6-3 0.49222 No mutation Start from city 3. Current Tour: 1-6-3 Possible links: 3-7, 3-8, 3-5 Random Number: 0.49506 Choose 3-8 0.10145 No mutation Start from city 8. Current Tour: 1—6—3-8 Possible links: 8-2, 8-4 Random Number: 0.48455 Choose 8—2 0.23505 No mutation /3'/7 Start from city 2. Current Tour: 1-6-3-8-2 Possible links: 2—8, 2-5, 2-5 Random Number: 0.90430 Choose 2-5 0.04180 Mutation Reject 2-5 and consider all other possible links:2-4, 2-7 Random Number: 0.24712 Choose 2-4 Start from city 4. Current Tour: 1-6-3-8-2-4 Possible links: 4—7 Random Number: 0.55799 Choose 4-7 0.60857 No mutation Start from city 7. Current Tour: 1-6-3-8—2-4-7 Possible links: 7-1 Therefore, C1 = 1-6-3-8-2-4-7-1 (c) ) P1: 1-5-7-4-6-2-3-8-1 and P2: 1-3-7-2-5-6-8-4-1 Start from city 1. Possible links: 1-5, 1-8, l—3, 1-4. Random Number: 0.09656 Choose 1-5 0.96657 No mutation Start from city 5. Current Tour: 1-5 Possible links: 5-7, 5-2, 5-6. Random Number: 0.64842 Choose 5-2 0.49222 No mutation Start from city 2. Current Tour: 1-5-2 Possible links: 2—6, 2—3, 2-7 Random Number: 0.49506 Choose 2-3 0.10145 No mutation Start from city 3. Current Tour: 1—5-2-3 Possible links: 3-8, 3-7 Random Number: 0.48455 Choose 3-8 0.23505 No mutation Start from city 8. Current Tour: 1-5-2-3-8 Possible links: 8-6, 8—4 Random Number: 0.90430 Choose 8-4 0.041 80 Mutation Reject 8-4 and consider all other possible 1inks:8-6, 8-7 Random Number: 0.24712 Choose 8-6 N'W Start from city 6. Current Tour: 1-5-2-3-8-6 Possible links: 6-4 Random Number: 0.55799 Choose 6-4 0.60857 No mutation Start from city 4. Current Tour: 1-5-2-3-8-6-4 Possible links: 7-1 Therefore, C1 = 1-5-2-3-8-6-4-7—1 Problem 13.4-3 (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 I-10364.0 |—19836.0 (b) Because of the randomness in the aigorithm, its output wi11 vary. 8-21 Problem 13.4-5 Because of the randomness in the algorithm, its output will vary. Problem 13.4-6 Because of the randomness in the algorithm, its output will vary. Problem 13.4-7 (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 13.4-8 (a) Genetic Algorithm: Iteration | Best Solution I Fitness | 1 I1-2—5—4-3-1 124.0 I Iteration 1: Member I Population I Fitness | | Member | Children | Fitness | l I 1-4-5-3-2-1 I 33.0 | | 10 I 1-3-4-5-2-1 I 24.0 2 | 1-2-5-4-3-1 I 24.0 I | 3 | 1-3-2—4-5-1 | 310 3 I 1-3-2-4-5-1 | 31.0 I I 5 | 1-2-3-4-5-1 I 330 4 I 1-4-5-3-2-1 I 33.0 | | 1 I 1-2-3-5-4-1 | 33 0 5 | 1-2-3-4-5-1 | 33.0 I I 2 | 1-5-2-4-3-1 I 280 6 I 1-3-4-2-5-1 I 28.0 | | 8 I 1-4-3-2-5-1 I 310 7 | 1-5-2-3-4-1 | 31.0 I 8 I 1-4-2-3-5-1 | 37.0 | 9 I 1-5-2-3-4-1 I 31.0 | 10 | 1-5-2-4-3-1 | 28.0 I (b) Because of the randomness in the algorithm, its output will vary. Problem 13.4-9 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 13.5-1 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 13.5-2 See the solution for Problem 13.2-6(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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This note was uploaded on 11/28/2010 for the course INDUSTRIAL 200821586 taught by Professor Choijin during the Fall '10 term at 아주대학교.

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Hill13 - Chapter 13 Problem 13.1-1 a b 0 Tours Distance...

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